MHB Unsolved statistics questions from other sites, part II

[chisigma]

Re: Unsolved statistic questions from other sites, part II

Posted on 03 11 2013 on [URL="http://www.artofproblemsolving.com"]www.artofproblemsolving.com[/URL] by the user hallohola and not yet solved...

Let random variables $\displaystyle X_{1} \ge X_{2} \ge … \ge X_{n}$ be independent uniformly distributed r.v. within [0,1] and let 0 < r < 1. Given that $\displaystyle X_{2} \ge r$ find the expected value of $X_{2}$...

The problem is solved 'step by step' and the first step is to find the expected value of the greatest of the r.v. $X_{1} \ge X_{2} \ge ... \ge X_{n}$ uniformly distributed in [0,1]. The probability that all the r.v. are in [0,x] is...

$\displaystyle P_{1} (x)= x^{n}$ (1)

... so that deriving (1) we obtain...

$\displaystyle f_{1} (x) = n\ x^{n-1}$ (2)

... and is...

$\displaystyle E \{X_{1}\}= \int_{0}^{1} n\ x^{n} dx = \frac{n}{n+1}$ (3)The second step is, given the same condition, to find the expected value of $X_{2}$. The probability that n-1 r.v. are in [0,x] and one r.v. is in [x,1] is...
$\displaystyle P_{2} = x^{n-1}$ (4)... so that we have...

$\displaystyle f_{2} (x) = (n-1)\ x^{n-2}$ (5)

$\displaystyle E \{X_{2}\}= \int_{0}^{1} (n-1)\ x^{n-1} dx = \frac{n-1}{n}$ (6)

The third step is to find the expected value of $X_{2}$ with the condition $X_{2} \ge r$ that is...

$\displaystyle E \{X_{2}\}_{X_{2} \ge r} = \frac{1}{1-r^{n-1}}\ \int_{r}^{1} (n-1)\ x^{n-1}\ dx = \frac{1-r^{n}}{1-r^{n-1}}\ \frac{n-1}{n}$ (7) Kind regards…

$\chi$ $\sigma$

 

[chisigma]

Re: Unsolved statistic questions from other sites, part II

Posted the 04 09 2013 on www.mathhelpforum.com by the user Kibbygirl and not jet solved...

According to a certain book, men have 14.334% chance of dying at age 90. A group of 90 year old men meet monthly to discuss philosophy and politics. They are discussing the possibility of some of their group dying this year. If 2 men are chosen at random from the group, what is the probability that both of the 2 men will die this year?...

Kind regards

$\chi$ $\sigma$​

 

[chisigma]

Re: Unsolved statistic questions from other sites, part II

Posted the 04 09 2013 on www.mathhelpforum.com by the user Kibbygirl and not jet solved...

According to a certain book, men have 14.334% chance of dying at age 90. A group of 90 year old men meet monthly to discuss philosophy and politics. They are discussing the possibility of some of their group dying this year. If 2 men are chosen at random from the group, what is the probability that both of the 2 men will die this year?...

​

This problem has been choosen not because it is 'difficult' but because it is a good example of conditional probability. Let suppose that the cumulative distribution of life time $\lambda$ is, normalizing to 1 a time of 100 years, something like...

$\displaystyle P \{\lambda \le x\} = x^{\alpha}$ (1)

... where $\alpha$ is an unknown parameter that we can evalutate knowing that...

$\displaystyle P \{.9 \le \lambda \le 1\}= .14334$ (2)

From (2) we derive easily $\displaystyle \alpha= 1.4684...$ so that we can go to the second part of the problem. For semplicity sake let suppose that the men are both 90 and we want to evaluate the probability that one of them dies between 90 and 91 once he has lived till 90 that is...

$\displaystyle P\{ .9 \le \lambda \le .91\} = \frac{.91^{\alpha} - .9^{\alpha}}{.14334} = .09776...$ (3)

The probability that both men die this year is the square of (3)...

Kind regards

$\chi$ $\sigma$

​

 

[chisigma]

Re: Unsolved statistic questions from other sites, part II

Posted the 10 07 2012 on www.scienzematematiche.it by the user pic [original in Italian...] and not yet solved...

Let be $X_{k}$ a sequence of independent r.v. uniformly distributed in [0,1]. How many variable in average do You need to be $\displaystyle X_{1} + X_{2} + ... + X_{n} > 1$?... Kind regards $\chi$ $\sigma$

 

[chisigma]

Re: Unsolved statistic questions from other sites, part II

Posted the 10 07 2012 on www.scienzematematiche.it by the user pic [original in Italian...] and not yet solved...

Let be $X_{k}$ a sequence of independent r.v. uniformly distributed in [0,1]. How many variable in average do You need to be $\displaystyle X_{1} + X_{2} + ... + X_{n} > 1$?...

The Laplace Transform of the p.d.f. of a single r.v. $X_{k}$ is...

$\displaystyle \mathcal{L} \{f_{1} (t)\} = \frac{1-e^{- s}}{s}$ (1)

... so that the Laplace Transform of the p.d.f. of the r.v. $\displaystyle X=\sum_{k=1}^{n} X_{k}$ is... $\displaystyle \mathcal{L} \{f_{n} (t)\} = \frac{1}{s^{n}}\ \sum_{k=0}^{n} (-1)^{k}\ \binom{n}{k}\ e^{- k s}$ (2)

... and from (2) ...

$\displaystyle f_{n} (t) = \frac{1}{(n-1)!}\ \sum_{k=0}^{n} (-1)^{k}\ \binom {n}{k}\ (t-k)^{n-1}\ \mathcal {U} (t-k)$ (3)

From (3) we derive the probability...

$\displaystyle P_{n} = P \{X < 1\} = \int_{0}^{1} \frac{t^{n-1}}{(n-1)!} \ dt = \frac{1}{n!}$ (4)

... so that is...

$\displaystyle E \{n\} = \sum_{n=0}^{\infty} n\ P_{n} = e$ (5)

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Re: Unsolved statistic questions from other sites, part II

Posted on 04 03 2013 on www.artofproblemsolving.com by the user newsun and not yet solved ...

Number of men and women arriving at a shop in an hour are independently Poisson distributed with mean $\lambda$ and $\mu$ respectively and suppose that there are 3 people arrived during the first hour. What is the probability that...

a) no one arrived during the first 20 minutes...

b) one man and one woman arrived during the first 20 minutes...

c) the first customer is woman...

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Re: Unsolved statistic questions from other sites, part II

Posted on 04 03 2013 on www.artofproblemsolving.com by the user newsun and not yet solved ...

Number of men and women arriving at a shop in an hour are independently Poisson distributed with mean $\lambda$ and $\mu$ respectively and suppose that there are 3 people arrived during the first hour. What is the probability that...

a) no one arrived during the first 20 minutes...

b) one man and one woman arrived during the first 20 minutes...

c) the first customer is woman...

The distributions are both Poisson so that the probability that k men arrived in one hour is $\displaystyle P_{m} (k) = e^{- \lambda}\ \frac{\lambda^{k}}{k!}$ and the probability that k women arrived in one hour is $\displaystyle P_{w} (k) = e^{- \mu}\ \frac{\mu^{k}}{k!}$, and we use them to compute the probability $P_{3}$ that 3 people arrived in the first hour...$\displaystyle P_{3} = \sum_{k=0}^{3} e^{-\lambda}\ e^{- \mu} \frac{\lambda^{k}\ \mu^{3-k}}{k!\ (3-k)!} = \frac{e^{- (\lambda+ \mu)}}{3!} \sum_{k=0}^{3} \binom{3}{k} \lambda^{k}\ \mu^{3-k} = \frac{e^{- (\lambda+ \mu)}}{3!}\ (\lambda + \mu)^{3}$ (1)

All the requested probabilities are conditioned to the event that 3 people arrived in the first hour the probability of which is the (1) and precisely...a) the probability that no men and no women arrived in the first 20 minutes is...$\displaystyle P_{0 + 0} = \frac{e^{- \frac{\lambda + \mu}{3}}}{P_{3}}$ (2)

b) the probability that one man and one woman arrived in the first 20 minutes is...$\displaystyle P_{1 + 1} = \lambda\ \mu\ \frac{e^{- \frac{\lambda + \mu}{3}}}{3\ P_{3}}$ (3)

The question c) is a little more subtle and requires further considerations...

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Re: Unsolved statistic questions from other sites, part II

The answer to question c) is the conditioned probability that the first was a woman and that means that the field of events is restricted to have a woman and she arrived first so that it is...

$\displaystyle P_{\text{woman first}} = \frac{e^{- (\lambda + \mu)}}{P_{3}}\ (\mu + \sum_{k=2}^{3} \frac{\mu^{k}\ \lambda^{3-k}}{k!\ (3-k)!})$ (1)

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Re: Unsolved statistic questions from other sites, part II

Posted on 04 03 2013 on www.artofproblemsolving.com by the user newsun and not yet solved ...

The number of cars entering a car park on a given day is Poisson distributed with mean 100 and the parking fee is 2 dollars per hour with minimum fee of 1 dollar. The amount of time a car is in the car park is exponential distributed with mean 30 minutes. Find the mean and the variance of amount of money that the car park takes in on a given day...

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Re: Unsolved statistic questions from other sites, part II

Posted on 04 03 2013 on www.artofproblemsolving.com by the user newsun and not yet solved ...

The number of cars entering a car park on a given day is Poisson distributed with mean 100 and the parking fee is 2 dollars per hour with minimum fee of 1 dollar. The amount of time a car is in the car park is exponential distributed with mean 30 minutes. Find the mean and the variance of amount of money that the car park takes in on a given day...

Assuming as unity time 30 minutes the fee is constant (1 dollar) for 0<t<1 and linearly increasing with t for t>1 so that the expected amount of dollars for each car is...

$\displaystyle E\{d\} = 1 + \int_{1}^{\infty} (t-1)\ e^{-t}\ dt = 1 + \frac{1}{e} = 1.367879...$ (1)

The number of cars enetering the park on a given day is Poisson distributed with mean 100, so that the expect daily gain of the park is $\displaystyle 100\ (1+\frac{1}{e}) = 136.7879...\ \text{dollars}\ $...

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Re: Unsolved statistic questions from other sites, part II

Posted the 04 14 2013 on www.mathhelpforum.com by the user mathmari and not yet solved...

We have a container that contains r balls that have numbers 1,...,r. We choose at random n of them without replacement. Let Y be the greatest and Z the smallest of the numbers of the balls we chose. Which are the probabilities P(Y<x) and P(Z>x)?...

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Re: Unsolved statistic questions from other sites, part II

Posted the 04 14 2013 on www.mathhelpforum.com by the user mathmari and not yet solved...

We have a container that contains r balls that have numbers 1,...,r. We choose at random n of them without replacement. Let Y be the greatest and Z the smallest of the numbers of the balls we chose. Which are the probabilities P(Y<x) and P(Z>x)?...

The core of the problem is the computation of the probability $\displaystyle P \{Y=k\}$ where k=1,2,...,r. The probability that all the selected numbers are less or equal to k is done by the so called Hypergeometric Distribution...

$\displaystyle P \{ Y = k \} = p_{k} = \frac{\binom{k}{n}\ \binom{r-k}{1}}{\binom{r}{n}} = (r-k)\ \frac{k!}{r!\ (k-n)!}$ (1)

Of course is...

$\displaystyle P \{Y > x\} = \sum_{k > x} p_{k}$ (2)

With similar procedure You find $\displaystyle P \{Z<x\}$...

Kind regards $\chi$ $\sigma$

 

[chisigma]

Re: Unsolved statistic questions from other sites, part II

Posted the 04 06 2013 on ww.talkstats.com by the user dante and not yet solved...

... a scientific theory supposes that mistakes in cell division occur according to a Poisson process with rate $\lambda = 2.5$ per year, and the individual dies when 196 such mistakes have occured... what's the mean lifetime of an individual?..

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Re: Unsolved statistic questions from other sites, part II

Posted the 04 06 2013 on ww.talkstats.com by the user dante and not yet solved...

... a scientific theory supposes that mistakes in cell division occur according to a Poisson process with rate $\lambda = 2.5$ per year, and the individual dies when 196 such mistakes have occured... what's the mean lifetime of an individual?..

The answer is clearly 'spontaneous' and it is based on a connection between Poisson and exponential distribution. More precisely if a sequence of events is Poisson distributed with parameter $\lambda$, then the time T between successive events is exponentially distributed with parameter $\lambda$, i.e. with PDF... $\displaystyle f(t) = \lambda\ e^{- \lambda\ t}$ (1)

According to the basic property of the exponential distribution, if the expected value of T in a single event is $\frac{1}{\lambda}$ the expected value of the sum of n independent exponentially distributed r.v. with the same $\lambda$ is $\frac{n}{\lambda}$, so that the requested mean life time is... $\displaystyle T_{l} = \frac{n}{\lambda} = \frac{196}{2.5} = 78.4\ \text{years}$ (2)Kind regards$\chi$ $\sigma$

 

[chisigma]

Re: Unsolved statistic questions from other sites, part II

Posted the 08 20 2008 [!] on http://www.scienzematematiche.it [original in Italian...] and [incredibly (Wasntme)...] not yet solved...At the time t=0 the vertex of a triangle are sized be three fleas. At the times t=1,2,... each flea jumps from a vetex to one other vertex of the triangle. What is the mean number of jumps after whom the three fleas come together?... Kind regards $\chi$ $\sigma$

 

[Nono713]

Re: Unsolved statistic questions from other sites, part II

Posted the 08 20 2008 [!] on http://www.scienzematematiche.it [original in Italian...] and [incredibly (Wasntme)...] not yet solved...At the time t=0 the vertex of a triangle are sized be three fleas. At the times t=1,2,... each flea jumps from a vetex to one other vertex of the triangle. What is the mean number of jumps after whom the three fleas come together?... Kind regards $\chi$ $\sigma$

Each flea must jump independently and has two possibilities, so the probability that all fleas land on the same vertex at any given time is just:

$$\left ( \frac{1}{2} \right )^3$$
Therefore the fleas will come together on average after:​

$$\frac{1}{\left ( \frac{1}{2} \right )^3} = 8 ~ ~ \text{jumps}$$
(or is it $8/2 = 4$? I'm not sure now, my statistics are rusty, can anyone confirm?)

 

[chisigma]

Re: Unsolved statistic questions from other sites, part II

Posted the 08 20 2008 [!] on http://www.scienzematematiche.it [original in Italian...] and [incredibly (Wasntme)...] not yet solved...

At the time t=0 the vertex of a triangle are sized be three fleas. At the times t=1,2,... each flea jumps from a vetex to one other vertex of the triangle. What is the mean number of jumps after whom the three fleas come together?...

Calling A,B and C the vertex of the triangle, the possible transictions are represented in the figure...

http://d16cgiik7nzsna.cloudfront.net/f3/32/i69153523._szw380h285_.jpg

We suppose for semplicity that each transition has probability ${1}{2}$. The problem can be structured as a Markov Chain represented by the following graph... The state 1 represents the three fleas on three different vertex, the state 2 represents two fleas on one vertex and one flea on one of the other vertex, the state 3 represents all the fleas on the same vertex. The next step is the computation of the transitions probabilities. If we are in state 1 the possible transitions are...

http://www.123homepage.it/u/i69185898._szw380h285_.jpg.jfif

A -> B A -> C A -> B A -> B A -> B A -> C A -> C A -> C
B -> C B -> A B -> A B -> C B -> A B -> C B -> A B -> C
C -> A C -> B C -> B c -> B C -> A C -> A C -> A C -> B (1)

... and from (1) we can easily verify that is $\displaystyle p_{1,1}= \frac{1}{4}$ and $\displaystyle p_{1,2}= \frac{3}{4}$. If we are in the state 2 with fleas in A,A and B the possible transitions are...

A -> B A -> B A -> B A -> B A -> C A -> C A -> C A -> C
A -> B A -> C A -> B A -> C A -> C A -> B A -> C A -> B
B -> A B -> A B -> C B -> C B -> C B -> A B -> A B -> C (2)

... and from (2) we can easily verify that is $\displaystyle p_{2,3}= \frac{1}{8}$, $\displaystyle p_{2,2}=\frac{5}{8}$ and $\displaystyle p_{2,1}=\frac{1}{4}$. Now we are able to write the transition matrix of the Markov Chain...

$\displaystyle P = \left | \begin{matrix} \frac{1}{4} & \frac{3}{4} & 0 \\ \frac{1}{4} & \frac{5}{8} & \frac{1}{8} \\ 0 & 0 & 1 \end{matrix} \right| $ (3)

... and proceed as in...

http://www.mathhelpboards.com/f23/expected-number-questions-win-game-4154/

Clearly the only adsorbing state is the state 3 so that we can write matrix Q...

$\displaystyle Q = \left | \begin{matrix} \frac{1}{4} & \frac{3}{4} \\ \frac{1}{4} & \frac{5}{8} \end{matrix} \right| $ (4)

... and then the fundamental matrix N...

$\displaystyle N = (I_{2} - Q)^{-1} = \left | \begin{matrix} 4 & 8 \\ \frac{8}{3} & 8 \end{matrix} \right| $ (5)

The (5) permits to conlude that, starting from state 1 the expected number of jumps is $T= 4 + 8 = 12$ and starting from the state 2 the expected number of jumps is $ T = \frac{8}{3} + 8 = \frac{32}{3}$ ... Kind regards $\chi$ $\sigma$

 

[chisigma]

Re: Unsolved statistic questions from other sites, part II

Posted the 02 22 2013 on www.artofproblemsolving.com by the user Phun_TZ and not yet solved... Problem : A stick is broken at a point, chosen at random, along its length. Find the probability that the ratio of the length of the shorter piece to the length of the longer piece is less than $r$. I am not sure whether my answer is correct, I've got $\displaystyle \frac{2\ \sqrt{2}\ r}{r+1}$. Could anyone check for me please?...Kind regards $\chi$ $\sigma$

 

[chisigma]

Re: Unsolved statistic questions from other sites, part II

Posted the 02 22 2013 on www.artofproblemsolving.com by the user Phun_TZ and not yet solved... Problem : A stick is broken at a point, chosen at random, along its length. Find the probability that the ratio of the length of the shorter piece to the length of the longer piece is less than $r$. I am not sure whether my answer is correct, I've got $\displaystyle \frac{2\ \sqrt{2}\ r}{r+1}$. Could anyone check for me please?...

Let's suppose that X is a r.v. uniformely distributed in $[0,\frac{1}{2}]$ , we have to compute the probability... $P \{ \frac{X}{1-X} < r\} = P \{ X < \frac{r}{r+1} \} = 2\ \int_{0}^{\frac{r}{r+1}} dx = 2\ \frac{r}{r+1}$ (1)

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Re: Unsolved statistic questions from other sites, part II

Posted the 04 21 2013 on www.artofproblemsolving.com by the user mynamearzo and not yet solved...

Let d be the distance between two points picked independently at random from a uniform distribution inside a disc of radius r. Show that $\displaystyle E \{d^{2}\} = r^{2}$...

Kind regards$\chi$ $\sigma$

 

[chisigma]

Re: Unsolved statistic questions from other sites, part II

Posted the 03 02 2012 on www.mathhelpforum.com by the user Suvadip and not yet solved [!]...

Let X have a Poisson distribution with parameter m. Show that... $\displaystyle P \{ X\ \text{even}\} = \frac{1+ e^{-2\ m}}{2}$

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Re: Unsolved statistic questions from other sites, part II

Posted the 03 02 2012 on www.mathhelpforum.com by the user Suvadip and not yet solved [!]...

Let X have a Poisson distribution with parameter m. Show that... $\displaystyle P \{ X\ \text{even}\} = \frac{1+ e^{-2\ m}}{2}$

If X is Poisson distributed with parameter m is...

$\displaystyle P \{ X=k\} = e^{- m}\ \frac{m^{k}}{k!}$ (1)

... so that is...

$\displaystyle P\{k\ \text{even}\} = e^{- m}\ \sum_{k\ \text{even}} \frac{m^{k}}{k!} = e^{- m}\ \cosh m = \frac{1+e^{-2 m}}{2}\ \text{ }\ $ (2)

The problem is 'simple' but very important. If You know that a procees is Poisson but the parameter m is unknown what to do?... You can do a 'large enough' number of observations, evaluate the probability of k even and use (2) to find m...Kind regards $\chi$ $\sigma$

 

[chisigma]

Re: Unsolved statistic questions from other sites, part II

Posted the 05 12 2013 on www.matematicamente.it by the user salvo 89 [original in Italien...] and not yet solved... Let's suppose to have a discrete r.v. X and a continuous r.v. Y, what is the p.d.f. of the r.v. Z= X Y?... Kind regards $\chi$ $\sigma$

 

[chisigma]

Re: Unsolved statistic questions from other sites, part II

Posted the 05 12 2013 on www.matematicamente.it by the user salvo 89 [original in Italien...] and not yet solved...

Let's suppose to have a discrete r.v. X and a continuous r.v. Y, what is the p.d.f. of the r.v. Z= X Y?...

Let’s suppose that X is in the range from 0 to + infinity and set $P_{n}= P \{ X=n\}$ . If $f_{y} (y)$ is the p.d.f. of Y we can write…

$$ P\{Z<z\} = \sum_{n=0}^{\infty} P_{n}\ \int_{- \infty}^{\frac{z}{n}} f_{y}(y)\ dy\ = P_{0} + \sum_{n=1}^{\infty} P_{n}\ \int_{- \infty}^{\frac{z}{n}} f_{y}(y)\ dy\ (1)$$

The p.d.f. of Z is the derivative of (1)...

$$ f_{z} (z) = \sum_{n=1}^{\infty} P_{n}\ f_{y} (\frac{z}{n})\ (2)$$

Let's consider as example the case where X is Poisson distributed with mean $\lambda$, so that is...

$$P_{n} = e^{- \lambda}\ \frac{\lambda^{n}}{n!}\ (3)$$

... and Y is uniformely distributed in (0,1) so that is...

$$ f_{y}(y) = \begin{cases} 1 & \text{if } 0<y<1\\ 0 & \text{otherwise } \end{cases}\ (4)$$

In such a case, applying (2), the p.d.f. of Z=X Y is...

$$f_{z}(z) = e^{- \lambda}\ \sum_{n=1}^{\infty} \frac{\lambda^{n}}{n!}\ u_{n} (z)\ (5)$$

... where...

$$ u_{n}(z) = \begin{cases} 1 & \text{if } 0<z<n\\ 0 & \text{otherwise } \end{cases}\ (6)$$

The $f_{z}(z)$ in the case $\lambda=1$ is represented in the figure...

http://www.123homepage.it/u/i70169162._szw380h285_.jpg.jfifKind regards

$\chi$ $\sigma$

 

[chisigma]

Re: Unsolved statistic questions from other sites, part II

Posted the 04 11 2013 on www.artofproblemsolving.com by the user Semigroups and not yet solved...

X and Y are independent r.v. subject to the uniform distribution on [0,3] and [0,4] respectively. Calculate P {X<Y}...

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Re: Unsolved statistic questions from other sites, part II

Posted the 04 11 2013 on www.artofproblemsolving.com by the user Semigroups and not yet solved...

X and Y are independent r.v. subject to the uniform distribution on [0,3] and [0,4] respectively. Calculate P {X<Y}...

Indicating with $f_{X}(*)$ and $f_{-Y}(*)$ the p.d.f. of the r.v. X and -Y is... $$\mathcal{L} \{f_{X} (t)\} = \frac{1- e^{-3\ s}}{3\ s}$$

$$\mathcal{L} \{f_{-Y} (t)\} = \frac{e^{4\ s}-1}{4\ s}\ (1)$$

... so that the L-Transform of the p.d.f. of the r.v. Z= X-Y is... $$ \mathcal{L} \{f_{Z} (t)\} = \mathcal{L} \{f_{X} (t)\}\ \mathcal{L} \{f_{-Y} (t)\} = \frac{e^{4\ s} - e^{s} + e^{-3\ s} -1}{12\ s^{2}}\ (2) $$

... so that is...

$$f_{Z} (t) = \begin{cases} \frac{1}{3} + \frac{t}{12} & \text{if }\ -4 < t< -1\\ \frac{1}{4} & \text{if}\ -1 < t < 0 \\ \frac{1}{4} - \frac{t}{12} & \text{if}\ 0 < t < 3 \\ 0 & \text{otherwise } \end{cases}\ (3) $$

... and the requested probability...

$$ P \{ X < Y\} = P \{ Z< 0\} = \int_{-4}^{0} f_{Z} (t)\ dt = \frac{5}{8}\ (4) $$

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Re: Unsolved statistic questions from other sites, part II

Posted the 05 22 2013 on www.talkstats.com by the user StatsMan04 and not yet solved...

A store has 80 modems in its inventory, 30 coming from Source A and the remainder from Source B. Of the modems from Source A, 20% are defective. Of the modems from Source B, 8% are defective. Calculate the probability that exactly 2 of 5 randomly sampled modems are defective...

Kind regards $\chi$ $\sigma$

 

[chisigma]

Re: Unsolved statistic questions from other sites, part II

Posted the 05 22 2013 on www.talkstats.com by the user StatsMan04 and not yet solved...

A store has 80 modems in its inventory, 30 coming from Source A and the remainder from Source B. Of the modems from Source A, 20% are defective. Of the modems from Source B, 8% are defective. Calculate the probability that exactly 2 of 5 randomly sampled modems are defective...

Setting $p_{A} = \frac{1}{5}$ the probability of defective modem of the source A and $p_{B} = \frac{2}{25}$ the probability of defective modem of the source B, the probability of defective modem in the whole stock is...

$$p= p_{A}\ P \{\text{modem from source A}\} + p_{B}\ P \{\text{modem from source B}\} = \frac{1}{5}\ \frac{30}{80} + \frac{2}{25}\ \frac{50}{80} = \frac {1}{8}\ (1)$$

The problem can be considered as sampling without substitution from a finite population which is regulated by the hypergeometric distribution probability...

$$ P\{X=k\} = \frac{\binom{K}{k}\ \binom {N-K}{n-k}}{\binom{N}{n}}\ (2)$$

... and in our case is N=80, K=10, n=5, k=2 so that is...

$$ P\{X=2\} = \frac{\binom{10}{2}\ \binom {70}{3}}{\binom{80}{5}}\ (3)$$

Of course the computation (3) is not very comfortable but in the web You can easiliy find adequate tools like that...

Hypergeometric Calculator

... and using it You can find...

$$ P\{X=2\} = 0.102466653932343...\ (4)$$

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Re: Unsolved statistic questions from other sites, part II

Posted the 06 02 2013 on www.artofproblemsolving.com by the user gaurav 1292 and not yet solved...

Assume that X,Y,Z are independent random variables having identical density functions...

$$ f(x)=\begin{cases} e^{- x}\ &\text{if}\ x \ge 0\\ 0\ &\text{if}\ x<0 \end{cases}$$

... then find the joint distribution of U = X+Y, V= X+Z, W=Y+Z, T = X + Y + Z

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Re: Unsolved statistic questions from other sites, part II

Posted the 06 02 2013 on www.artofproblemsolving.com by the user gaurav 1292 and not yet solved...

Assume that X,Y,Z are independent random variables having identical density functions...

$$ f(x)=\begin{cases} e^{- x}\ &\text{if}\ x \ge 0\\ 0\ &\text{if}\ x<0 \end{cases}$$

... then find the joint distribution of U = X+Y, V= X+Z, W=Y+Z, T = X + Y + Z

This type of problem has been solved several times, given the r.v. $X_{1},X_{2},...,X_{n}$ with p.d.f. $f_{1} (x),f_{2}(x),...,f_{n}(x)$ the p.d.f. nof the r.v. $X= X_{1} + X_{2} + ... + X_{n}$ is $f(x)= f_{1}(x) * f_{2}(x) * ... * f_{n}(x)$ where '*" means convolution. In our case X, Y and Z have p.d.f. ... $$ f(x)=\begin{cases} e^{- x}\ &\text{if}\ x \ge 0\\ 0\ &\text{if}\ x<0 \end{cases}\ (1)$$

Now is...

$$\mathcal{L} \{f(x)\} = \frac{1}{s + 1}\ (2)$$

... so that the r.v. U,V and W have p.d.f. ... $$ g(x) = \mathcal{L}^{-1} \{\frac{1}{(s+1)^{2}}\} = \begin{cases} x\ e^{- x}\ &\text{if}\ x \ge 0\\ 0\ &\text{if}\ x<0 \end{cases}\ (2)$$

... and the r.v. T has p.d.f. ...

$$ h(x) = \mathcal{L}^{-1} \{\frac{1}{(s+1)^{3}}\} = \begin{cases} 2\ x^{2}\ \ e^{- x}\ &\text{if}\ x \ge 0\\ 0\ &\text{if}\ x<0 \end{cases}\ (3)$$

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Re: Unsolved statistic questions from other sites, part II

Posted the 07 01 2010 on www.artofproblemsolving.com by the user hsiljak and not yet solved...

... each of 3600 subscribers of a telephone exchange calls it once an hour on average. What is the probability that in a given second 5 or more calls are received?... estimate the mean interval of time between such seconds...

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Re: Unsolved statistic questions from other sites, part II

Posted the 07 01 2010 on www.artofproblemsolving.com by the user hsiljak and not yet solved...

... each of 3600 subscribers of a telephone exchange calls it once an hour on average. What is the probability that in a given second 5 or more calls are received?... estimate the mean interval of time between such seconds...

We have a Poisson process with mean $\lambda=1$ so is...

$$P \{ N(t+1) - N(t) = n\} = \frac{e^{-1}}{n!}\ (1)$$

... and the probability to have 5 or more calls in 1 second is...

$$ p = 1 - e^{-1}\ \sum_{n=0}^{4} \frac{1}{n!} = 1 - e^{-1}\ \frac{65}{24} \sim 3.66\ 10^{-3}\ (2)$$

The (2) allows us to computed the mean time between such seconds...

$$E \{ \Delta t\} = \sum_{n=1}^{\infty} n\ p\ (1-p)^{n-1} = \frac{p}{p^{2}} = \frac{1}{p} \sim 273.224\ (3)$$

Of course the result (3) is not a surprise...

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Re: Unsolved statistic questions from other sites, part II

Posted the 07 01 2010 on www.artofproblemsolving.com by the user hsiljak and not yet solved...

Calculate the mean value of the solid angle by which the disc $x^{2} + y^{2} \le 1$ lying in the plane z=0 is seen from points of the sphere $x^{2} + y^{2} + (z-2)^{2} = 1$...

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Re: Unsolved statistic questions from other sites, part II

Posted the 07 01 2010 on www.artofproblemsolving.com by the user hsiljak and not yet solved...

Calculate the mean value of the solid angle by which the disc $x^{2} + y^{2} \le 1$ lying in the plane z=0 is seen from points of the sphere $x^{2} + y^{2} + (z-2)^{2} = 1$...

Fortunately the problem is rotationally invariant and it can be engaged in two dimension, referring to the following pitcure...

http://ddpozwy746ijz.cloudfront.net/78/2d/i71708024._szw380h285_.jpg

If $\theta$ is the angle of the vector connecting the point [0,2] to the point A on the circle, then we have...$$|A B| = a = \sqrt{(1 + \cos \theta )^{2} + (2 + \sin \theta)^{2}} = \sqrt{6 + 2\ \cos \theta + 4\ \sin \theta}\ (1)$$

$$|A C| = b = \sqrt{(1 - \cos \theta )^{2} + (2 + \sin \theta)^{2}} = \sqrt{6 - 2\ \cos \theta + 4\ \sin \theta}\ (2)$$

$$|B C|= c = 2\ (3)$$

Now we apply the so called 'law of cosines' to the triangle A B C ...

$$ c^{2} = a^{2} + b^{2} - 2\ a\ b\ \cos \alpha\ (4)$$

... and from (4) we derive the explicit expression for...$$\alpha = \cos ^{-1} \frac{2\ (1+ \sin \theta)}{\sqrt{3 + \cos \theta + 2\ \sin \theta}\ \sqrt{3 - \cos \theta + 2\ \sin \theta}}\ (5)$$

Now the expected value of $\alpha$ is required and that can be computed directly solving the integral...

$$E \{\alpha\} = \frac{1}{\pi}\ \int_{- \frac{\pi}{2}}^{\frac{\pi}{2}} \cos ^{-1} \frac{2\ (1+ \sin \theta)}{\sqrt{3 + \cos \theta + 2\ \sin \theta}\ \sqrt{3 - \cos \theta + 2\ \sin \theta}}\ d \theta\ (6)$$

The integral (6) can [probably...] only numerically computed and 'Monster Wolfram' gives the result $E \{\alpha\} \sim .927294\ \text {radians}$. Now You can go from two to three dimensions and, indicating with $\Omega$ the solid angle, You obtain its mean value multiplying the angle by two $E \{ \Omega \} \sim 1.854588\ \text{Steradians}$...

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Re: Unsolved statistic questions from other sites, part II

Posted the 02 01 2013 on www.artofproblemsolving.com by the user hsj0660 and not yet solved...

You own n cats. One day, all of your n cats run out of the house into the yard. You go around sequentially to each cat and ask them to return to the house. When asked, the cat will return to the house with probability $\frac{1}{k+1}$ where k is the number of cats currently in the house. What's the expected value of number of cats in the house once you ask all the n cats to go in the house?...

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Re: Unsolved statistic questions from other sites, part II

Posted the 02 01 2013 on [URL="http://www.artofproblemsolving.com"]www.artofproblemsolving.com[/URL] by the user hsj0660 and not yet solved...You own n cats. One day, all of your n cats run out of the house into the yard. You go around sequentially to each cat and ask them to return to the house. When asked, the cat will return to the house with probability $\frac{1}{k+1}$ where k is the number of cats currently in the house. What's the expected value of number of cats in the house once you ask all the n cats to go in the house?...

The problem is Markov type whith the following transition diagram...http://d1r9jua05newpd.cloudfront.net/ea/d2/i71946986._szw380h285_.jpg
The probability transition matrix is...$\displaystyle P = \left | \begin{matrix} \frac{1}{2} & \frac{1}{2} & 0 & 0 & \ & \ & \ & \ & 0 & 0 \\ 0 & \frac{2}{3} & \frac{1}{3} & 0 & \ & \ & \ & \ & 0 & 0 \\ 0 & 0 & \frac{3}{4} & \frac{1}{4} & \ & \ & \ & \ & 0 & 0 \\ \ & \ & \ & \ & \ & \ & \ & \ & 0 & 0 \\ \ & \ & \ & \ & \ & \ & \ & \ & 0 & 0\\ \ & \ & \ & \ & \ & \ & \ & \ & 0 & 0 \\ \ & \ & \ & \ & \ & \ & \ & \ & 0 & 0 \\ \ & \ & \ & \ & \ & \ & \ & \ & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \frac{n-1}{n} & \frac{1}{n} \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \end{matrix} \right| $ (1)
Setting $p_{k}$ the probability to arrive from the state 1 to the state k after n-1 steps, the $p_{k}, k=1,2,...,n$ are the first line of the matrix $P^{n-1}$ and the required expected number of cats is...$$E \{k\} = \sum_{k=1}^{n} k\ p_{k}\ (2)$$
Let's start with small value of n...

$$n = 2; p_{1}= \frac{1}{2}, p_{2}= \frac {1}{2}; E \{k\} = \frac{3}{2}$$

$$n=3; p{1}= \frac{1}{4}, p_{2} = \frac{7}{12}, p_{3}= \frac{1}{6}; E \{k\} = \frac{23}{12}$$

$$n=4; p_{1} = \frac{1}{8}, p_{2} = \frac{37}{72}, p_{3} = \frac{23}{72}, p_{4}= \frac{1}{24}; E\{k\} = \frac {41}{18}$$
$$n=5;p_{1}=\frac{1}{16},p_{2}=\frac{175}{432},p_{3}= \frac{355}{864},p_{4}=\frac{163}{1440},p_{5}=\frac{1}{120}; E\{k\} = \frac{11231}{4320}$$$$n=6;p_{1}=\frac{1}{32},p_{2}=\frac{781}{2592},p_{3}= \frac{4595}{10368},p_{4}= \frac{16699}{86400},p_{5}= \frac{71}{2400},p_{6}= \frac{1}{720}; E \{k\}= \frac{749813}{259200}$$

 

[chisigma]

Re: Unsolved statistic questions from other sites, part II

I opologize for the fact that my previous post was not complete, but I had graphics problems...

Clearly the complexity increases considerably with n increasing. For n from 1 to 6 it seems to be $E\{k\} \approx n^{.6}$ but some more analysis is requested...

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Re: Unsolved statistic questions from other sites, part II

Posted the 06 10 2013 on Statistics Help @ Talk Stats Forum by the user lunxcell and not yet solved...

... X and Y are exponential distributed and independend with the parameter $\lambda > 0$. Which distribution has X+Y?...

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Re: Unsolved statistic questions from other sites, part II

Posted the 06 10 2013 on Statistics Help @ Talk Stats Forum by the user lunxcell and not yet solved...

... X and Y are exponential distributed and independend with the parameter $\lambda > 0$. Which distribution has X+Y?...

X and Y have identical p.d.f. ...

$\displaystyle f_{X}(x)= f_{Y}(x)=\lambda\ e^{- \lambda\ x}\ H(x)\ (1)$

... where H(*) is the so called 'Heaviside Step Function', so Z= X + Y has p.d.f. ...

$\displaystyle f_{Z}(x)= f_{X}(x) * f_{Y}(x) = \lambda^{2}\ x\ e^{- \lambda\ x}\ H(x)\ (2)$

... where * means convolution...

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Re: Unsolved statistic questions from other sites, part II

Posted the 06 05 2013 on www.mathhelpforum.com by the user Nazarin and not yet solved...

Consider a machine which is turned off when there is no work. It is turned on and restarts work when enough orders, say N, arrived to the machine. The setup times are negligible. The processing times are exponentially distributed with mean 1/mu and the average number of orders arriving per unit time is lambda (<mu ). Suppose that lambda = 20 orders per hour, (1/mu) = 2 minutes and that the setup cost is 54 dollar. In operation the machine costs 12 dollar per minute. The waiting cost is 1 dollar per minute per order. Determine, for given threshold N, the average cost per unit time...

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Re: Unsolved statistic questions from other sites, part II

Posted the 06 05 2013 on www.mathhelpforum.com by the user Nazarin and not yet solved...

Consider a machine which is turned off when there is no work. It is turned on and restarts work when enough orders, say N, arrived to the machine. The setup times are negligible. The processing times are exponentially distributed with mean 1/mu and the average number of orders arriving per unit time is lambda (<mu ). Suppose that lambda = 20 orders per hour, (1/mu) = 2 minutes and that the setup cost is 54 dollar. In operation the machine costs 12 dollar per minute. The waiting cost is 1 dollar per minute per order. Determine, for given threshold N, the average cost per unit time...

The average cost per unit time is the sum of three components...

a) the setup cost $C_{s}$...

b) the operating cost $C_{o}$...

c) the waiting time order cost $C_{w}$...

If $\lambda$ is the mean number of orders arriving in the unit time [we supposed that u.t. is 1 hour...], then the mean number of pochets of N orders arriving in the unit time is $\displaystyle \frac{\lambda}{N}$, so that is $\displaystyle C_{s}= 54\ \frac{\lambda}{N}$. The operating time to process N orders is $\displaystyle \frac{N}{\mu}$, so that is $\displaystyle C_{o}= 720\ \frac{N}{\mu}$. The mean waiting time order is $\displaystyle \frac{1}{2}\ (\frac{N}{\lambda} + \frac{N}{\mu})$ so that the mean waiting time cost is $\displaystyle 30\ N^{2}\ (\frac{1}\lambda + \frac{1}{\mu})$. The total mean cost per hour will be... $\displaystyle C=C_{s} + C_{o} + C_{w} = 54\ \frac{\lambda}{N} + 720\ \frac{N}{\mu} + 30\ N^{2}\ (\frac{1}{\lambda} + \frac{1}{\mu})\ (1)$

C can easily be computed setting in (1) $\lambda = 20\ $ and $\mu=30\ $ ... Kind regards

$\chi$ $\sigma$

 

[chisigma]

Re: Unsolved statistic questions from other sites, part II

Posted the 05 17 2013 on www.artofproblemsolving.com by the user doom48 and not yet solved...

Given that $\displaystyle Z \sim N (0,1)$ and n is a positive integer, prove that... $\displaystyle E \{Z^{2 n}\} = \frac{(2 n)!}{n!\ 2^{n}}$



Kind regards

$\chi$ $\sigma$

 

[chisigma]

Re: Unsolved statistic questions from other sites, part II

Posted the 05 17 2013 on www.artofproblemsolving.com by the user doom48 and not yet solved...

Given that $\displaystyle Z \sim N (0,1)$ and n is a positive integer, prove that... $\displaystyle E \{Z^{2 n}\} = \frac{(2 n)!}{n!\ 2^{n}}$

The moment generating function of $\displaystyle N(0,1)$ is...

$\displaystyle m(x)= E \{e^{Z\ x}\} = \frac{1}{\sqrt{2 \pi}}\ \int_{- \infty}^{+ \infty} e^{x\ z}\ e^{- \frac{z^{2}}{2}}\ dz = e^{\frac{x^{2}}{2}}\ (1)$

... and by definition is...

$\displaystyle E \{Z^{2 n}\} = \lim_{x \rightarrow 0} \frac {d^{2 n}}{d x^{2 n}} m(x)\ (2)$

Remembering the definition of Hermite Polynomial of order k...

$\displaystyle He_{k} (x)= (-1)^{k}\ e^{\frac{x^{2}}{2}}\ \frac {d^{k}}{d x^{k}} e^{- \frac{x^{2}}{2}}\ (3)$

... with symple steps we derive that is...

$\displaystyle E \{ z^{2 n}\} = He_{2 n} (0) = (2 n -1)! = \frac{(2 n)!}{2^{n}\ n!}\ (4)$

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Re: Unsolved statistic questions from other sites, part II

Posted the 07 01 2010 on www.artofproblemsolving by the user hsiljak and not yet solved…

… a particle performing a random walk on the integer points of the semi-axis $x \ge 0$ moves a distance 1 to the right with probability a, and to the left with probability b, and stands still in the remaining cases [if x=0 it stands still instead of moving to the left]. Determine the steady-state probability distribution, and also the expectation of $x$ and $x^2$ over a long time, if the particle starts at the point 0…

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Re: Unsolved statistic questions from other sites, part II

Posted the 07 01 2010 on www.artofproblemsolving by the user hsiljak and not yet solved…

… a particle performing a random walk on the integer points of the semi-axis $x \ge 0$ moves a distance 1 to the right with probability a, and to the left with probability b, and stands still in the remaining cases [if x=0 it stands still instead of moving to the left]. Determine the steady-state probability distribution, and also the expectation of $x$ and $x^2$ over a long time, if the particle starts at the point 0…

The position of the particle after n steps can be schematized as a Markov chain with the following state diagram...

http://dmqg0yef478ix.cloudfront.net/87/ad/i73182599._szw380h285_.jpg

The probability transition matrix is...$\displaystyle P = \left | \begin{matrix} 1-a & a & 0 & 0 & \ & \ & \ & 0 & 0 & 0 \\ b & 1-a-b & a & 0 & \ & \ & \ & 0 & 0 & 0 \\ 0 & b & 1 – a - b & a & \ & \ & \ & 0 & 0 & 0 \\ \ & \ & \ & \ & \ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \ & \ & \ & \ \\ 0 & 0 & 0 & 0 & \ & \ & \ & b & 1 – a - b & a \\ 0 & 0 & 0 & 0 & \ & \ & \ & 0 & 0 & 1 \end{matrix} \right| $ (1)

The expected value of X after n steps starting from 0 is...

$\displaystyle E \{ X\} = \sum_{k=1}^{n} k\ p_{k}\ (2)$

... where $\displaystyle p_{k}$ is the k-th element of the row 0 of $\displaystyle P^{n}$. Let’s evaluate some expected values…

$\displaystyle n=1;\ p_{1}=a;\ E\{X\}_{n=1} = a$

$\displaystyle n=2;\ p_{1}= 2\ a - a\ b -2\ a^{2},\ p_{2}= a^{2};\ E\{X\}_{n=2} = 2\ a - a\ b\ $

$\displaystyle n=3;\ p_{1}= 3\ a^{3} + (5\ b -6)\ a^{2} + (b^{2} - 3\ b + 3)\ a,\ p_{2}= -3\ a^{3} + (3 - 2\ b)\ a^{2},\ p_{3}= a^{3};\ E \{X\}_{n=3} = b\ a^{2} + (b^{2} -3\ b + 3)\ a\ $

Clearly if n increases the task becomes more and more hard because the n-th power of the matrix P has to be computed. In next post we will try a different approach to the problem...

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Re: Unsolved statistic questions from other sites, part II

Posted the 05 27 2013 on www.artofproblemsolving.com by the user juliancasaa and not yet solved... 1- At the airport in some city the number of flights arriving at a time is a random variable with Poisson distribution with mean of 5. What is the probability that the time duration until you get the third flight is at most two hours?...

2.The time in years elapsed from the time of removal until the death of the employee, in a factory, it is a random variable with exponential distribution with mean time B of 7 years. If it pensionan 10 employees of the factory, what is the probability that at most seven are still alive at the end of 10 years?...

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Re: Unsolved statistic questions from other sites, part II

Posted the 05 27 2013 on www.artofproblemsolving.com by the user juliancasaa and not yet solved... 1- At the airport in some city the number of flights arriving at a time is a random variable with Poisson distribution with mean of 5. What is the probability that the time duration until you get the third flight is at most two hours?...

If the mean number of arriving flights in one hour is 5, in two hours it is 10. The requested probability is...

$\displaystyle P = 1 - e^{-10}\ \sum_{k=0}^{3} \frac{10^{k}}{k!} = .98966...$Kind regards $\chi$ $\sigma$

 

[chisigma]

Re: Unsolved statistic questions from other sites, part II

Posted the 05 27 2013 on www.artofproblemsolving.com by the user juliancasaa and not yet solved... 2.The time in years elapsed from the time of removal until the death of the employee, in a factory, it is a random variable with exponential distribution with mean time B of 7 years. If it pensionan 10 employees of the factory, what is the probability that at most seven are still alive at the end of 10 years?...

The probability that one pensioned is dead after 10 years is... $\displaystyle p = 1 - \frac{1}{7}\ e^{- \frac{10}{7}} = .76...\ (1)$ The requested probability is... $\displaystyle P = 1 - \sum_{k=0}^{3} \binom{10}{k} p^{k}\ (1-p)^{10-k}\ (2)$ The effective computation of (2) can be performed with a standard calculator and it is left to the reader...

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Posted the 08 17 2013 on www.artofproblemsolving.com by the user mathemath and not yet solved...

If characteristic function of a random variable X is $\varphi (t)$, what is the characteristic function of 1/X?...

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Posted the 08 17 2013 on www.artofproblemsolving.com by the user mathemath and not yet solved...

If characteristic function of a random variable X is $\varphi (t)$, what is the characteristic function of 1/X?...

The problem is very interesting!... first step is to remember that the definition of characteristic function of a r.v. X with p.d.f. $f_{X} (x)$ is... $\displaystyle \varphi_{X} (t)= \int_{- \infty}^{+ \infty} f(x)\ e^{i\ t\ x}\ dx\ (1)$

... and that $\displaystyle f_{X} (x)$ can be obtained from $\displaystyle \varphi_{X} (t)$ with the inversion formula...

$\displaystyle f_{X} (x) = \frac{1}{2 \pi}\ \int_{- \infty}^{+ \infty} \varphi_{X} (t)\ e^{- i\ t\ x}\ dt\ (2)$

The second step is, setting $\displaystyle Y= \frac{1}{X}$ to verify that the p.d.f. of Y is...

$\displaystyle f_{Y} (x) = \frac{1}{x^{2}}\ f_{X} (\frac{1}{x})\ = \frac{1}{2 \pi x^{2}}\ \int_{- \infty}^{+ \infty} \varphi_{X} (t)\ e^{- i\ \frac{t}{x}}\ dt\ (3)$

The third step is the use of (1) to compute...

$\displaystyle \varphi_{Y} (t) = \frac{1}{2 \pi}\ \int_{- \infty}^{+ \infty}\ \int_{- \infty}^{+ \infty}\ \frac{\varphi_{X} (t)}{x^{2}}\ e^{i\ t\ (x-\frac{1}{x})}\ dt\ dx\ (4)$

It has to be evaluated if is possible to write the expression (4) in a more easy form...

Kind regards

$\chi$ $\sigma$