MHB Unsolved statistics questions from other sites, part II

[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

$\chi$ $\sigma$