MHB Unsolved statistics questions from other sites....

[chisigma]

Posted on 04 23 2012 on http://www.scienzematematiche.it/ by the user whitefang [original in Italian…] and not yet properly solved…

We are shooting at a target over a two-dimension plane. The horizontal and vertical distances of the hits respect to the target are normal r.v. with $\mu=0$ and $\sigma=4$. R is the distance between the hit and the target. Find $E\{R\}$...

That is material for a basic course of probability. If X and Y are normal r.v. with mean 0 and variance $\sigma$, then $R=\sqrt{X^{2}+Y^{2}}$ is Rayleigh distributed, i.e. its p.d.f. is...

$\displaystyle f(r) =\frac{r}{\sigma^{2}}\ e^{- \frac{r^{2}}{2\ \sigma^{2}}}$ (1)

... and the expected value of D is...

$\displaystyle E\{R\}= \sigma\ \sqrt{\frac{\pi}{2}}$ (2)

See for more details...

http://mathworld.wolfram.com/RayleighDistribution.html

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Posted on 04 19 2012 on www.mathhelpforum.com by the member cjtdevil and not yet solved…

How do you find a generating function for S(n,2) {S is the 2nd Stirling function} as a ratio of polynomials?...

That is not properly a probability question, even if it has been posted in the 'Advaced Statistic' section. Anyway it is interesting...

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Posted on 04 19 2012 on www.mathhelpforum.com by the member cjtdevil and not yet solved…

How do you find a generating function for S(n,2) {S is the 2nd Stirling function} as a ratio of polynomials?...

That is not properly a probability question, even if it has been posted in the 'Advaced Statistic' section. Anyway it is interesting...

Kind regards

$\chi$ $\sigma$

We can start from the definition of Second Kind Stirling Numbers...

$\displaystyle S(n,k)= \frac{1}{k!}\ \sum_{i=0}^{k} (-1)^{k-i}\ \binom{k}{i}\ i^{n}$ (1)

... that obey to the recursive relation...

$\displaystyle S(n+1,k)= k\ S(n,k)+ S(n,k-1)$ (2)

From (1) we derive $\displaystyle S(n,1)= 1-\delta_{n}$ and $\displaystyle S(n,2)=1 - 2^{n-1}- \frac{\delta_{n}}{2}$ so that from (2) we have...

$\displaystyle S(n+1,2)= 2^{n} \implies S(n,2)= 2^{n-1}$ (3)

... and its generating function is...

$\displaystyle f(x)=\frac{1}{2}\ \sum_{n=0}^{\infty} (2 x)^{n}= \frac{1}{2\ (1-2 x)}$ (4)

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Posted on 05 10 2012 on www.talkstat.com by the member rogersticks and not yet solved…

A game is played as follows: A pile contains 1 dollar and a coin is flipped. Each time a heads occurs, the amount in the piled is doubled and if a tail appears, the pile is given to the player. How much money should be payed to play this game?...

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Posted on 05 10 2012 on www.talkstat.com by the member rogersticks and not yet solved…

A game is played as follows: A pile contains 1 dollar and a coin is flipped. Each time a heads occurs, the amount in the piled is doubled and if a tail appears, the pile is given to the player. How much money should be payed to play this game?...

The probability that a tail appears after n-1 consecutive heads is $\displaystyle p_{n}= 2^{-n}$ and the cost has been $\displaystyle c_{n}= 2^{n-1}-1$ so that the expected value of the cost is...

$\displaystyle C= \sum_{n=1}^{\infty} c_{n}\ p_{n} = \sum_{n=1}^{\infty} (\frac{1}{2} - 2^{-n})$ (1)

Now the series (1) diverges, so that the expected cost is unlimited. That is a 'paradox' of the same type of the 'Saint Petersburg Paradox'...

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Posted the o6 30 2012 on www.matematicamente.it by superfox [original in Italian…] and not yet properly solved…

How to demonstrate that, given the r.v. X and Y uniformly distributed in (0,1), is $\displaystyle P \{X^{2}+Y^{2} <1 \}= \frac{\pi}{4}$ ?...

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Posted the o6 30 2012 on www.matematicamente.it by superfox [original in Italian…] and not yet properly solved…

How to demonstrate that, given the r.v. X and Y uniformly distributed in (0,1), is $\displaystyle P \{X^{2}+Y^{2} <1 \}= \frac{\pi}{4}$ ?...

If $X$ is uniformly distributed in (0,1), then the p.d.f. $f(x)$ of $X^{2}$ can be found as follows...

$\displaystyle P\{X^{2}<x \}= \int_{0}^{\sqrt{x}} d \xi = \sqrt{x} \implies f(x)=\begin{cases}\frac{1}{2\ \sqrt{x}} &\text{if}\ 0<x<1\\ 0 &\text{otherwise} \end{cases} $ (1)

The Laplace Transform of (1) is...

$\displaystyle F(s)= \mathcal {L}\{f(x) \}= \frac{\sqrt{\pi}}{2}\ \frac{1-e^{-s}}{\sqrt{s}}$ (2)

... so that the Laplace Transform of the p.d.f. $g(x)$ of the r,v, $Z=X^{2}+Y^{2}$ is...

$\displaystyle G(s)=F^{2}(s)= \frac{\pi}{4}\ \frac{(1-e^{-s})^{2}}{s}$ (3)

Now we are interested to the integral from 0 to 1 of the $g(x)$ that is...

$\displaystyle P\{Z<1\}= \mathcal{L}^{-1} \{\frac{G(s)}{s}\}_{x=1} = \frac{\pi}{4}$ (4)

Kind regards

$\chi$ $\sigma$

 

[CaptainBlack]

If $X$ is uniformly distributed in (0,1), then the p.d.f. $f(x)$ of $X^{2}$ can be found as follows...

$\displaystyle P\{X^{2}<x \}= \int_{0}^{\sqrt{x}} d \xi = \sqrt{x} \implies f(x)=\begin{cases}\frac{1}{2\ \sqrt{x}} &\text{if}\ 0<x<1\\ 0 &\text{otherwise} \end{cases} $ (1)

The Laplace Transform of (1) is...

$\displaystyle F(s)= \mathcal {L}\{f(x) \}= \frac{\sqrt{\pi}}{2}\ \frac{1-e^{-s}}{\sqrt{s}}$ (2)

... so that the Laplace Transform of the p.d.f. $g(x)$ of the r,v, $Z=X^{2}+Y^{2}$ is...

$\displaystyle G(s)=F^{2}(s)= \frac{\pi}{4}\ \frac{(1-e^{-s})^{2}}{s}$ (3)

Now we are interested to the integral from 0 to 1 of the $g(x)$ that is...

$\displaystyle P\{Z<1\}= \mathcal{L}^{-1} \{\frac{G(s)}{s}\}_{x=1} = \frac{\pi}{4}$ (4)

Kind regards

$\chi$ $\sigma$

Making hard work of a trivial problem: the question is asking for the area of the unit circle in the first quadrant, so without calculation the answer must be \(\pi/4\)

CB

 

[chisigma]

Posted on 07 05 2012 on www.mathhelpforum.com by the member Len and not yet solved…

Let $X_{1}$, $X_{2}$ and $X_{3}$ be i.i.d. r.v.'s with common probability density function...

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

Find $\displaystyle P \{ X_{1}<X_{3} -X_{2} \}$...

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Posted on 07 05 2012 on www.mathhelpforum.com by the member Len and not yet solved…

Let $X_{1}$, $X_{2}$ and $X_{3}$ be i.i.d. r.v.'s with common probability density function...

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

Find $\displaystyle P \{ X_{1}<X_{3} -X_{2} \}$...

The requested probability of course is $\displaystyle P \{X_{1} + X_{2} - X_{3} <0 \}$. The r.v. $X_{1}$ and $X_{2}$ have p.d.f. $ f(x)=\begin{cases} e^{-x} &\text{if}\ x \ge 0\\ 0 &\text{otherwise} \end{cases}$ and the r.v. $X_{3}$ has p.d.f $f(x)=\begin{cases} e^{x} &\text{if}\ x \le 0\\ 0 &\text{otherwise} \end{cases}$ so that the Fourier Transform of the p.d.f. of the r.v. $Z=X_{1} + X_{2} - X_{3}$ is...

$\displaystyle F(i\ \omega) = \frac{1}{4}\ \frac{1}{1+i\ \omega} + \frac{1}{2}\ \frac{1}{(1+i\ \omega)^{2}} + \frac{1}{4}\ \frac{1}{1-i\ \omega}$ (1)

From (1) we derive immediately that is $P\{X_{1} + X_{2} - X_{3} <0 \}= \frac{1}{4}$ ...

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Posted on 07 06 2012 on www.matematicamente.it by the member simeon [original in Italian language...] and not yet solved...

In 80000 trials we have 50000 'favorable results'... what is the 'error' if we estimate the probability as $p=\frac{50000}{80000}$?...

Although not 'rigorously expressed' that is a 'top probability problem' ...

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Posted on 07 06 2012 on www.matematicamente.it by the member simeon [original in Italian language...] and not yet solved...

In 80000 trials we have 50000 'favorable results'... what is the 'error' if we estimate the probability as $p=\frac{50000}{80000}$?...

Although not 'rigorously expressed' that is a 'top probability problem' ...

The probability of k 'successes' in n trials is given by...

$\displaystyle P_{n,k} = \binom{n}{k}\ p^{k}\ (1-p)^{n-k}$ (1)

... and for n 'large enough', according to the De Moivre-Laplace theorem is...

$\displaystyle P_{n,k} \sim \frac{1}{\sqrt {2\ \pi\ n\ p\ (1-p)}}\ e^{- \frac{(k-n\ p)^{2}}{2\ n\ p\ (1-p)}}$ (2)

If we consider the r.v. $X= \frac{k}{n} -p$ for n 'large enough' the (2) shows that X is normal distributed with $\mu=0$ and $\sigma^{2}= \frac{p\ (1-p)}{n}$...

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Posted on 07 05 2012 on www.artofproblemsolving.com by the member tian_275461 and not yet solved…

Indicating with $\{*\}$ the function 'fractional part of', prove that $\displaystyle \{ \frac {e^{n}}{2\ \pi} \}$ is uniformly distributed in [0,1)

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Posted on 07 05 2012 on www.artofproblemsolving.com by the member tian_275461 and not yet solved…

Indicating with $\{*\}$ the function 'fractional part of', prove that $\displaystyle \{ \frac {e^{n}}{2\ \pi} \}$ is uniformly distributed in [0,1)

This is probably the most 'difficult' problem posted in this section till now. According to Weyl's criterion a sequence $a_{n}$ is uniformly distributed modulo 1 if and only if for any integer $k \ne 0$ is...

$\displaystyle \lim_{m \rightarrow \infty} \frac{1}{m}\ \sum_{n=1}^{m} e^{2\ \pi\ i\ k\ a_{n}} =0$ (1)

On the basis of (1) Weyls has demonstrated that if the sequence is a polynomial p(n) with at least one irrational coefficient then the sequence is uniformly distributed modulo 1. If that holds also for a polynomial with infinite coefficients, then is...

$\displaystyle \frac{e^{n}}{2\ \pi} = \sum_{k=0}^{\infty} \frac{n^{k}}{2\ \pi\ k!}$ (2)

and the sequence $\displaystyle \frac{e^{n}}{2\ \pi}$ is uniformly distributed modulo 1. A demonstration of that however does't seem comfortable...

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Posted on 07 12 2012 on www.talkstat.com by the member kzeldon and not yet solved…

Working a problem from a text that asks if you have n balls and place them in n bins, what is the probability that exactly one bin remains empty. The text gives the answer as C(n 2)*n!/n^n. I wish to understand where this expression came from...

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Posted on 07 12 2012 on www.talkstat.com by the member kzeldon and not yet solved…

Working a problem from a text that asks if you have n balls and place them in n bins, what is the probability that exactly one bin remains empty. The text gives the answer as C(n 2)*n!/n^n. I wish to understand where this expression came from...

The solution requires the concept of multinomial distribution, illustrated for example in...

http://mathworld.wolfram.com/MultinomialDistribution.html

In general if we have a set of n r.v. $\displaystyle X_{1},X_{2},...,X_{n}$ with $\displaystyle \sum_{k=1}^{n} X_{k}=N$, the r.v. are multinomial distributed if their probability function is...

$\displaystyle P\{X_{1}=x_{1},X_{2}=x_{2},...,X_{n}=x_{n}\}= N!\ \prod_{k=1}^{n} \frac{\theta_{k}^{x_{k}}}{x_{k}!}$ (1)

... where all the $\theta_{k}$ are positive and is $\displaystyle \sum_{k=1}^{n} \theta_{k}=1$. In our case is $\displaystyle N=n$, $\displaystyle \theta_{k}=\frac{1}{n}$ and let's start with the set $\displaystyle x_{1}=0, x_{2}=2, x_{2}=x_{3}=...=x_{n}=1$. From (1) we obtain...

$\displaystyle P\{X_{1}=0, X_{2}=2, X_{3}=X_{4}=...=X_{n}=1\} = \frac{n!}{2\ n^{n}}$ (2)

Since the set satisfying the same condition are $\displaystyle n\ (n-1)$ the probability that only one bin remains empty is...

$\displaystyle P= \frac{n\ (n-1)\ n!}{2\ n^n}$ (3)

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Posted on 07 11 2012 on www.talkstat.com by the member wuid and not yet properly solved…

In a gambling game , you can win 1 dollar in each round with probability 0.6 or lose 2 dollars in probability 0.4 and suppose you start with 100 dollars. Find the probability that after 10 rounds you have between 93 to 107 dollars...

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Posted on 07 11 2012 on www.talkstat.com by the member wuid and not yet properly solved…

In a gambling game , you can win 1 dollar in each round with probability 0.6 or lose 2 dollars in probability 0.4 and suppose you start with 100 dollars. Find the probability that after 10 rounds you have between 93 to 107 dollars...

Setting u the number of 'up jumps', d the number of 'down jumps' and $\displaystyle \Delta=u-2\ d$, for $\displaystyle -7 \le \Delta \le 7$ we have the following possible cases...

$\displaystyle \Delta=-5 \implies u=5,\ d=5$

$\displaystyle \Delta=-2 \implies u=6,\ d=4$

$\displaystyle \Delta=1 \implies u=7,\ d=3$

$\displaystyle \Delta=4 \implies u=8,\ d=2$

$\displaystyle \Delta=7 \implies u=9,\ d=1$ (1)

From (1) we derive that the requested probability is...

$\displaystyle P= \sum_{u=5}^{9} \binom {10}{u}\ p^{u}\ (1-p)^{10-u}$ (2)

... where $p=.6$ is the probability of 'up jump'. The effective computation of (2) can be made with a simple calculator and it is left to the reader...

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Posted on 07 20 2012 on www.talkstats.com by the user goodmorningda and not yet solved...

View attachment 274

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Posted on 07 20 2012 on www.talkstats.com by the user goodmorningda and not yet solved...

https://www.physicsforums.com/attachments/274

a) Mean and variance are computed in standard fashion...

$\displaystyle E\{X\}= \mu = \int_{0}^{1} 2\ x\ (1-x)\ dx = \frac{1}{3}$ (1)

$\displaystyle E\{(X-\mu)^{2}\}= \sigma^{2}= \int_{0}^{1} 2\ (x-\frac{1}{3})^{2}\ (1-x)\ dx= \frac{1}{18} \implies \sigma=\frac{1}{3\sqrt{2}}$ (2)

... so that is $\sigma=\mu$...

b) the Laplace Transform of the p.d.f. of X is...

$\displaystyle \mathcal{L} \{f(x)\} = F(s)= \int_{0}^{1} 2\ (1-x)\ e^{- s\ x}\ dx = 2\ \frac{s-1+ e^{-s}}{s^{2}}$ (3)

... so that the Laplace Transform of the p.d.f. g(x) of the r.v. Y is...

$\displaystyle \mathcal{L} \{g(x)\} = F^{2}(s)= 4\ \{(\frac{1}{s^{2}} - \frac{2}{s^{3}} + \frac{1}{s^{4}}) + (\frac{1}{s^{3}} -\frac{2}{s^{4}})\ e^{-s} + \frac{1}{s^{4}}\ e^{-2\ s}\}$ (4)

... and g(x) is found computing the inverse Laplace Transform of (4)...

$\displaystyle g(x)= \mathcal{L}^{-1} \{F^{2}(s)\}= \begin{cases}4\ (x - x^{2} + \frac{1}{6}\ x^{3}) &\ \text{if }\ 0<x<1\\ 4\ (\frac{4}{3} -2\ x + x^{2} -\frac{1}{6}\ x^{3}) &\text {if } 1<x<2\\ 0 &\text{otherwise}\end{cases}$ (5)

d) from (5) we derive quickly...

$\displaystyle P\{1<Y<2\}= 1-P\{0<Y<1\} = 1- 4\ \int_{0}^{1} (x-x^{2}+ \frac{x^{3}}{6})\ dx = 1-\frac{5}{6} = \frac{1}{6}$ (6)

The points c) and e) will be attacked in a successive post...

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Posted on 07 20 2012 on www.talkstats.com by the user goodmorningda and not yet solved...

https://www.physicsforums.com/attachments/274

The solution of the points c) requires the formulas that the French mathematician Ire'ne'e Jules Bienayme' found about hundred and fifty years ago about the mean and variance of the r.v. $\displaystyle X= \sum_{k=1}^{n} X_{k}$ where the $X_{k}$ are a set of n independent r.v. with mean value $\mu_{k}=E\{X_{k}\}$ and mean square error $\sigma^{2}_{k}= E\{(X_{k}-\mu_{k})^{2}\}$...

$\displaystyle E\{X\}=\mu = \sum_{k=1}^{n} \mu_{k}$

$\displaystyle E\{(X-\mu)^{2}\}=\sigma^{2} = \sum_{k=1}^{n} \sigma^{2}_{k}$ (1)

In our case is $\displaystyle n=2$, $\displaystyle \mu_{1}=\mu_{2}= \frac{1}{3}$ and $\displaystyle \sigma^{2}_{1}=\sigma^{2}_{2}= \frac{1}{18}$, so that it would be $\displaystyle \mu=\frac{2}{3}$ and $\displaystyle \sigma^{2}= \frac{1}{9} \implies \sigma=\frac{1}{3}$. Bienayme’ however didn’t give rigorousdemonstration of the fact that the (1) are valid for any n and the discussion is still open so that now we verify these results using the p.d.f. of X we found in last post…

$\displaystyle \mu= 4\ \{\int_{0}^{1} x\ (x-x^{2}+ \frac{1}{6}\ x^{3})\ dx + \int_{1}^{2} x\ (\frac{4}{3} -2\ x + x^{2} -\frac{1}{6}\ x^{3})\ dx \} = \frac{28}{60} + \frac{4}{20}= \frac{2}{3}$ (2)

$\displaystyle \sigma^{2}= 4\ \{\int_{0}^{1} (x-\frac{2}{3})^{2}\ (x-x^{2}+ \frac{1}{6}\ x^{3})\ dx + \int_{1}^{2} (x-\frac{2}{3})^{2}\ (\frac{4}{3} -2\ x + x^{2} -\frac{1}{6}\ x^{3})\ dx \} = \frac{8}{135} + \frac{28}{540}= \frac{1}{9}$ (3)

The solution of e) is also very interesting and it will be made in a successive post...

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Posted on 07 20 2012 on www.talkstats.com by the user goodmorningda and not yet solved...

https://www.physicsforums.com/attachments/274

The solution of the last question implies the use of the so called 'Central Limit Theorem'. Formulated in the years 1920-1930, the CLT has an enormous number of 'daddies', but in my [very modest] opinion it is the natural development of the Bienayme' result of the previous century. In simply words the CLT establishes that, given n independent r.v. $\displaystyle X_{1},\ X_{2},..., X_{n}$ each of them has an arbitrary p.d.f. with mean $\displaystyle \mu_{k}$ and variance $\displaystyle \sigma^{2}_{k}$, then the r.v. $\displaystyle X= \sum_{k=1}^{n} X_{k}$ for n 'large enough' has a p.d.f that approaches ...

$\displaystyle f(x)= \frac{1}{\sigma \sqrt{2\ \pi}}\ e^{-\frac{(x-\mu)^{2}}{2\ \sigma^{2}}}$ (1)

... where...

$\displaystyle \mu=\sum_{k=1}^{n} \mu_{k}$

$\displaystyle \sigma^{2}= \sum_{k=1}^{n} \sigma^{2}_{k}$ (2)

In our case is n=100 [we suppose it 'large enough'...], $\displaystyle \mu_{k}= \frac{1}{3}$, $\displaystyle \sigma^{2}_{k}= \frac{1}{18}$ so that is $\displaystyle \mu=\frac{100}{3}$ and $\displaystyle \sigma^{2}= \frac{100}{18}$. Now the probability $\displaystyle P\ \{X>x\}$ is given by...

$\displaystyle P\ \{X>x\}= \frac{1}{2}\ \text{erfc}\ (\frac{z}{\sqrt{2}})$ (3)

... where is $\displaystyle z=\frac{x-\mu}{\sigma}$. For x=34 is...

​$\displaystyle P\ \{X>34\}= \frac{1}{2}\ \text{erfc}\ (\frac{1}{5})= .388649... $ (4)

Kind regards

$\chi$ $\sigma$

 

[CaptainBlack]

The solution of the last question implies the use of the so called 'Central Limit Theorem'. Formulated in the years 1920-1930, the CLT has an enormous number of 'daddies'

1820-30's by Laplace and Poisson.

CB

 

[chisigma]

1820-30's by Laplace and Poisson.

CB

A good description of the historical course of the CLT is the following...

http://www.sal.tkk.fi/vanhat_sivut/Opinnot/Mat-2.108/pdf-files/emet03.pdf

As curious tail I can add what I read in a Alan Turing's biography written by Andrew Hodges: a 'proof' of the CLT was the subject of Alan Turing's 1934 Fellowship Dissertation for King's College at the University of Cambridge. Only after submitting the work did Turing learn it had already been proved in 1922 by Jarl Waldemar Lindenberg and consequently Turing's dissertation was never published...

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Posted on 07 26 2012 on www.mathhelpforum.com by the member salohcin and not yet solved…

Suppose you have some random variable X that is distributed according to a Pareto distribution...

$\displaystyle f(x)=\begin{cases} k\ \frac{m^{k}}{x^{k-1}} &\text{if}\ x>m\\ 0 &\text{if}\ x<m \end{cases}$

... and then you have a transformation $Y = c\ X$ where c is a constant. I want to find out the standard deviation of the natural log of Y. I have been trying for several hours but have had no success...

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Posted on 07 26 2012 on www.mathhelpforum.com by the member salohcin and not yet solved…

Suppose you have some random variable X that is distributed according to a Pareto distribution...

$\displaystyle f(x)=\begin{cases} k\ \frac{m^{k}}{x^{k-1}} &\text{if}\ x>m\\ 0 &\text{if}\ x<m \end{cases}$

... and then you have a transformation $Y = c\ X$ where c is a constant. I want to find out the standard deviation of the natural log of Y. I have been trying for several hours but have had no success...

I apologize with the community of MHB but the 'Pareto distribution' I posted is wrong. The correct expression is...

$\displaystyle f(x)=\begin{cases} k\ \frac{m^{k}}{x^{k+1}} &\text{if}\ x>m\\ 0 &\text{if}\ x<m \end{cases}$ (1)

Very sorry! (Thinking) ...

Kind regards

$\chi$ $\sigma$

 

[chisigma]

I apologize with the community of MHB but the 'Pareto distribution' I posted is wrong. The correct expression is...

$\displaystyle f(x)=\begin{cases} k\ \frac{m^{k}}{x^{k+1}} &\text{if}\ x>m\\ 0 &\text{if}\ x<m \end{cases}$ (1)

Very sorry! ...

Now that we have the effective 'Pareto distribution' we are able to solve the question. Because is $\displaystyle \ln (c\ X)= \ln X + \ln c$ and the variance is required, we can set c=1 and operate on the r.v. $\displaystyle Z= \ln X$. First we compute the probability...

$\displaystyle P \{\ln m <Z< z \} = P \{m< X< e^{z} \} = \int_{m}^{e^{z}} k\ \frac{m^{k}}{x^{k-1}}\ dx=$
$\displaystyle = \begin{cases} 1-(m\ e^{-z})^{k} &\text{if}\ z> \ln m \\ 0 &\text{if}\ z< \ln m \end{cases}$ (1)

The p.d.f. of the r.v. $\ln X$ is found deriving (1)...

$\displaystyle g(z) = \begin{cases} k\ m^{k}\ e^{-k\ z} &\text{if}\ z> \ln m \\ 0 &\text{if}\ z< \ln m \end{cases}$ (2)

The mean value of $\ln X$ is...

$\displaystyle E \{ \ln X\} = \mu = \int_{\ln m}^{\infty} z\ g(z)\ dz= \frac{1}{k} + \ln m $ (3)

... and the variance ...

$\displaystyle \text{Var}\ \{\ln X \}= \sigma^{2}= \int_{\ln m}^{\infty} (z-\mu)^{2}\ g(z)\ dz= \frac{1}{k^{2}}$ (4)

... so that the final result is [that is not a surprise...] quite simple ...

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Posted the o7 29 2012 on www.matematicamente.it by Martessa [original in Italian…] and not yet solved…

Alice writes on a paper an integer between 1 and n and asks Bob to guess it. Bob can try indefinitely until he guess the number. What is the expected number of efforts of Bob to have success?...

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Posted the o7 29 2012 on www.matematicamente.it by Martessa [original in Italian…] and not yet solved…

Alice writes on a paper an integer between 1 and n and asks Bob to guess it. Bob can try indefinitely until he guess the number. What is the expected number of efforts of Bob to have success?...

The probability that Bob guess the number at the k-th trial is...

$\displaystyle P_{k,n}= \frac{n-1}{n}\ \frac{n-2}{n-1}\ ...\ \frac{1}{n-k+1}= \frac{1}{n}$ (1)

... so that the expected number of trials is...

$\displaystyle E\{k\}= \frac{1}{n}\ \sum_{k=1}^{n} k = \frac{n+1}{2}$ (2)

Very easy!...

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Posted on 07 26 2012 on www.mathhelpforum.com by the member salohcin and not yet solved…

Suppose you have some random variable X that is distributed according to a Pareto distribution...

$\displaystyle f(x)=\begin{cases} k\ \frac{m^{k}}{x^{k-1}} &\text{if}\ x>m\\ 0 &\text{if}\ x<m \end{cases}$

... and then you have a transformation $Y = c\ X$ where c is a constant. I want to find out the standard deviation of the natural log of Y. I have been trying for several hours but have had no success...

This thread, that I really hope has been appreciated by You, now has a too large size so that it will be terminated and a new thread with the same scope will be open. Before to do that however I would to spent some word about the figure of the Italian engineer, sociologist, economist, political scientist and philosopher Vilfredo Pareto...

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Pareto was born in Paris in 1848 of a nobile Genoese family and in 1870 he earned the degree of civil engineering in Polytechnic of Turin. For some years he worked in the Italian Railway Company and after he decided to dedicate his interest to the economy. In my opinion his greatest credit is to have been one of the first to use Math instruments in economical studies. His name is tied to the Pareto principle [know as '80-20 principle'...] according to that the most stable economical status of a country is when, independently form historical and geographical contests, 80% of the wealth is owned by 20% of the population. How present is the Pareto's thought You can understand considering that the present global status of economy is near 90-10 and that , according to Pareto, sooner or later will produce explosive consequences...

Kind regards

$\chi$ $\sigma$