Unsolved statistics questions from other sites, part II

[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

$\chi$ $\sigma$

 

[chisigma]

Posted the 8 25 2013 on www.artofproblemsolving.com by the user adrianomeis and not yet solved...

Three sectors are chosen at random from circle C, having angles $\displaystyle \frac{\pi}{10}, \frac{2 \pi}{10}, \frac{3 \pi}{10}$ respectively. What is the probability that these sectors have no points (other than the centre) in common?...

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Posted the 8 25 2013 on www.artofproblemsolving.com by the user adrianomeis and not yet solved...

Three sectors are chosen at random from circle C, having angles $\displaystyle \frac{\pi}{10}, \frac{2 \pi}{10}, \frac{3 \pi}{10}$ respectively. What is the probability that these sectors have no points (other than the centre) in common?...

The problem is [relatively] easy if we condider that, calling $\displaystyle S_{k}$ the sector the angle of which is $\displaystyle \frac{k}{10} \pi$, ...

a) $\displaystyle S_{3}$ can be considered without limitations the sector for which is $\displaystyle 0< \theta < \frac{3}{10} \pi$...

b) all the favourable cases are represented by the distict sequence $\displaystyle S_{3}-S_{2}- S_{1}$ because the other sequence $\displaystyle S_{3}-S_{1}-S_{2}$ is the same swapinng $\displaystyle \theta$ and $\displaystyle - \theta$...

Under these assumptions the requested probability is...

$\displaystyle P = \frac{1}{4 \pi^{2}}\ \int_{\frac{3}{10} \pi}^{\frac{17}{10} \pi} d x \int_{x + \frac{2}{10} \pi}^{\frac{19}{10} \pi} d y = \frac{49}{200}$

Kind regards

$\chi$ $\sigma$

P.S. : as explained in the post # 105, the result is wrong and the correct result should be $\displaystyle P= \frac{49}{100}$

 

[chisigma]

Posted on 8 27 2013 on www.artofproblemsolving.com by the user aktyw19 and not yet solved...

Points A, B and C are randomly chosen inside a circle. A fourth point, O is chosen. What is the probability that O lies inside triangle ABC?...

Kind regards

$\chi$ $\sigma$

 

[zzephod]

The problem is [relatively] easy if we condider that, calling $\displaystyle S_{k}$ the sector the angle of which is $\displaystyle \frac{k}{10} \pi$, ...

a) $\displaystyle S_{3}$ can be considered without limitations the sector for which is $\displaystyle 0< \theta < \frac{3}{10} \pi$...

b) all the favourable cases are represented by the distict sequence $\displaystyle S_{3}-S_{2}- S_{1}$ because the other sequence $\displaystyle S_{3}-S_{1}-S_{2}$ is the same swapinng $\displaystyle \theta$ and $\displaystyle - \theta$...

Under these assumptions the requested probability is...

$\displaystyle P = \frac{1}{4 \pi^{2}}\ \int_{\frac{3}{10} \pi}^{\frac{17}{10} \pi} d x \int_{x + \frac{2}{10} \pi}^{\frac{19}{10} \pi} d y = \frac{49}{200}$

Kind regards

$\chi$ $\sigma$

Which is wrong, since your second integral does not allow for the \(\displaystyle S_1\) sector to fall in the gap between \(\displaystyle S_3\) and \(\displaystyle S_2\) also the upper limit in the first integral should be \(\displaystyle {\small{\frac{18}{10}}}\pi\).

Monte-Carlo simulation (which could have error but ..) gives a probability of no overlap of \(\displaystyle 0.75\) with SE roughly \(\displaystyle 1.5 \times 10^{-4}\).

.

 

[chisigma]

The problem is [relatively] easy if we condider that, calling $\displaystyle S_{k}$ the sector the angle of which is $\displaystyle \frac{k}{10} \pi$, ...

a) $\displaystyle S_{3}$ can be considered without limitations the sector for which is $\displaystyle 0< \theta < \frac{3}{10} \pi$...

b) all the favourable cases are represented by the distict sequence $\displaystyle S_{3}-S_{2}- S_{1}$ because the other sequence $\displaystyle S_{3}-S_{1}-S_{2}$ is the same swapinng $\displaystyle \theta$ and $\displaystyle - \theta$...

Under these assumptions the requested probability is...

$\displaystyle P = \frac{1}{4 \pi^{2}}\ \int_{\frac{3}{10} \pi}^{\frac{17}{10} \pi} d x \int_{x + \frac{2}{10} \pi}^{\frac{19}{10} \pi} d y = \frac{49}{200}$

What I said in the point b) is 'half correct' and 'half wrong', in the sense that all the favourable cases are represented by both the $\displaystyle S_{3}-S_{2}- S_{1}$ and $\displaystyle S_{3}-S_{1}-S_{2}$ sequences. To understand that let suppose that the sectors have angles $\displaystyle \theta_{3} = \theta_{2}= \theta_{1}=0$, so that the no overlapping probability is of course P=1. If we consider only the sequence $\displaystyle S_{3}-S_{2}- S_{1}$ and proceed we obtain... $\displaystyle P = \frac{1}{4 \pi^{2}}\ \int_{0}^{2 \pi} d x \int_{x}^{2 \pi} d y = \frac{1}{2}$

... and that demonstrates that also the sequence $\displaystyle S_{3}-S_{1}-S_{2}$ must be taken into account. Proceeding along this way we obtain...

$\displaystyle P = \frac{1}{4 \pi^{2}}\ \int_{\frac{3}{10} \pi}^{\frac{17}{10} \pi} d x \int_{x + \frac{2}{10} \pi}^{\frac{19}{10} \pi} d y + \frac{1}{4 \pi^{2}}\ \int_{\frac{3}{10} \pi}^{\frac{17}{10} \pi} d x \int_{x + \frac{1}{10} \pi}^{\frac{18}{10} \pi} d y = 2\ \frac{49}{200} = \frac{49}{100}$

Kind regards

$\chi$ $\sigma$