# A couple of questions

• Sep 30th 2008, 12:03 PM
Xan21
A couple of questions
Hi i was working through my Mathematical Statistics book and i came across a couple of examples that have got me stumped.

A Die is cast independently N times until a six appears on the up side of the die.
a) Find probability p(N) and prove it is a discrete distribution
b) Find p(N=2,4,6,8,...)
c) Find p(N=3,6,9,12,...)
d) Find the cdf for random variable N

I know that the pmf is $\displaystyle p(N) = ( {5} / {6})^{N-1} ({1} /{6})$
And it gives the example of p(N=1,3,5,7,...) = $\displaystyle ({6}/{11})$

Edit:
For a) is this proof that it is a discrete distribution?
$\displaystyle p(N) = \sum_{x=1}^{\infty} ({5} / {6})^{N-1} ({1} /{6}) = {1}$

Second Question:
Let probability density f(x)=x/2 for 0<x<2 and 0 elsewhere. Compute E(1/X), compute cumulative distribution function and probability density function of Y=1/X, find the expected value and variance of Y.

Heres what i did so far:

$\displaystyle f(x) = \frac{x}{2}$
$\displaystyle E(\frac{1}{x}) = \int_{-\infty}^{\infty} (\frac{1}{x}) f(x) dx = \int_{0}^{2} (\frac{1}{x}) (\frac{x}{2}) dx = {1}$

As for Y=1/X i have no idea how to get it. Is it a transformation? My book is terrible and i am lost.

cdf =
{0 if x < 0}
{x if 0 $\displaystyle \leq$ x < 1}
{1 if x $\displaystyle \geq$ 1}

pdf = (i have no idea)
• Sep 30th 2008, 01:15 PM
Plato
I am not sure about your question. But here is the example part .
$\displaystyle P\left( {\left\{ {1,3,5, \cdots } \right\}} \right) = \sum\limits_{k = 0}^\infty {\left( {\frac{5} {6}} \right)^{2k} \left( {\frac{1} {6}} \right) = \frac{{\frac{1} {6}}} {{1 - \left( {\frac{5} {6}} \right)^2 }} = \frac{6} {{11}}}$
• Sep 30th 2008, 01:50 PM
Xan21
Thanks. However i dont understand how you came to get
$\displaystyle \frac{\frac{1}{6}}{1-({\frac{5}{6}})^2}$

but I got this for b)
$\displaystyle p(2,4,6,8,...) = \sum_{x=1}^{\infty} ({5} / {6})^{2x-1} ({1} /{6}) = ({5}/{11})$

and c)
$\displaystyle p(3,6,9,12,...) = \sum_{x=1}^{\infty} ({5}/{6})^{3x} ({1}/{6}) = ({125}/{546})$

as for d) i have no idea how to do a cdf if there is no limit...
• Sep 30th 2008, 02:13 PM
Plato
Quote:

Originally Posted by Xan21
$\displaystyle p(2,4,6,8,...) = \sum_{x=1}^{\infty} ({5} / {6})^{2x} ({1} /{6}) = ({5}/{11})$

While $\displaystyle ({5}/{11})$ is correct for part a, $\displaystyle \sum_{x=1}^{\infty} ({5} / {6})^{2x} ({1} /{6}) = \frac {25}{66} \not= \frac{5}{11})$.
How did you get that sum?

The correct sum is $\displaystyle p(2,4,6,8,...) = \sum_{x=1}^{\infty} ({5} / {6})^{2x-1} ({1} /{6}) = ({5}/{11})$
• Sep 30th 2008, 02:35 PM
Xan21
Oh i did put the "-1" in there.... just missed it when i was typing it up. (Giggle)
• Sep 30th 2008, 02:36 PM
icemanfan
The cdf for random variable N is simply equal to P(1,2,3,...N), which is $\displaystyle \sum _{k=1} ^N (1/6)(5/6)^{k-1} = (1/6)\cdot \frac{1 - (5/6)^N}{1 - 5/6} = 1 - (5/6)^N$.
• Sep 30th 2008, 02:59 PM
Xan21
Quote:

Originally Posted by icemanfan
The cdf for random variable N is simply equal to P(1,2,3,...N), which is $\displaystyle \sum _{k=1} ^N (1/6)(5/6)^{k-1} = (1/6)\cdot \frac{1 - (5/6)^N}{1 - 5/6} = 1 - (5/6)^N$.

ok i see... thanks. Added another question to the top!
• Sep 30th 2008, 03:11 PM
icemanfan
Regarding the added question: I'm not sure about this, but I think $\displaystyle E(1/X) = \frac{1}{E(X)}$.
• Sep 30th 2008, 03:56 PM
Plato
Quote:

Originally Posted by icemanfan
Regarding the added question: I'm not sure about this, but I think $\displaystyle E(1/X) = \frac{1}{E(X)}$.

Think about this: $\displaystyle \sum\limits_{k = 1}^\infty {x^k } = \frac{x} {{1 - x}}\quad \Rightarrow \quad \sum\limits_{k = 1}^\infty {kx^{k - 1} } = \frac{1} {{\left( {1 - x} \right)^2 }}$
• Oct 1st 2008, 11:41 AM
Xan21
Quote:

Originally Posted by Plato
Think about this: $\displaystyle \sum\limits_{k = 1}^\infty {x^k } = \frac{x} {{1 - x}}\quad \Rightarrow \quad \sum\limits_{k = 1}^\infty {kx^{k - 1} } = \frac{1} {{\left( {1 - x} \right)^2 }}$

Is that to me or him? Cuz i thought i did the E(1/X) part right... thats how they do it in the book....