# Joint Probability Mass Functions

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• Apr 29th 2010, 10:16 AM
shmounal
Joint Probability Mass Functions
Hi, I'm revising for an upcoming test and have a past paper with answers but I can't understand where they are coming from. If anyone could try explaining this too me it would be greatly appreciated.

The number of eggs laid by an insect is a random variable $\displaystyle N$ which has a poisson distribution with expectation$\displaystyle \lambda$ so $\displaystyle P(N=n) = \lambda^ne^{-\lambda}/n!$ for $\displaystyle n=0,1,2,...$ Independently for each egg laid there is probability $\displaystyle 1/2$ each that it develops into a male insect or a female insect. Let $\displaystyle X$ denote the number of male offspring, and $\displaystyle Y$ the number of female offspring.

i) Find the joint probability mass function of $\displaystyle (N,X)$

ii) Show that the marginal distribution of $\displaystyle X$, the number of male offspring, is Poisson with expectation $\displaystyle \lambda/2$

The answers I've been given are

i) $\displaystyle nx2^{-n}\lambda^ne^{-\lambda} /n!$ for non negative intergers $\displaystyle x \leq n$

ii)For non-negative integers $\displaystyle x$,
$\displaystyle P_{.x}= \sum_{n=x}^{\infty}P_{nx} = [e^{-\lambda}(\lambda/2)^x/x!] \sum_{n\ge x}(\lambda/2)^{n-x}/(n-x)! = e^{-\lambda/2}(\lambda/2)^x /x!$ as the sum is $\displaystyle \sum_{m\ge0}(\lambda/2)^m/m! = e^{\lambda/2}$

If anyone can explain how you get to these answers it would be a massive help,

Cheers
• May 3rd 2010, 12:29 AM
CaptainBlack
Quote:

Originally Posted by shmounal
Hi, I'm revising for an upcoming test and have a past paper with answers but I can't understand where they are coming from. If anyone could try explaining this too me it would be greatly appreciated.

The number of eggs laid by an insect is a random variable $\displaystyle N$ which has a poisson distribution with expectation$\displaystyle \lambda$ so $\displaystyle P(N=n) = \lambda^ne^{-\lambda}/n!$ for $\displaystyle n=0,1,2,...$ Independently for each egg laid there is probability $\displaystyle 1/2$ each that it develops into a male insect or a female insect. Let $\displaystyle X$ denote the number of male offspring, and $\displaystyle Y$ the number of female offspring.

i) Find the joint probability mass function of $\displaystyle (N,X)$

ii) Show that the marginal distribution of $\displaystyle X$, the number of male offspring, is Poisson with expectation $\displaystyle \lambda/2$

The answers I've been given are

i) $\displaystyle nx2^{-n}\lambda^ne^{-\lambda} /n!$ for non negative intergers $\displaystyle x \leq n$

ii)For non-negative integers $\displaystyle x$,
$\displaystyle P_{.x}= \sum_{n=x}^{\infty}P_{nx} = [e^{-\lambda}(\lambda/2)^x/x!] \sum_{n\ge x}(\lambda/2)^{n-x}/(n-x)! = e^{-\lambda/2}(\lambda/2)^x /x!$ as the sum is $\displaystyle \sum_{m\ge0}(\lambda/2)^m/m! = e^{\lambda/2}$

If anyone can explain how you get to these answers it would be a massive help,

Cheers

The number of males has a Poisson distribution with mean $\displaystyle \lambda/2$, as does the number of females born.

So:

$\displaystyle Pr(N,X)=p(X,\lambda/2)p(N-X,\lambda/2)$

CB
• May 5th 2010, 06:07 AM
shmounal
Ah ok thanks, i think part of what was confusing me is that my answers had $\displaystyle nx$ when it should have been$\displaystyle \binom{n}{x}$. With realising that and your help i think ive got it!!