# Joint Probability Mass Functions

• Apr 29th 2010, 11:16 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 $N$ which has a poisson distribution with expectation $\lambda$ so $P(N=n) = \lambda^ne^{-\lambda}/n!$ for $n=0,1,2,...$ Independently for each egg laid there is probability $1/2$ each that it develops into a male insect or a female insect. Let $X$ denote the number of male offspring, and $Y$ the number of female offspring.

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

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

The answers I've been given are

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

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

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

Cheers
• May 3rd 2010, 01:29 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 $N$ which has a poisson distribution with expectation $\lambda$ so $P(N=n) = \lambda^ne^{-\lambda}/n!$ for $n=0,1,2,...$ Independently for each egg laid there is probability $1/2$ each that it develops into a male insect or a female insect. Let $X$ denote the number of male offspring, and $Y$ the number of female offspring.

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

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

The answers I've been given are

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

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

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 $\lambda/2$, as does the number of females born.

So:

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

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