# Thread: Joint probability mass function

1. ## Joint probability mass function

Suppose X and Y are independent Poisson random variables with parameters a and b respectively. Find the joint mass function

P[X = k, Y = m]

I got

(exp(-a).a^k)/k! x (exp(-b).b^m)/m!

Find P[X + Y = n]

I got = P[X = k, Y = n - k]

= sum over k to n (exp(-a).a^k)/k! x (exp(-b).b^n - k)/(n-k)!

But I'm not sure how to complete

is this right so far?

thanks

2. Originally Posted by James0502
Suppose X and Y are independent Poisson random variables with parameters a and b respectively. Find the joint mass function

P[X = k, Y = m]

I got

(exp(-a).a^k)/k! x (exp(-b).b^m)/m!

Find P[X + Y = n]

I got = P[X = k, Y = n - k]

= sum over k to n (exp(-a).a^k)/k! x (exp(-b).b^n - k)/(n-k)!

But I'm not sure how to complete
So you got (correctly) $P(X+Y=n)=\sum_{k=0}^n e^{-a}\frac{a^k}{k!}e^{-b}\frac{b^{n-k}}{(n-k)!}=e^{-(a+b)}\sum_{k=0}^n \frac{1}{k!(n-k)!}a^k b^{n-k}$. This should remind you the binomial formula : remember that ${n\choose k}=\frac{n!}{k!(n-k)!}$...

3. Originally Posted by James0502
Suppose X and Y are independent Poisson random variables with parameters a and b respectively. Find the joint mass function

P[X = k, Y = m]

I got

(exp(-a).a^k)/k! x (exp(-b).b^m)/m!

Find P[X + Y = n]

I got = P[X = k, Y = n - k]

= sum over k to n (exp(-a).a^k)/k! x (exp(-b).b^n - k)/(n-k)! Isomorphism: Not from k to n, but k varying from 0 to n.

But I'm not sure how to complete

is this right so far?

thanks
First of all observe the correction in red.

What you have done is not wrong... but it is not very insightful..

$P(X+Y = n) = \sum_{k = 0}^{k=n} P(X = k, Y = n - k) = \sum_{k = 0}^{k=n} P(X = k) P(Y = n - k) =$ $\sum_{k = 0}^{k=n}\left(\frac{e^{-a} a^k}{k!}\right)\left(\frac{e^{-b}.b^{n - k}}{(n-k)!}\right)
$

Lets continue with what you should have done:

$\sum_{k = 0}^{k=n}\left(\frac{e^{-a} a^k}{k!}\right)\left(\frac{e^{-b}.b^{n - k}}{(n-k)!}\right) = e^{-(a+b)} b^n \sum_{k = 0}^{k=n} \dfrac{\left(\frac{a}{b}\right)^k}{k! (n-k)!}$

Now we can write $\frac1{k! (n-k)!} = \frac1{n!}\frac{n!}{k! (n-k)!} = \frac{\binom{n}{k}}{n!}$

So:

$e^{-(a+b)} b^n \sum_{k = 0}^{k=n} \dfrac{\left(\frac{a}{b}\right)^k}{k! (n-k)!} = e^{-(a+b)} b^n \sum_{k = 0}^{k=n} \dfrac{\binom{n}{k}\left(\frac{a}{b}\right)^k}{n!} =$ $e^{-(a+b)} b^n \dfrac{\sum_{k = 0}^{k=n} \binom{n}{k}\left(\frac{a}{b}\right)^k}{n!}$

Remember Binomial Theorem? It says $\sum_{k = 0}^{k=n} \binom{n}{k}\left(\frac{a}{b}\right)^k = \left(1 + \frac{a}{b}\right)^n$

This means that

$e^{-(a+b)} b^n \dfrac{\sum_{k = 0}^{k=n} \binom{n}{k}\left(\frac{a}{b}\right)^k}{n!} = e^{-(a+b)} b^n \dfrac{\left(1 + \frac{a}{b}\right)^n}{n!} =e^{-(a+b)} b^n\dfrac{(a+b)^n}{n! b^n} = \frac{e^{-(a+b)} (a+b)^n }{n!}$

But wait... $\frac{e^{-(a+b)} (a+b)^n }{n!}$ is actually the distribution of a Poisson R.V with parameter a+b.

Thus if X and Y are independent Poisson random variables with parameters a and b respectively, then X+Y is Poisson with parameter a+b !!

P.S: There is a slick proof of this result using moment generating functions. Try it!

4. ignore me - I misread what u'd written.. just gonna go over it in detail now.. thanks both! I'll let you know how I get on!

wow.. that's awesome.. thankyou both so much! I have quite a bit of the rest of this question to go.. I'll give it a good go and get back here if there's any probs!

5. Ok, the next part asks for
P[X = k | X + Y = n]

I got

[(n choose k)a^k . b^(n-k)]/[(a+b)^n]

I am asked to state the distribution.. is this right and is it binomially distributed or does the [(a+b)^n] denominator affect this?

many thanks

6. James,

$P( X = k | X+Y = n) = \frac{P(X = k, X+Y = n)}{P(X+Y = n)} = \frac{P(X = k) P(Y = n - k)}{P(X+Y = n)}$

And $\frac{P(X = k) P(Y = n - k)}{P(X+Y = n)} = \dfrac{\frac{e^{-a} a^k}{k!}\frac{e^{-b} b^{(n-k)}}{(n-k)!}}{\frac{e^{-(a+b)} (a+b)^n}{n!}}$ $= \frac{\binom{n}{k} a^k b^{(n-k)}}{(a+b)^n} = \binom{n}{k} \left(\frac{a}{a+b}\right)^k \left(\frac{b}{a+b}\right)^{(n-k)}$

Thus the R.V X conditioned on $X+Y = n$ is $\text{Binomial }\left(n, \frac{a}{a+b}\right)$ distributed