Proving a variable has a poission distribution

Printable View

August 12th 2011, 05:02 AM
yellowcarrotz

Proving a variable has a poission distribution

Hi :)

I am given that:

P(Y=k)= $\sum$ $\binom{n}{k}$ p^k.(1-p)^(n-k).e^-x.(x^n/n!)

where the sum is from n=k to infinity

and I need to show that Y is poission distributued with parameter $\lambda$ p

I'm not really sure where to start, should I take out all the variables which are not raised to the power of n out of the sum and change $\binom{n}{k}$ to n!/(n-k)!k!?

Any tips to get me started would be much appreciated :)
August 14th 2011, 12:41 AM
Moo

Re: Proving a variable has a poission distribution

Hello,

Yes you can take out the variables where n is not in the expression.

You can also, in a first time, change the summation from n=0 to infinity (since this is the range of a Poisson distribution). It should simplify quite a lot, since in the $n \choose k$ , the n! simplifies, the (n-k)! becomes 0!=1 and k! can be pulled out from the sum.
August 15th 2011, 04:20 AM
yellowcarrotz

Re: Proving a variable has a poission distribution

Thanks so much for your reply :)

The question is:
P(Y=k)= $\sum$ $\binom{n}{k}$ $p^k$ (1-p)^(n-k) $e^{-x}$ $\frac{x^n}{n!}$

where the sum runs from n=k to infinity (sorry i'm rubbish at using latex)

is this right?

P(Y=k)= $\sum$ x^n/(n-k)!k! $p^k$ (1-p)^(n-k) $e^{-x}$

P(Y=k)= $p^k$ $e^{-x}$ $\sum$ x^n/(n-k)!k! . (1-p)^(n-k)

then change the sum so that it runs from n=0 to infinity

P(Y=k)=(p^k)/k! $e^{-x}$ $\sum$
then I'm not sure what I'll have left inside? x^n...

any hints appreciated
August 15th 2011, 05:45 AM
girdav

Re: Proving a variable has a poission distribution

After the change of summation, we have $P(Y=k) = \frac{p^k}{k!}e^{-x}\sum_{j=0}^{+\infty}\frac{x^{j+k}}{j!}(1-p)^j = \frac{p^k}{k!}e^{-x}x^k\sum_{j=0}^{+\infty}\frac{((1-p)x)^j}{j!}$ .
We recognize the series of $\exp$ , and now you an conclude.
August 15th 2011, 08:05 AM
Moo

Re: Proving a variable has a poission distribution

Quote:

Originally Posted by yellowcarrotz

Thanks so much for your reply :)

The question is:
P(Y=k)= $\sum$ $\binom{n}{k}$ $p^k$ (1-p)^(n-k) $e^{-x}$ $\frac{x^n}{n!}$

where the sum runs from n=k to infinity (sorry i'm rubbish at using latex)

is this right?

P(Y=k)= $\sum$ x^n/(n-k)!k! $p^k$ (1-p)^(n-k) $e^{-x}$

P(Y=k)= $p^k$ $e^{-x}$ $\sum$ x^n/(n-k)!k! . (1-p)^(n-k)

then change the sum so that it runs from n=0 to infinity

P(Y=k)=(p^k)/k! $e^{-x}$ $\sum$
then I'm not sure what I'll have left inside? x^n...

any hints appreciated

Replace n-k by j, that's a way to change the summation index :)
And if there's just an n, n=j+k
August 16th 2011, 12:36 AM
yellowcarrotz

Re: Proving a variable has a poission distribution

yep i've done it, thank you so much for all your help :)