Marginal distribution function

[_joey]

Given:

$$P(X=x, Y=y)=\frac{a^ye^{-2a}}{x!(y-x)!}$$ where $$x=0,1,2,...y$$ and $$y=0,1,2...\infty$$, and $$a>0$$

Find $$P(X=x)$$ and $$P(Y=y)$$

An example is provided in a book on books.google.com
Page 96
http://books.google.com.au/books?id...AEwCTgK#v=onepage&q=marginal discrete&f=false

Here is my attempted solution
$$p_{X}(x)=\Sigma_{y=0}^{\infty}\frac{a^ye^{-2a}}{x!(y-x)!}=e^{-2a}+ae^{-2a}+\frac{a^2e^{-2a}}{x!(2-x)!}+...+\frac{a^ne^{-2a}}{x!(n-x)!}$$

And then I cannot simplify this serie. Any comments and suggestions will be very much appreciated

 

[_joey]

I solved the problem.

 

[issacnewton]

what's the answer you got since your summation over y is wrong... y should go as
x, x+1,x+2... infty. for given x

 

[_joey]

what's the answer you got since your summation over y is wrong... y should go as
x, x+1,x+2... infty. for given x

maybe it is wrong. Elaborate why 'y should go as x, x+1, x+2...infty'

 

[_joey]

I've got another question: $$E[X|Y=y]=?$$

My answer is $$E[X|Y=y]=\frac{y}{2}e^{2a-2y}$$

I am not sure if it is correct.

Thanks!

 

[issacnewton]

just plot the points (x,y) on the graph.lets say y=0, then x can take value x=0
so the plotted point would be (0,0) . now if y=1 then x=0,1 so the plotted points would be
(0,1) , (1,1). if y=2 then x=0,1,2. so the plotted points would be (0,2),(1,2),(2,2) and so on.
now the marginal distribution for x is for some given x , that means we fix the value of x and then look for the values that y can take. so on this graph, if you fix value of x to be say 5,then y must take values 5 onwards. y can't take value less than 5 because then x=5 would not be possible. if you take value of x to be 7, then y must take values 7 onwards. this is easy to see
if you draw a vertical line from some fixed value of x. you can see that , y can take values
x onwards for that fixed value of x.

once you do that, the summation is very easy.

 

[issacnewton]

now

$$E[X|Y=y]=\sum_{x\in X} x\, \frac{P[X=x,Y=y]}{P[Y=y]}$$

so first you have to calculate $$P(Y=y)$$ here for a fixed value of y, x can take values from 0 to y.

$$P(Y=y)\,=\, \sum_{x=0}^y \, \frac{a^y e^{-2a}}{x!\, (y-x)!}$$

$$= a^y e^{-2a}\, \sum_{x=0}^y \, \frac{1}{x!(y-x)!}$$

now here you do a small trick. you multiply and divide by y!. since y! is not dependent on x, you pull out the denominator y! and you have y! in the numerator. then what you have is
a binomial coefficient.

$$\sum_{x=0}^y \, \binom{y}{x}=\, \sum_{x=0}^y \, \frac{y!}{x!(y-x)!}$$

the above sum is just $$2^y$$. so

$$P(Y=y)\, = \, \frac{a^y e^{-2a}}{y!} \, 2^y$$

then use this to do the required summation. with the above expression, final answer I get is

$$E[X|Y=y]=\frac{y}{2}$$