Marginal Probability function?Anyone

[Mathemag1c1an]

Marginal Probability function?Anyone

I have this question which I cannot seem to solve:
The joint probability mass function p(x, y) of two discrete random variables X and Y is given by.
p(x,y) = ([5^x][7^y][e^-5])/x!(y-x)!
x and y are non-negative integers and x <= y
(i) Find the marginal probability mass functions of X and Y.

 

[mathman]

Sum p(x,y) over x from 0 to y to get the marginal for Y. Sum p(x,y) over y from x to oo to get the marginal for X.

 

[Mathemag1c1an]

but how do we integrate the x!(y-x)!

 

[mathman]

but how do we integrate the x!(y-x)!

Remember we are dealing with integers, so we have to carry out summations.

x summation to get marginal for Y: x terms are 5^x/[!(y-x)!] the sum from 0 to y can be gotten from the binomial expansion of (1+5)^y = sum 5^x[yCx], where yCx is the combinatorial symbol =y!/[x!(y-x)!].
Put this together and you have P(Y=y) = (e^-5)(42^y)/y!
This is wrong, since the total probability is not 1. I suggest you examine your original description.

y summation to get marginal for X: sum from x to oo of 7^y/(y-x)!
which is simply (7^x)(e^7).
This leads to P(X=x) = (e^2)(35^x)/x!. This also is wrong.

These could be corrected if e^-5 is replaced by e^-42.

 

[Mathemag1c1an]

I knew something was wrong with the question:

 

[Mathemag1c1an]

Thanks alot

 

[mathman]

p(x,y) = ([5^x][7^y][e^-5])/x!(y-x)!

If you write it in the form p(x,y) = ([a^x][b^y][e^-c])/x!(y-x)!, you need b(a+1)=c for it to be valid.