What is the Expected Value of (X/Y)?

[Samwise_geegee]

Homework Statement

Let x and y be discrete random variables with joint probability density function

P(X,Y)= 2^X-Y+1/9 for x=1,2 and y=1,2
0 Otherwise

Calculate E[X/Y]

Homework Equations

E[XY]= ∫∫XYP(X,Y)dxdy

The Attempt at a Solution

I can't find a property of Expected value to make E[X/Y] solvable. This is my best guess.
E[X/Y]= ∫∫X*1/YP(X,1/Y)dxdy

E[X/Y]=∫∫X*1/Y*(2^X-1/Y+1/9)dxdy from 1 to 2 on the first integral and 1 to 2 on the second integral

 

[Simon Bridge]

looks like you have a discrete probability function there.
there are only 4 states, so why not work out X/Y for each state and find the expectation by weighted sum?

 

[Samwise_geegee]

I'm just not sure how to calculate E(X/Y). Is what I wrote right?

 

[Simon Bridge]

You can answer that one yourself!

Consider: what you wrote involves some integrals.
Are integrals associated with continuous or discrete probability functions?
What kind do you have?

 

[Ray Vickson]

I'm just not sure how to calculate E(X/Y). Is what I wrote right?

What you wrote is wrong. You seem to be suffering from the disease of writing formulas without knowing what they mean or when they should be used, and the prognosis of that disease is not good. Integrals are used with continuous random variables having probability density functions. Do you see any such random variables in your problem?