Expectation of a function of X and Y

[gjones89]

Homework Statement

Find E[X^Y], where X and Y are independent random variables which are uniform on [0,1].

Homework Equations

The Attempt at a Solution

I know that to get E[f(x)] for a function of one continuous random variable X, you integrate xf(x) between minus and plus infinity.

I tried to generalise this method for a function of two variables and integrate [x*y*f(x,y)] as a double integral for x and y between 0 and 1. So I have a double integral with [x*y*f(x,y) = (x^(y+1))*y] under the integral signs. When I solved this as a double integral (integrating with respect to y first), I ended up with an answer of 2ln(2/3) + 1. Am I using the right method?

 

[fzero]

I know that to get E[f(x)] for a function of one continuous random variable X, you integrate xf(x) between minus and plus infinity.

This is incorrect. Suppose that X is a random variable, uniform on [0,1], with pdf $$f_X$$. Then the expectation value of a function $$g(X)$$ is

$$E(g(X)) = \int_0^1 dx~g(x) f_X(x).$$

There's no factor of $$x$$ inserted.

I tried to generalise this method for a function of two variables and integrate [x*y*f(x,y)] as a double integral for x and y between 0 and 1. So I have a double integral with [x*y*f(x,y) = (x^(y+1))*y] under the integral signs. When I solved this as a double integral (integrating with respect to y first), I ended up with an answer of 2ln(2/3) + 1. Am I using the right method?

No, if you use the correct formula above,

$$E(X^Y) = \int_0^1 dx \int_0^1 dy~ x^y f_X(x) f_Y(y).$$

Since $$x,y$$ are uniform, $$f_X=f_Y=1$$. The answer will still have a logarithm in it.