# Expected Value Question

Hello everyone,

I have the following question. Suppose that X and Y are independent and f(x,y) is nonnegative. Put g(x)=E[f(x,Y)] and show E[g(X)]=E[f(X,Y)]. Show more generally that Integral over X in A of g(X) dP = Integral over X in A of f(X,Y) dP. Extend to f that may be negative. I've had no issues, except with the extension to negative f part. Any suggestions?

chiro
Hello everyone,

I have the following question. Suppose that X and Y are independent and f(x,y) is nonnegative. Put g(x)=E[f(x,Y)] and show E[g(X)]=E[f(X,Y)]. Show more generally that Integral over X in A of g(X) dP = Integral over X in A of f(X,Y) dP. Extend to f that may be negative. I've had no issues, except with the extension to negative f part. Any suggestions?

Hello empyreandance and welcome to the forums.

I'm a little confused with your question. You mention that you have X and Y which are independent.

When you take expectation, it has to be with respect to a particular variable, especially with regard to situations where you have more than one variable.

In this situation you could do an expectation with respect to Y and integrate (or summate) out the Y terms to get a function of X, and then calculate the expectation of this to get an actual value, but you are not doing this.

Also your differential term is dP. Does dP refer to dYdX?

Also with regard to f(X,Y) being non-negative, this must be the case for any valid probability distribution (since probabilities are always between 0 and 1 for a discrete value or for an interval on the continuous distribution).

So based on the above comments, your question does not make sense.

Hi chiro,

I am working with the following definition:

E[X] = Integral of X dP (P is the probability measure on the space) = Integral X(ω)P(dω)

f(X,Y) is not the density function; it is simply a function applied to random variables, giving a composition of functions when one regards a random variable as a real-valued function.

Does this clear up the confusion?

When you write g(x)=E[f(x,Y)], do you mean g(x)=E[f(y|x)], where f(y|x) is the marginal distribution of y as a function of x?

Mute
Homework Helper
When you write g(x)=E[f(x,Y)], do you mean g(x)=E[f(y|x)], where f(y|x) is the marginal distribution of y as a function of x?

As the OP stated in the post above yours, f(x,y) is an arbitrary function of the random variables. It is not a density function whatsoever.

With regard to the question, I'll admit I am not really familiar with the measure-theory notation. In usual notation, I would say that the random variable Y has a probability density $\rho_Y(y)$ and the rv X has density $\rho_X(x)$. Then,

$$\mathbb{E}_Y[f(x,y] = \int_{-\infty}^\infty dy~f(x,y)\rho_Y(y) \equiv g(x),$$

and

$$\mathbb{E}_X[g(x)] = \int_{-\infty}^\infty dx~\rho_X(x) g(x) = \int_{-\infty}^\infty dx\int_{-\infty}^\infty dy~\rho_X(x)\rho_Y(y)f(x,y) = \mathbb{E}_{XY}[f(x,y)]$$

So, this is what I would have done to show E[g(x)] = E[f(x,y)], where I added subscripts to make it clear what variables were being averaged over. For the statement,

"Show more generally that Integral over X in A of g(X) dP = Integral over X in A of f(X,Y) dP"

I am afraid I don't know what this is trying to ask. It seems to me like it is the same as the first part.

Lastly, with regard to f(x,y) having negative values, I never really used the fact that f(x,y) was non-negative above. I don't know how different a measure-theoretic argument would be, so I can't guess at where the non-negativity of f(x,y) came into play.