Final answer:
To show that E[h(X)Y], one uses conditional expectation by calculating the expectation of Y given the values of X and multiplying by the distribution of X. The law of total expectation formalizes this as E[h(X)Y] = Σ h(x)E[Y|X=x]P(X=x) for discrete X, and integrates when X is continuous.
Step-by-step explanation:
To show that E[h(X)Y], we want to demonstrate how the expected value of the product of a function of X (represented by h(X)) and another random variable Y can be determined. This involves the idea of conditional expectation, which means we calculate the expectation of Y given the values that X can take, multiplied by the probability distribution of X.
First, define the random variable X and the function h as given in the question. Next, let's denote the conditional expectation of Y given X by E[Y|X]. By the law of total expectation, which tells us that the expectation of a conditional expectation is equal to the expectation of the unconditioned random variable, we write E[h(X)Y] = E[E[h(X)Y|X]]. Then, we break down the conditional expectation E[h(X)Y|X] as the product of h(X) and E[Y|X] since h(X) is a function of X only and can be treated as a constant when conditioned on X. Hence, we have E[h(X)Y|X] = h(X)E[Y|X]. We then take the expectation of this product concerning the distribution of X, which formally becomes E[h(X)Y] = Σ h(x)E[Y|X=x]P(X=x), assuming X is discrete. If X is continuous, the sum would be replaced by an integral.
Finally, we've shown that by conditioning on X, we can express the expectation of the product h(X)Y. This approach is particularly useful in cases where X and Y may have some dependency and direct computation of E[h(X)Y] is complicated.