Question

Let X, Y, and Z be jointly continuous random variables. Assume that all conditional PDFs and expectations are well defined. E.g., when conditioning on X = x assume that is such that f_{X}(x) > 0 For each one of the following formulas, state whether it is true for all choices of the function g or false (i.e., not true for all choices of g)

1 E[g(Y)| X = x ]= int g(y)f Y|X (y|x)dy

Select an option

2. E[g(y)| X = x ]= int g(y)f Y|X (y|x)dy

Select an option

Answer 1

Let's evaluate each statement one at a time.

### Statement 1:
\[ E[g(Y)| X = x ]= \int g(y) f_{Y|X}(y|x) \, dy \]

This statement is **True**.

**Explanation:** The formula states that the expected value of a function \(g(Y)\), given that \(X=x\), is equal to the integral of \(g(y)\) weighted by the conditional probability density function (PDF) of \(Y\) given \(X\). This is indeed a standard result in probability theory and holds for all choices of the function \(g\).

### Statement 2:
\[ E[g(y)| X = x ]= \int g(y) f_{Y|X}(y|x) \, dy \]

This statement is **False**.

**Explanation:** The expected value \(E[g(Y) | X = x]\) is a function of \(Y\) given the specific value of \(X=x\). Therefore, the left-hand side is not correctly represented by just \(g(y)\) but rather by the random variable \(g(Y)\), which must be evaluated conditionally based on the distribution of \(Y\) given \(X=x\). The correct form should actually reflect the expectation with respect to the conditional distribution of \(Y\) given \(X=x\). The expected value of \(g(Y)\) would depend on the distribution of \(Y\) and not just substitute \(y\) in \(g(y)\). Thus, it doesn't work for all choices of \(g\).

In summary:
1. **True**
2. **False**