Question

Let \(X_{1}\) and \(X_{2}\) be two random variables such that the conditional distributions and means exist. Show that: (a) \(E\left(X_{1}+X_{2} \mid X_{2}\right)=E\left(X_{1} \mid X_{2}\right)+X_{2}\) (b) \(E\left(u\left(X_{2}\right) \mid X_{2}\right)=u\left(X_{2}\right)\)

Short answer

The properties of conditional expected values can be demonstrated as follows: (a) \(E(X_{1} + X_{2} | X_{2}) = E(X_{1} | X_{2}) + X_{2}\) and (b) \(E(u(X_{2}) | X_{2}) = u(X_{2})\). These results exploit the fact that the expectation of a random variable given itself is itself and that the expectation of a measurable function of a sigma-algebra generating random variable, given that random variable, is equal to the function of that random variable.

Step-by-step solution

Demonstrating Property for the Sum of Two Random Variables

The expectation of the sum of two random variables given one of them can be broken down as follows: \(E(X_{1} + X_{2} | X_{2}) = E(X_{1} | X_{2}) + E(X_{2} | X_{2})\). According to properties of conditional expectation, \(E(X_{2} | X_{2}) = X_{2}\), as a random variable conditioned on itself is itself. So, you have \(E(X_{1} + X_{2} | X_{2}) = E(X_{1} | X_{2}) + X_{2}\), which demonstrates the property.

Demonstrating Property for a Function of a Random Variable

For any Borel-measurable function \(u\), the expectation of \(u(X_{2})\) given \(X_{2}\) can be simplified as follows: \(E(u(X_{2}) | X_{2})\). According to properties of conditional expectation, the expectation of a measurable function of a sigma-algebra generating random variable, conditioned on that random variable, equals the function of that random variable. Thus, \(E(u(X_{2}) | X_{2}) = u(X_{2})\), which demonstrates the required property.