Question

Show that (a) \(\quad E[X Y \mid Y=y]=y E[X \mid Y=y]\) (b) \(E[g(X, Y) \mid Y=y]=E[g(X, y) \mid Y=y]\) (c) \(E[X Y]=E[Y E[X \mid Y]]\)

Short answer

In summary, we have proved the following properties of conditional expectation: (a) \(E[XY \mid Y = y] = y E[X \mid Y = y]\) (b) \(E[g(X, Y) \mid Y = y] = E[g(X, y) \mid Y = y]\) (c) \(E[XY] = E[Y E[X \mid Y]]\)

Step-by-step solution

Proof for Part (a)

: We want to show that \(E[XY \mid Y = y] = y E[X \mid Y = y]\). By definition of conditional expectation, we have: \(E[XY \mid Y = y] = \sum_x x\cdot y \cdot P(X = x \mid Y = y)\) But, since the value of y is constant in this expression, we can write: \(E[XY \mid Y = y] = y \sum_x x \cdot P(X = x \mid Y = y) = y E[X \mid Y = y]\) This proves (a).

Proof for Part (b)

: We want to show that \(E[g(X, Y) \mid Y = y] = E[g(X, y) \mid Y = y]\). By definition of conditional expectation, we have: \(E[g(X, Y) \mid Y = y] = \sum_x g(x, y) \cdot P(X = x \mid Y = y)\) Notice that the function g(X, Y) is now independent of Y because we replace Y with y when we calculate the particular value g(x, y). Thus, we can write: \(E[g(X, Y) \mid Y = y] = \sum_x g(x, y) \cdot P(X = x \mid Y = y) = E[g(X, y) \mid Y = y]\) This proves (b).

Proof for Part (c)

: We want to show that \(E[XY] = E[Y E[X \mid Y]]\). First, let's find the value of \(E[XY | Y]\). By definition of conditional expectation, we have: \(E[XY \mid Y] = \sum_x x P(X = x \mid Y = y) \cdot Y = Y \sum_x x \cdot P(X = x \mid Y = y)\) Now, using the Law of Iterated Expectation, we have: \(E[XY] = E[E[XY \mid Y]]\) \(E[XY] = E[Y \sum_x x \cdot P(X = x \mid Y = y)] = E[Y E[X \mid Y]]\) This proves (c).