Question

1. Suppose X and Y are independent continuous random variables. Show that
E[X|Y = y] = E[X] for all y

2. The joint density of X and Y is
f (x, y) = (y^2 − x^ 2) * e^(−y), 0 < y < ∞, −y <= x <= y
Show that E[X|Y = y] = 0.

Answer 1

Answer on Question #70766 – Math – Statistics and Probability

Question

1. Suppose $X$ and $Y$ are independent continuous random variables. Show that $E[X|Y = y] = E[X]$ for all $y$.

Solution

$$E[X|Y = y] = |by\ definition| = \sum_{i} x_{i} P\{X = x_{i}|Y = y\} = \sum_{i} x_{i} P\{X = x_{i}\} = E[X].$$

Question

2. The joint density of $X$ and $Y$ is

$$f(x, y) = (y^{2} - x^{2}) e^{-y}, \quad 0 < y < \infty, \quad -y \leq x \leq y.$$

Show that $E[X|Y = y] = 0$.

Solution

$$E[X|Y = y] = \int_{-\infty}^{\infty} x \frac{f(x, y)}{f_{Y}(y)} dx.$$ $$f_{Y}(y) = \int_{-\infty}^{\infty} f(x, y) dx = \int_{-y}^{y} (y^{2} - x^{2}) e^{-y} dx = x y^{2} e^{-y} \Big|_{-y}^{y} - \frac{x^{3} e^{-y}}{3} \Big|_{-y}^{y} = \frac{4}{3} e^{-y} y^{3}.$$

Hence,

$$\begin{array}{l} E[X|Y = y] = \frac{3}{4} \int_{-\infty}^{\infty} \frac{x (y^{2} - x^{2}) e^{-y}}{e^{-y} y^{3}} dx = \frac{3}{4y} \int_{-\infty}^{\infty} x dx - \frac{3}{4y^{3}} \int_{-\infty}^{\infty} x^{3} dx \\ = |the\ integral\ of\ the\ odd\ function\ in\ a\ symmetric\ boundary\ is\ zero| = 0. \end{array}$$

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