The joint density of X and Y is f(x, y) = (y2 − x2)/8 .e−y, 0 < y < ∞, −y x y Show that E[X|Y = y] = 0.

We are tasked with showing that the conditional expectation \( E[X | Y = y] = 0 \) given the joint density \( f(x, y) = \frac{y^2 - x^2}{8} e^{-y} \), where \( 0 < y < \infty \) and \( -y < x < y \).

### 1. **Understanding the Problem**

The conditional expectation \( E[X | Y = y] \) is defined as:

\[
E[X | Y = y] = \int_{-\infty}^{\infty} x f_{X|Y}(x | y) \, dx
\]

where \( f_{X|Y}(x | y) \) is the conditional density of \( X \) given \( Y = y \), obtained by normalizing the joint density \( f(x, y) \) over \( x \), for a fixed \( y \).

We know from the joint density that \( X \) is supported on \( -y < x < y \). So, the conditional expectation becomes:

\[
E[X | Y = y] = \int_{-y}^{y} x f_{X|Y}(x | y) \, dx
\]

### 2. **Finding the Conditional Density \( f_{X|Y}(x | y) \)**

The conditional density \( f_{X|Y}(x | y) \) is given by:

\[
f_{X|Y}(x | y) = \frac{f(x, y)}{f_Y(y)}
\]

where \( f_Y(y) \) is the marginal density of \( Y \).

#### a. **Marginal Density \( f_Y(y) \)**

To find \( f_Y(y) \), we integrate the joint density over \( x \):

\[
f_Y(y) = \int_{-y}^{y} f(x, y) \, dx = \int_{-y}^{y} \frac{y^2 - x^2}{8} e^{-y} \, dx
\]

This simplifies as follows:

\[
f_Y(y) = \frac{e^{-y}}{8} \int_{-y}^{y} (y^2 - x^2) \, dx
\]

\[
\int_{-y}^{y} (y^2 - x^2) \, dx = 2 \int_0^{y} (y^2 - x^2) \, dx
\]

The integral evaluates to:

\[
2 \left[ y^2 x - \frac{x^3}{3} \right]_0^y = 2 \left( y^3 - \frac{y^3}{3} \right) = 2 \times \frac{2y^3}{3} = \frac{4y^3}{3}
\]

Thus, the marginal density becomes:

\[
f_Y(y) = \frac{e^{-y}}{8} \times \frac{4y^3}{3} = \frac{y^3 e^{-y}}{6}
\]

#### b. **Conditional Density \( f_{X|Y}(x | y) \)**

Now, we can write the conditional density as:

\[
f_{X|Y}(x | y) = \frac{f(x, y)}{f_Y(y)} = \frac{\frac{y^2 - x^2}{8} e^{-y}}{\frac{y^3 e^{-y}}{6}} = \frac{3(y^2 - x^2)}{4y^3}
\]

### 3. **Computing the Conditional Expectation**

Now that we have the conditional density, we can compute the conditional expectation \( E[X | Y = y] \):

\[
E[X | Y = y] = \int_{-y}^{y} x \cdot \frac{3(y^2 - x^2)}{4y^3} \, dx
\]

This integral breaks into two parts:

\[
E[X | Y = y] = \frac{3}{4y^3} \int_{-y}^{y} x(y^2 - x^2) \, dx
\]

Now, we compute the integral \( \int_{-y}^{y} x(y^2 - x^2) \, dx \). Since the integrand is an odd function (i.e., \( x \cdot (y^2 - x^2) \) is odd in \( x \)), the integral over the symmetric interval \( [-y, y] \) is zero:

\[
\int_{-y}^{y} x(y^2 - x^2) \, dx = 0
\]

Thus:

\[
E[X | Y = y] = 0
\]

### Conclusion

We have shown that the conditional expectation \( E[X | Y = y] = 0 \).

---

Would you like further details or explanations?

Here are 5 related questions you might explore:
1. How would we compute \( E[Y] \) for the given joint density?
2. What is the variance of \( X \) given \( Y = y \)?
3. How can we generalize this result for different joint distributions?
4. What role does the symmetry of the density function play in this result?
5. How does the marginal density \( f_Y(y) \) change if the limits for \( x \) change?

**Tip:** When working with joint densities, it's crucial to normalize the conditional densities properly to compute expectations effectively.

The joint density of X and Y is f(x, y) = (y^2 − x^2)/8 e^{-y}, 0 < y < ∞, −y < x < y. Show that E[X|Y = y] = 0.

Explore how to prove that E[X|Y = y] = 0 for the joint density function f(x, y) = (y^2 - x^2)/8 e^{-y}, where 0 < y < ∞ and -y < x < y. This solution covers key probability concepts like conditional expectation, marginal distribution, and the role of odd functions in integrals. Suitable for undergraduate students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation