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

To show that \( E[X \mid Y = y] = 0 \), we need to calculate the conditional expectation of \( X \) given \( Y = y \). 

### Step 1: Determine the Joint Density Function

The joint density function of \( X \) and \( Y \) is given by:

\[
f(x, y) = (y^2 - x^2) \cdot 8 e^{-y}, \quad 0 < y < \infty, \, -y < x < y.
\]

### Step 2: Find the Marginal Density of \( Y \)

To find the marginal density \( f_Y(y) \), integrate the joint density over all possible values of \( X \):

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

Simplifying the expression inside the integral:

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

Next, let's compute the integral:

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

This is a standard integral that can be solved as:

\[
\int_{-y}^{y} (y^2 - x^2) \, dx = \left[y^2 x - \frac{x^3}{3}\right]_{-y}^{y} = \left[y^3 - \frac{y^3}{3}\right] - \left[-y^3 + \frac{(-y)^3}{3}\right].
\]

\[
= \left(y^3 - \frac{y^3}{3}\right) - \left(-y^3 + \frac{y^3}{3}\right) = \frac{4y^3}{3}.
\]

So, the marginal density of \( Y \) is:

\[
f_Y(y) = 8 e^{-y} \cdot \frac{4y^3}{3} = \frac{32 y^3}{3} e^{-y}.
\]

### Step 3: Find the Conditional Density of \( X \mid Y = y \)

The conditional density of \( X \) given \( Y = y \) is given by:

\[
f_{X \mid Y}(x \mid y) = \frac{f(x, y)}{f_Y(y)} = \frac{(y^2 - x^2) \cdot 8 e^{-y}}{\frac{32 y^3}{3} e^{-y}}.
\]

Simplifying:

\[
f_{X \mid Y}(x \mid y) = \frac{(y^2 - x^2) \cdot 8}{\frac{32 y^3}{3}} = \frac{3 (y^2 - x^2)}{4 y^3}.
\]

\[
f_{X \mid Y}(x \mid y) = \frac{3 (y^2 - x^2)}{4 y^3}, \quad -y < x < y.
\]

### Step 4: Calculate \( E[X \mid Y = y] \)

To find \( E[X \mid Y = y] \), we use:

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

Simplifying the integral:

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

Notice that the integrand is an odd function of \( x \) because it is \( x \times (y^2 - x^2) \), and we are integrating it over a symmetric interval \( [-y, y] \). Therefore, the integral of an odd function over a symmetric interval is zero:

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

Thus:

\[
E[X \mid Y = y] = 0.
\]

This confirms that:

\[
E[X \mid Y = y] = 0.
\]

Would you like any further details or have any questions about this proof?

### Related Questions:

1. How do you find the variance of \(X \mid Y = y\)?
2. What is the significance of the expectation being zero in terms of symmetry?
3. How can we verify the marginal density of \( Y \) is correctly normalized?
4. What is the joint distribution of \(X\) and \(Y\) if it were to represent a physical phenomenon?
5. How would the results change if the bounds on \(X\) were different?

**Tip:** When dealing with joint density functions, always ensure that the function is normalized correctly to make valid probability statements!

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 the proof that E[X|Y = y] = 0 given the joint density f(x, y) = (y^2 − x^2) 8 e^−y. This solution involves calculating the marginal density, conditional expectation, and applying symmetry of functions. Suitable for advanced high school or undergraduate-level probability students.

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