The joint density of X and Y is given by f (x,y) = (e−x/ye−y)/ y, Show E[X|Y = y]=y.

To show that \( E[X | Y = y] = y \), we need to compute the conditional expectation of \( X \) given \( Y = y \) using the joint probability density function (pdf) of \( X \) and \( Y \).

The joint pdf of \( X \) and \( Y \) is given as:

\[
f(x, y) = \frac{e^{-x/y} e^{-y}}{y}, \quad x > 0, y > 0
\]

### Step 1: Find the conditional pdf \( f_{X|Y=y}(x|y) \)
The conditional pdf of \( X \) given \( Y = y \) is given by:

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

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

### Step 2: Find the marginal pdf \( f_Y(y) \)
To find \( f_Y(y) \), we integrate the joint pdf \( f(x, y) \) over all possible values of \( X \):

\[
f_Y(y) = \int_0^\infty f(x,y) \, dx = \int_0^\infty \frac{e^{-x/y} e^{-y}}{y} \, dx
\]

Since \( e^{-y} \) is independent of \( x \), it can be factored out:

\[
f_Y(y) = e^{-y} \int_0^\infty \frac{e^{-x/y}}{y} \, dx
\]

Now compute the integral. The integral of \( e^{-x/y} \) is:

\[
\int_0^\infty e^{-x/y} \, dx = y
\]

So, the marginal pdf of \( Y \) becomes:

\[
f_Y(y) = e^{-y}
\]

### Step 3: Write the conditional pdf \( f_{X|Y=y}(x|y) \)
Now that we have \( f_Y(y) \), we can substitute it into the expression for the conditional pdf:

\[
f_{X|Y=y}(x|y) = \frac{\frac{e^{-x/y} e^{-y}}{y}}{e^{-y}} = \frac{e^{-x/y}}{y}, \quad x > 0, y > 0
\]

Thus, the conditional pdf is:

\[
f_{X|Y=y}(x|y) = \frac{e^{-x/y}}{y}
\]

### Step 4: Compute \( E[X | Y = y] \)
The conditional expectation of \( X \) given \( Y = y \) is:

\[
E[X | Y = y] = \int_0^\infty x f_{X|Y=y}(x|y) \, dx = \int_0^\infty x \frac{e^{-x/y}}{y} \, dx
\]

We can factor out \( 1/y \):

\[
E[X | Y = y] = \frac{1}{y} \int_0^\infty x e^{-x/y} \, dx
\]

Now, we use the following known result from integration by parts:

\[
\int_0^\infty x e^{-ax} \, dx = \frac{1}{a^2}
\]

In our case, \( a = \frac{1}{y} \), so:

\[
\int_0^\infty x e^{-x/y} \, dx = y^2
\]

Thus, we get:

\[
E[X | Y = y] = \frac{1}{y} \cdot y^2 = y
\]

### Conclusion:
We have shown that \( E[X | Y = y] = y \), as required.

---

Do you want more details or have any questions about the steps? Here are some related questions to consider:

1. How is the marginal density function \( f_Y(y) \) derived in more detail?
2. Can you explain why the integral result \( \int_0^\infty x e^{-x/y} dx = y^2 \) holds?
3. How does the concept of conditional expectation extend to multivariate distributions?
4. What are other examples of joint probability density functions where conditional expectations simplify?
5. How would we compute \( \text{Var}(X | Y = y) \) based on this setup?

**Tip**: When computing conditional expectations, breaking down the problem into manageable steps (finding the marginal and conditional densities) simplifies the overall process.

The joint density of X and Y is given by f(x,y) = (e−x/ye−y)/ y, Show E[X|Y = y]=y.

Explore the step-by-step solution to finding the conditional expectation E[X|Y = y] for the joint density function f(x, y) = (e−x/ye−y)/ y. This problem covers key probability concepts like joint probability density functions and integration techniques, suitable for undergraduate students.

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