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.

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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.