Question

The joint density of \(X\) and \(Y\) is given by $$ f(x, y)=\frac{e^{-x / y} e^{-y}}{y}, \quad 0

Short answer

In summary, we found the conditional probability density function of X given Y = y as \(f(x|Y=y) = \frac{e^{-x / y}}{y}\), and then calculated the expected value E[X|Y=y] by integrating x * f(x|Y=y) dx over the range of x (0 to ∞). This resulted in E[X|Y=y] = y.

Step-by-step solution

Find the conditional probability density function of X, given Y = y

To find the conditional probability density function, we need to divide the joint density by the marginal density of Y (GetY). 1. Joint density function: f(x, y) = \(\frac{e^{-x / y} e^{-y}}{y}\) 2. So now we need to find the marginal density of Y.

Step 1.1: Find the marginal density of Y (GetY)

To find the marginal density of Y, we need to integrate the joint density function f(x, y) with respect to x from 0 to ∞. \[ \text{GetY}(y) = \int_{0}^{\infty} f(x,y) dx \] \[ \text{GetY}(y) = \int_{0}^{\infty} \frac{e^{-x / y} e^{-y}}{y} dx \]

Step 1.2: Integrate with respect to x

Now, let's integrate the expression above with respect to x. \[ \text{GetY}(y) = \frac{e^{-y}}{y} \int_{0}^{\infty} e^{-x / y} dx \] We can perform the integration using a substitution: Let u = -x/y, so du = -dx/y. \[ \text{GetY}(y) = \frac{e^{-y}}{y} \int_{-\infty}^{0} -ye^u du \] \[ \text{GetY}(y) = e^{-y} \int_{-\infty}^{0} e^u du \] \[ \text{GetY}(y) = e^{-y} (e^0 - e^{-\infty}) \] \[ \text{GetY}(y) = e^{-y} \] Now we have the marginal density of Y: \(\text{GetY}(y) = e^{-y}\)

Step 1.3: Find the conditional probability density function of X given Y = y

Now, let's find the conditional probability density function f(x|Y=y) by dividing the joint density function by the marginal density of Y. \[ f(x|Y=y) = \frac{f(x, y)}{\text{GetY}(y)} \] \[ f(x|Y=y) = \frac{\frac{e^{-x / y} e^{-y}}{y}}{e^{-y}} \] \[ f(x|Y=y) = \frac{e^{-x / y}}{y} \]

Compute the expected value E[X|Y=y]

Now, let's compute the expected value of X given Y = y using the following formula: \[ E[X|Y=y] = \int_{0}^{\infty} x * f(x|Y=y) dx \] \[ E[X|Y=y] = \int_{0}^{\infty} x * \frac{e^{-x / y}}{y} dx \] Now, let's integrate the expression above with respect to x. We can use integration by parts: \(u = x\), and \(dv = \frac{e^{-x / y}}{y} dx\) \(du = dx\), and \(v = -ye^{-x / y}\) \[ E[X|Y=y] = -yxe^{-x / y}\Big|_{0}^{\infty} + \int_{0}^{\infty} y e^{-x / y} dx \] The first term in the expression above is 0 since x approaches infinity and e^(−x/y) approaches 0. Therefore, we have: \[ E[X|Y=y] = \int_{0}^{\infty} y e^{-x / y} dx \] Using the same substitution as before: \(u = -x/y\), and \(du = -dx/y\) \[ E[X|Y=y] = \int_{-\infty}^{0} -y^2 e^u du \] \[ E[X|Y=y] = y^2 \int_{-\infty}^{0} e^u du \] \[ E[X|Y=y] = y^2 (e^0 - e^{-\infty}) \] \[ E[X|Y=y] = y^2 (1 - 0) \] \[ E[X|Y=y] = y \] Thus, we have shown that E[X|Y=y] = y.