Problem 50 - A First Course in Probability

Question

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

Step-by-Step Solution

Answer

The conditional expectation \(E[X^2|Y=y] = 2y^3\).

Step 1: Find the conditional PDF of X given Y=y

In order to find the conditional PDF of \(X\) given \(\displaystyle Y=y\), we will use the following formula: \(f_{X|Y}(x|y) = \dfrac{f(x, y)}{f_Y(y)}\) First, we need to find the marginal PDF of \(Y\), denoted by \(\displaystyle f_Y( y)\). To find this, we need to integrate \(\displaystyle f( x,\ y)\) with respect to \(\displaystyle x\) over its limits: \(f_Y(y) = \int_{0}^{\infty} f(x, y)dx\) Now, plug in the given joint density function: \(f_Y(y) = \int_{0}^{\infty} \dfrac{e^{-x/y}e^{-y}}{y} dx \) To find the integral, perform a substitution: Set \(\displaystyle u=\dfrac{x}{y}\), which means that \(\displaystyle x=uy\), and \(\displaystyle dx=y\ du\) Substitute the new variables into the integral: \(f_Y(y) = \int_{0}^{\infty} \dfrac{e^{-u}e^{-y}}{y} \cdot y \ du = \int_{0}^{\infty} e^{-u}e^{-y} du\) Now integrate with respect to u: \(f_Y(y) = e^{-y} \left[ -e^{-u} \right]_{0}^{\infty} = e^{-y}(0 - (-1)) = e^{-y}\) Now that we have the marginal PDF of \(Y\), we can find the conditional PDF of \(X\) given \(Y=y\): \(f_{X|Y}(x|y) = \dfrac{f(x, y)}{f_Y(y)} = \dfrac{e^{-x/y} \cdot e^{-y}}{y \cdot e^{-y}} = \dfrac{e^{-x/y}}{y}\)

Step 2: Compute E[X^2 | Y=y]

Now, we can use this conditional PDF to compute the expectation \(E[X^2 | Y=y]\): \(E[X^2|Y=y] = \int_{0}^{\infty} x^2 \cdot f_{X|Y}(x|y) dx = \int_{0}^{\infty} x^2 \cdot \dfrac{e^{-x/y}}{y} dx\) To compute this integral, we will first do integration by parts: Set \(\displaystyle v=x^2\), and \(\displaystyle dw = \dfrac{e^{-x/y}}{y} dx\) Now we will need to find \(\displaystyle dv\) and \(\displaystyle w\): \(\displaystyle dv = 2x\ dx\), and \(\displaystyle w = -ye^{-x/y}\) Using integration by parts, we have: \(\displaystyle E[X^2|Y=y] = x^2(-ye^{-x/y})\Big|_0^{\infty} - \int_{0}^{\infty} (-2xye^{-x/y})(-y) dx\) Simplify the expression to: \(\displaystyle E[X^2|Y=y] = 0 + 2y^2\int_{0}^{\infty} xe^{-x/y} dx \) Now, we can perform another integration by parts, with a similar substitution as before: \(\displaystyle u=\dfrac{x}{y}\), which means that \(\displaystyle x=uy\), and \(\displaystyle dx=y\ du\) Our integral simplifies to: \(\displaystyle E[X^2|Y=y] = 2y^2\int_{0}^{\infty} (uy)e^{-u} (y\ du) = 2y^3\int_{0}^{\infty} ue^{-u} du \) Now perform integration by parts for the last time: Let \(\displaystyle v=u\), and \(\displaystyle dw=e^{-u} du\) Now find \(\displaystyle dv\) and \(\displaystyle w\): \(\displaystyle dv=du\), and \(\displaystyle w=-e^{-u}\) Using integration by parts, we have: \(E[X^2|Y=y] = 2y^3\left(-ue^{-u}\Big|_{0}^{\infty} + \int_{0}^{\infty} e^{-u} du\right)\) Simplify the expression: \(E[X^2|Y=y] = 2y^3\left(0 + \left[-e^{-u}\right]_{0}^{\infty}\right)\) Finally, calculate the integral: \(E[X^2|Y=y] = 2y^3\left(0 - (-1)\right) = 2y^3\) So, the conditional expectation \(E[X^2|Y=y] = 2y^3\).

Key Concepts

Joint Probability Density Function

A joint probability density function (PDF) describes the likelihood of two continuous random variables, say \( X \) and \( Y \), occurring simultaneously. It captures the probability distribution over a range of values for both variables combined. Mathematically, it is denoted by \( f(x, y) \), where \( x \) and \( y \) are potential values that the variables can take. In our case, we're given the joint PDF as \( f(x, y) = \frac{e^{-x / y} e^{-y}}{y} \), applicable for \( x > 0 \) and \( y > 0 \). This function tells us the likelihood of \( X \) taking a value like \( x \) while simultaneously \( Y \) takes a value like \( y \). These functions are crucial for determining related probabilities and expectations across multidimensional distributions. They can also be used to derive other probability functions, like marginal and conditional PDFs.

Marginal Probability Density Function

The marginal probability density function (PDF) of a random variable represents the distribution of that variable in a joint distribution, ignoring the effects of other variables. It is essentially the projection or summary of the joint PDF onto the axis of the variable of interest. For instance, to find the marginal PDF of \( Y \), denoted as \( f_Y(y) \), we integrate the joint PDF \( f(x, y) \) over all values of \( x \):

• \( f_Y(y) = \int_{0}^{\infty} f(x, y) dx \)

In our situation, this involves the substitution \( u = \frac{x}{y} \), leading to the solution \( f_Y(y) = e^{-y} \). This tells us how the probability is distributed over the values of \( Y \) alone, decoupling it from \( X \). Understanding marginal PDFs is crucial for analyzing the individual behavior of variables within a joint distribution.

Integration by Parts

Integration by parts is a technique used in calculus to simplify complex integrals. It is particularly useful when dealing with integrals involving products of functions. The main formula utilized is:

• \( \int u \ dv = uv - \int v \ du \)

To solve for \( E[X^2 | Y=y] \) using our conditional PDF, integration by parts is applied multiple times to evaluate the complex integrals. Specifically, we set \( v = x^2 \) and \( dw = \frac{e^{-x/y}}{y} dx \). This transformation allows us to swap complexity between the terms and simplify them into a more manageable form. Integration by parts is often repeated, as seen in our exercise, to further reduce the integral until it reaches a solvable form. This method is invaluable for tackling integrals found in probability and expectations calculations.

Conditional Probability Density Function

The conditional probability density function (PDF) provides the probability distribution of one random variable given that another random variable takes a specific value. It is computed using the joint PDF and the marginal PDF. For \( X \) given \( Y = y \), it is calculated as:

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

In our example, after deriving the marginal PDF \( f_Y(y) \), we plug it back into the equation to find \( f_{X|Y}(x|y) = \frac{e^{-x/y}}{y} \). This reveals how \( X \) behaves in relation to specific values of \( Y \), effectively showing \( X \)‘s distribution restricted by \( Y \)‘s condition. Understanding conditional PDFs is critical for calculating expectations and making inferences about dependent random variables.