Multivariable Version Change of Variables Formula for Continuous Random Variable

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5.2.4 Functions of Two Continuous Random Variables

So far, we have seen several examples involving functions of random variables. When we have two continuous random variables $g(X,Y)$, the ideas are still the same. First, if we are just interested in $E[g(X,Y)]$, we can use LOTUS:

LOTUS for two continuous random variables:

\begin{align}\label{eq:LOTUS-2D-Cont} E[g(X,Y)]=\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} g(x,y)f_{XY}(x,y) \hspace{5pt} dxdy \hspace{40pt} (5.19) \end{align}

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Example
Let $X$ and $Y$ be two jointly continuous random variables with joint PDF \begin{equation} \nonumber f_{XY}(x,y) = \left\{ \begin{array}{l l} x+y & \quad 0 \leq x,y \leq 1 \\ & \quad \\ 0 & \quad \text{otherwise} \end{array} \right. \end{equation} Find $E[XY^2]$.

• Solution
  • We have \begin{align} \nonumber E[XY^2]&=\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} (xy^2)f_{XY}(x,y) \hspace{5pt} dxdy\\ \nonumber &=\int_{0}^{1} \int_{0}^{1} xy^2(x+y) \hspace{5pt} dxdy\\ \nonumber &=\int_{0}^{1} \int_{0}^{1} x^2y^2+xy^3 \hspace{5pt} dxdy\\ \nonumber &=\int_{0}^{1} \left(\frac{1}{3}y^2+\frac{1}{2}y^3\right) dy\\ \nonumber &=\frac{17}{72}. \end{align}

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If $Z=g(X,Y)$ and we are interested in its distribution, we can start by writing \begin{align} \nonumber F_Z(z)&=P(Z \leq z)\\ \nonumber &=P(g(X,Y) \leq z)\\ \nonumber &=\iint\limits_D f_{XY}(x,y) \hspace{5pt} dxdy, \end{align} where $D=\{(x,y)|g(x,y)<z\}$. To find the PDF of $Z$, we differentiate $F_Z(z)$.

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Example
Let $X$ and $Y$ be two independent $Uniform(0,1)$ random variables, and $Z=XY$. Find the CDF and PDF of $Z$.

• Solution
  • First note that $R_Z=[0,1]$. Thus, \begin{align} \nonumber F_Z(z)&=0, \hspace{30pt} \textrm{ for }z \leq 0,\\ \nonumber F_Z(z)&=1, \hspace{30pt} \textrm{ for }z \geq 1. \end{align} For $0<z<1$, we have \begin{align} \nonumber F_Z(z)&=P(Z \leq z)\\ \nonumber &=P(XY \leq z)\\ \nonumber &=P\left(X \leq \frac{z}{Y}\right). \end{align} Just to get some practice, we will show you two ways to calculate $P(X \leq \frac{z}{Y})$ for $0<z<1$. The first way is just integrating $f_{XY}(x,y)$ in the region $x \leq \frac{z}{y}$. We have \begin{align} \nonumber P\left(X \leq \frac{z}{Y}\right)&=\int_{0}^{1} \int_{0}^{\frac{z}{y}}f_{XY}(x,y) \hspace{5pt} dxdy\\ \nonumber &=\int_{0}^{1} \int_{0}^{\min(1,\frac{z}{y})} 1 \hspace{5pt} dxdy\\ \nonumber &=\int_{0}^{1} \min\left(1,\frac{z}{y}\right) \hspace{5pt} dy. \end{align} Note that if we let $g(y)=\min\left(1,\frac{z}{y}\right)$, then \begin{equation} \nonumber g(y) = \left\{ \begin{array}{l l} 1 & \quad \textrm{for } 0<y<z\\ & \quad \\ \frac{z}{y} & \quad \textrm{for } z \leq y\leq 1 \end{array} \right. \end{equation} Therefore, \begin{align} \nonumber P\left(X \leq \frac{z}{Y}\right)&=\int_{0}^{1} g(y) \hspace{5pt} dy\\ \nonumber &=\int_{0}^{z} 1 \hspace{5pt} dy+ \int_{z}^{1} \frac{z}{y} \hspace{5pt} dy\\ \nonumber &=z-z \ln z. \end{align} The second way to find $P(X \leq \frac{z}{Y})$ is to use the law of total probability. We have \begin{align} \nonumber P(X \leq \frac{z}{Y})&=\int_{0}^{1} P(X \leq \frac{z}{Y}|Y=y) f_Y(y) \hspace{5pt} dy\\ \nonumber &=\int_{0}^{1} P\left(X \leq \frac{z}{y}\right) f_Y(y) \hspace{5pt} dy &\textrm{ (since $X$ and $Y$ are independent)} \hspace{20pt} (5.20) \end{align} Note that \begin{equation} \nonumber P\left(X \leq \frac{z}{y}\right) = \left\{ \begin{array}{l l} 1 & \quad \textrm{for } 0<y<z\\ & \quad \\ \frac{z}{y} & \quad \textrm{for } z \leq y\leq 1 \end{array} \right. \end{equation} Therefore, \begin{align} \nonumber P\left(X \leq \frac{z}{Y}\right)&=\int_{0}^{1} P\left(X \leq \frac{z}{y}\right) f_Y(y) \hspace{5pt} dy\\ \nonumber &=\int_{0}^{z} 1 \hspace{5pt} dy+ \int_{z}^{1} \frac{z}{y} \hspace{5pt} dy\\ \nonumber &=z-z \ln z. \end{align} Thus, in the end we obtain \begin{equation} \nonumber F_Z(z) = \left\{ \begin{array}{l l} 0 & \quad z \leq 0 \\ z-z \ln z & \quad 0<z<1 \\ 1 & \quad z \geq 1 \end{array} \right. \end{equation} You can check that $F_Z(z)$ is a continuous function. To find the PDF, we differentiate the CDF. We have \begin{equation} \nonumber f_Z(z) = \left\{ \begin{array}{l l} - \ln z & \quad 0<z<1 \\ 0 & \quad \textrm{otherwise} \end{array} \right. \end{equation}

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The Method of Transformations:

When we have functions of two or more jointly continuous random variables, we may be able to use a method similar to Theorems 4.1 and 4.2 to find the resulting PDFs. In particular, we can state the following theorem. While the statement of the theorem might look a little confusing, its application is quite straightforward and we will see a few examples to illustrate the methodology.

Theorem
Let $X$ and $Y$ be two jointly continuous random variables. Let $(Z,W)=g(X,Y)=(g_1(X,Y),g_2(X,Y))$, where $g:\mathbb{R}^2 \mapsto \mathbb{R}^2$ is a continuous one-to-one (invertible) function with continuous partial derivatives. Let $h=g^{-1}$, i.e., $(X,Y)=h(Z,W)=(h_1(Z,W),h_2(Z,W))$. Then $Z$ and $W$ are jointly continuous and their joint PDF, $f_{ZW}(z,w)$, for $(z,w) \in R_{ZW}$ is given by \begin{align} \nonumber f_{ZW}(z,w)=f_{XY}(h_1(z,w),h_2(z,w)) |J|, \end{align} where $J$ is the Jacobian of $h$ defined by \begin{align} \nonumber J= \det \begin{bmatrix} \frac{\partial h_1}{\partial z} & \frac{\partial h_1}{\partial w} \\ & \\ \frac{\partial h_2}{\partial z} & \frac{\partial h_2}{\partial w} \\ \end{bmatrix} =\frac{\partial h_1}{\partial z}.\frac{\partial h_2}{\partial w}-\frac{\partial h_2}{\partial z}\frac{\partial h_1}{\partial w}. \end{align}

The following examples show how to apply the above theorem.

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Example
Let $X$ and $Y$ be two independent standard normal random variables. Let also \begin{equation} \nonumber \left\{ \begin{array}{l} Z=2X-Y \\ W=-X+Y \end{array} \right. \end{equation} Find $f_{ZW}(z,w)$.

• Solution
  • $X$ and $Y$ are jointly continuous and their joint PDF is given by \begin{align}%\label{} \nonumber f_{XY}(x,y)=f_X(x)f_Y(y)=\frac{1}{2 \pi} \exp\left\{-\frac{x^2+y^2}{2}\right\}, \hspace{30pt} \textrm{ for all }x,y \in \mathbb{R}. \end{align} Here, the function $g$ is defined by $(z,w)=g(x,y)=(g_1(x,y),g_2(x,y))=(2x-y,-x+y)$. Solving for $x$ and $y$, we obtain the inverse function $h$: \begin{equation} \nonumber \left\{ \begin{array}{l} x=z+w=h_1(z,w) \\ y=z+2w=h_2(z,w) \end{array} \right. \end{equation} We have \begin{align} \nonumber f_{ZW}(z,w)&=f_{XY}(h_1(z,w),h_2(z,w)) |J|\\ \nonumber &=f_{XY}(z+w,z+2w) |J|, \end{align} where \begin{align} \nonumber J= \det \begin{bmatrix} \frac{\partial h_1}{\partial z} & \frac{\partial h_1}{\partial w} \\ & \\ \frac{\partial h_2}{\partial z} & \frac{\partial h_2}{\partial w} \\ \end{bmatrix} =\det \begin{bmatrix} 1 & 1 \\ & \\ 1 & 2 \\ \end{bmatrix} =1. \end{align} Thus, we conclude that \begin{align} \nonumber f_{ZW}(z,w)&=f_{XY}(z+w,z+2w) |J|\\ \nonumber &=\frac{1}{2 \pi} \exp\left\{-\frac{(z+w)^2+(z+2w)^2}{2}\right\} \\ \nonumber &=\frac{1}{2 \pi} \exp\left\{-\frac{2z^2+5w^2+6zw}{2}\right\}. \end{align}

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Example
Let $X$ and $Y$ be two random variables with joint PDF $f_{XY}(x,y)$. Let $Z=X+Y$. Find $f_{Z}(z)$.

• Solution
  • To apply Theorem 5.1, we need two random variables $Z$ and $W$. We can simply define $W=X$. Thus, the function $g$ is given by \begin{equation} \nonumber \left\{ \begin{array}{l} z=x+y \\ w=x \end{array} \right. \end{equation} Then, we can find the inverse transform: \begin{equation} \nonumber \left\{ \begin{array}{l} x=w \\ y=z-w \end{array} \right. \end{equation} Then, we have \begin{align} \nonumber |J|= | \det \begin{bmatrix} 0 & 1 \\ & \\ 1 & -1 \\ \end{bmatrix}| =|-1|=1. \end{align} Thus, \begin{align} \nonumber f_{ZW}(z,w)=f_{XY}(w,z-w). \end{align} But since we are interested in the marginal PDF, $f_Z(z)$, we have \begin{align} \nonumber f_Z(z)=\int_{-\infty}^{\infty} f_{XY}(w,z-w)dw. \end{align} Note that, if $X$ and $Y$ are independent, then $f_{XY}(x,y)=f_X(x)f_Y(y)$ and we conclude that \begin{align} \nonumber f_Z(z)=\int_{-\infty}^{\infty} f_{X}(w)f_Y(z-w)dw. \end{align}

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The above integral is called the convolution of $f_X$ and $f_Y$, and we write \begin{align} \nonumber f_Z(z)&=f_X(z)\ast f_Y(z)\\ \nonumber &=\int_{-\infty}^{\infty} f_{X}(w)f_Y(z-w)dw=\int_{-\infty}^{\infty} f_{Y}(w)f_X(z-w)dw. \end{align}

If $X$ and $Y$ are two jointly continuous random variables and $Z=X+Y$, then \begin{align} \nonumber f_Z(z)=\int_{-\infty}^{\infty} f_{XY}(w,z-w)dw=\int_{-\infty}^{\infty} f_{XY}(z-w,w)dw. \end{align} If $X$ and $Y$ are also independent, then \begin{align} \nonumber f_Z(z)&=f_X(z)\ast f_Y(z)\\ \nonumber &=\int_{-\infty}^{\infty} f_{X}(w)f_Y(z-w)dw=\int_{-\infty}^{\infty} f_{Y}(w)f_X(z-w)dw. \end{align}

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Example
Let $X$ and $Y$ be two independent standard normal random variables, and let $Z=X+Y$. Find the PDF of $Z$.

• Solution
  • We have \begin{align} \nonumber f_Z(z)&=f_X(z)\ast f_Y(z)\\ \nonumber &=\int_{-\infty}^{\infty} f_{X}(w)f_Y(z-w)dw\\ \nonumber &=\int_{-\infty}^{\infty} \frac{1}{2 \pi}e^{-\frac{w^2}{2}}e^{-\frac{(z-w)^2}{2}}dw\\ \nonumber &=\frac{1}{\sqrt{4 \pi}} e^{\frac{-z^2}{4}}\int_{-\infty}^{\infty} \frac{1}{\sqrt{\pi}}e^{-(w-\frac{z}{2})^2}dw\\ \nonumber &=\frac{1}{\sqrt{4 \pi}} e^{\frac{-z^2}{4}}, \end{align} where $\int_{-\infty}^{\infty} \frac{1}{\sqrt{\pi}}e^{-(w-\frac{z}{2})^2}dw=1$ because it is the integral of the PDF of a normal random variable with mean $\frac{z}{2}$ and variance $\frac{1}{2}$. Thus, we conclude that $Z \sim N(0,2)$. In fact, this is one of the interesting properties of the normal distribution: the sum of two independent normal random variables is also normal. In particular, similar to our calculation above, we can show the following:

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Theorem
If $X \sim N(\mu_X,\sigma^2_X)$ and $Y \sim N(\mu_Y,\sigma^2_Y)$ are independent, then \begin{align}%\label{} \nonumber X+Y \hspace{5pt} \sim \hspace{5pt} N\bigg(\mu_X+\mu_Y, \sigma_X^2+\sigma_Y^2\bigg). \end{align}

We will see an easier proof of Theorem 5.2 when we discuss moment generating functions.

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Source: https://www.probabilitycourse.com/chapter5/5_2_4_functions.php

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