5.2.5 Solved Problems
Let $X$ and $Y$ be jointly continuous random variables with joint PDF \begin{equation} \nonumber f_{X,Y}(x,y) = \left\{ \begin{array}{l l} cx+1 & \quad x,y \geq 0, x+y<1 \\ & \quad \\ 0 & \quad \text{otherwise} \end{array} \right. \end{equation}
- Show the range of $(X,Y)$, $R_{XY}$, in the $x-y$ plane.
- Find the constant $c$.
- Find the marginal PDFs $f_X(x)$ and $f_Y(y)$.
- Find $P(Y<2X^2)$.
- Solution
-
- Figure 5.8(a) shows $R_{XY}$ in the $x-y$ plane.
The figure shows (a) $R_{XY}$ as well as (b) the integration region for finding $P(Y<2X^2)$ for Solved Problem 1.
- To find the constant $c$, we write \begin{align}%\label{} \nonumber 1&=\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f_{XY}(x,y)dxdy\\ \nonumber &=\int_{0}^{1} \int_{0}^{1-x} cx+1 \hspace{5pt} dydx\\ \nonumber &=\int_{0}^{1} (cx+1)(1-x) \hspace{5pt} dx\\ \nonumber &=\frac{1}{2}+\frac{1}{6}c. \end{align} Thus, we conclude $c=3$.
- We first note that $R_X=R_Y=[0,1]$. \begin{align}%\label{} \nonumber f_X(x)&=\int_{-\infty}^{\infty} f_{XY}(x,y)dy\\ \nonumber &=\int_{0}^{1-x} 3x+1 \hspace{5pt} dy\\ \nonumber &=(3x+1)(1-x), \hspace{10pt} \textrm{ for }x \in [0,1]. \end{align} Thus, we have \begin{equation} \nonumber f_X(x) = \left\{ \begin{array}{l l} (3x+1)(1-x) & \quad 0 \leq x \leq 1 \\ & \quad \\ 0 & \quad \text{otherwise} \end{array} \right. \end{equation} Similarly, we obtain \begin{align}%\label{} \nonumber f_Y(y)&=\int_{-\infty}^{\infty} f_{XY}(x,y)dx\\ \nonumber &=\int_{0}^{1-y} 3x+1 \hspace{5pt} dx\\ \nonumber &=\frac{1}{2}(1-y)(5-3y), \hspace{10pt} \textrm{ for }y \in [0,1]. \end{align} Thus, we have \begin{equation} \nonumber f_Y(y) = \left\{ \begin{array}{l l} \frac{1}{2}(1-y)(5-3y) & \quad 0 \leq y \leq 1 \\ & \quad \\ 0 & \quad \text{otherwise} \end{array} \right. \end{equation}
- To find $P(Y<2X^2)$, we need to integrate $f_{XY}(x,y)$ over the region shown in Figure 5.8(b). We have \begin{align}%\label{} \nonumber P(Y<2X^2)&=\int_{-\infty}^{\infty} \int_{-\infty}^{2x^2} f_{XY}(x,y)dydx\\ \nonumber &=\int_{0}^{1} \int_{0}^{\min(2x^2,1-x)} 3x+1 \hspace{5pt} dydx\\ \nonumber &=\int_{0}^{1} (3x+1) \min(2x^2,1-x) \hspace{5pt} dx\\ \nonumber &=\int_{0}^{\frac{1}{2}} 2x^2(3x+1) \hspace{5pt} dx + \int_{\frac{1}{2}}^1 (3x+1)(1-x) \hspace{5pt} dx\\ \nonumber &=\frac{53}{96}. \end{align}
- Figure 5.8(a) shows $R_{XY}$ in the $x-y$ plane.
-
Problem
Let $X$ and $Y$ be jointly continuous random variables with joint PDF \begin{equation} \nonumber f_{X,Y}(x,y) = \left\{ \begin{array}{l l} 6e^{-(2x+3y)} & \quad x,y \geq 0 \\ & \quad \\ 0 & \quad \text{otherwise} \end{array} \right. \end{equation}
- Are $X$ and $Y$ independent?
- Find $E[Y|X>2]$.
- Find $P(X>Y)$.
- Solution
-
- We can write \begin{align} \nonumber f_{X,Y}(x,y) =f_X(x)f_Y(y), \end{align} where \begin{align} \nonumber f_X(x)=2e^{-2x}u(x), \hspace{20pt} f_Y(y)=3e^{-3y}u(y). \end{align} Thus, $X$ and $Y$ are independent.
- Since $X$ and $Y$ are independent, we have $E[Y|X>2]=E[Y]$. Note that $Y \sim Exponential(3)$, thus $EY=\frac{1}{3}$.
- We have \begin{align} \nonumber P(X>Y)&=\int_{0}^{\infty} \int_{y}^{\infty} 6e^{-(2x+3y)} dx dy \\ \nonumber &=\int_{0}^{\infty} 3e^{-5y} dy \\ \nonumber &=\frac{3}{5}. \end{align}
-
Problem
Let $X$ be a continuous random variable with PDF \begin{equation} \nonumber f_X(x) = \left\{ \begin{array}{l l} 2x & \quad 0 \leq x \leq 1 \\ & \quad \\ 0 & \quad \text{otherwise} \end{array} \right. \end{equation} We know that given $X=x$, the random variable $Y$ is uniformly distributed on $[-x,x]$.
- Find the joint PDF $f_{XY}(x,y)$.
- Find $f_Y(y)$.
- Find $P(|Y|<X^3)$.
- Solution
-
- First note that, by the assumption \begin{equation} \nonumber f_{Y|X}(y|x) = \left\{ \begin{array}{l l} \frac{1}{2x} & \quad -x \leq y \leq x \\ & \quad \\ 0 & \quad \text{otherwise} \end{array} \right. \end{equation} Thus, we have \begin{equation} \nonumber f_{XY}(x,y)=f_{Y|X}(y|x)f_X(x) = \left\{ \begin{array}{l l} 1 & \quad 0 \leq x \leq 1, -x \leq y \leq x \\ & \quad \\ 0 & \quad \text{otherwise} \end{array} \right. \end{equation} Thus, \begin{equation} \nonumber f_{XY}(x,y) = \left\{ \begin{array}{l l} 1 & \quad |y| \leq x \leq 1 \\ & \quad \\ 0 & \quad \text{otherwise} \end{array} \right. \end{equation}
- First, note that $R_Y=[-1,1]$. To find $f_Y(y)$, we can write \begin{align} \nonumber f_Y(y)&=\int_{-\infty}^{\infty}f_{XY}(x,y)dx\\ \nonumber &=\int_{|y|}^{1} 1 dx\\ \nonumber &=1-|y|. \end{align} Thus, \begin{equation} \nonumber f_{Y}(y) = \left\{ \begin{array}{l l} 1-|y| & \quad |y| \leq 1 \\ & \quad \\ 0 & \quad \text{otherwise} \end{array} \right. \end{equation}
- To find $P(|Y|<X^3)$, we can use the law of total probability (Equation 5.16): \begin{align}%\label{} \nonumber P(|Y|<X^3)&=\int_0^1 P(|Y|<X^3|X=x) f_X(x) dx \\ \nonumber &= \int_0^1 P(|Y|<x^3|X=x) 2x dx \\ \nonumber &= \int_0^1 \left(\frac{2x^3}{2x}\right) 2x dx \hspace{20pt} \textrm{since }Y|X=x \hspace{5pt} \sim \hspace{5pt} Uniform(-x,x)\\ \nonumber &=\frac{1}{2}. \end{align}
-
Problem
Let $X$ and $Y$ be two jointly continuous random variables with joint PDF \begin{equation} \nonumber f_{X,Y}(x,y) = \left\{ \begin{array}{l l} 6xy & \quad 0 \leq x \leq 1, 0 \leq y \leq \sqrt{x} \\ & \quad \\ 0 & \quad \text{otherwise} \end{array} \right. \end{equation}
- Show $R_{XY}$ in the $x-y$ plane.
- Find $f_X(x)$ and $f_Y(y)$.
- Are $X$ and $Y$ independent?
- Find the conditional PDF of $X$ given $Y=y$, $f_{X|Y}(x|y)$.
- Find $E[X|Y=y]$, for $0 \leq y \leq 1$.
- Find $\textrm{Var}(X|Y=y)$, for $0 \leq y \leq 1$.
- Solution
-
- Figure 5.9 shows $R_{XY}$ in the $x-y$ plane.
Figure 5.9: The figure shows $R_{XY}$ for Solved Problem 4.
- First, note that $R_X=R_Y=[0,1]$. To find $f_X(x)$ for $0 \leq x \leq 1$, we can write \begin{align} \nonumber f_X(x)&=\int_{-\infty}^{\infty}f_{XY}(x,y) \hspace{5pt} dy\\ \nonumber &=\int_{0}^{\sqrt{x}} 6xy \hspace{5pt} dy\\ \nonumber &=3x^2. \end{align} Thus, \begin{equation} \nonumber f_X(x) = \left\{ \begin{array}{l l} 3x^2 & \quad 0 \leq x \leq 1 \\ & \quad \\ 0 & \quad \text{otherwise} \end{array} \right. \end{equation} To find $f_Y(y)$ for $0 \leq y \leq 1$, we can write \begin{align} \nonumber f_Y(y)&=\int_{-\infty}^{\infty}f_{XY}(x,y) \hspace{5pt} dx\\ \nonumber &=\int_{y^2}^{1} 6xy \hspace{5pt} dx\\ \nonumber &=3y(1-y^4). \end{align} \begin{equation} \nonumber f_Y(y) = \left\{ \begin{array}{l l} 3y(1-y^4) & \quad 0 \leq y \leq 1 \\ & \quad \\ 0 & \quad \text{otherwise} \end{array} \right. \end{equation}
- $X$ and $Y$ are not independent, since $f_{XY}(x,y)\neq f_x(x)f_Y(y)$.
- We have \begin{align} \nonumber f_{X|Y}(x|y)&=\frac{f_{XY}(x,y)}{f_Y(y)}\\ \nonumber &=\left\{ \begin{array}{l l} \frac{2x}{1-y^4} & \quad y^2 \leq x \leq 1 \\ & \quad \\ 0 & \quad \text{otherwise} \end{array} \right. \end{align}
- We have \begin{align} \nonumber E[X|Y=y]&=\int_{-\infty}^{\infty}xf_{X|Y}(x|y) \hspace{5pt} dx\\ \nonumber &=\int_{y^2}^{1} x\frac{2x}{1-y^4} \hspace{5pt} dx\\ \nonumber &=\frac{2(1-y^6)}{3(1-y^4)}. \end{align}
- We have \begin{align} \nonumber E[X^2|Y=y]&=\int_{-\infty}^{\infty}x^2f_{X|Y}(x|y) \hspace{5pt} dx\\ \nonumber &=\int_{y^2}^{1} x^2\frac{2x}{1-y^4} \hspace{5pt} dx\\ \nonumber &=\frac{1-y^8}{2(1-y^4)}. \end{align} Thus, \begin{align} \nonumber \textrm{Var}(X|Y=y)&=E[X^2|Y=y]-(E[X|Y=y])^2\\ \nonumber &=\frac{1-y^8}{2(1-y^4)}-\left(\frac{2(1-y^6)}{3(1-y^4)}\right)^2. \end{align}
- Figure 5.9 shows $R_{XY}$ in the $x-y$ plane.
-
Problem
Consider the unit disc \begin{align}%\label{} \nonumber D=\{(x,y)|x^2+y^2 \leq 1\}. \end{align} Suppose that we choose a point $(X,Y)$ uniformly at random in $D$. That is, the joint PDF of $X$ and $Y$ is given by \begin{equation} \nonumber f_{XY}(x,y) = \left\{ \begin{array}{l l} \frac{1}{\pi} & \quad (x,y) \in D \\ & \quad \\ 0 & \quad \text{otherwise} \end{array} \right. \end{equation} Let $(R,\Theta)$ be the corresponding polar coordinates as shown in Figure 5.10. The inverse transformation is given by \begin{equation} \nonumber \left\{ \begin{array}{l} X=R \cos \Theta \\ Y=R \sin \Theta \end{array} \right. \end{equation} where $R \geq 0$ and $-\pi < \Theta \leq \pi$. Find the joint PDF of $R$ and $\Theta$.
Figure 5.10: Polar Coordinates
- Solution
- Here $(X,Y)$ are jointly continuous and are related to $(R,\Theta)$ by a one-to-one relationship. We use the method of transformations (Theorem 5.1). The function $h(r,\theta)$ is given by \begin{equation} \nonumber \left\{ \begin{array}{l} x=h_1(r,\theta)=r \cos \theta \\ y=h_2(r,\theta)=r \sin \theta \end{array} \right. \end{equation} Thus, we have \begin{align} \nonumber f_{R\Theta}(r,\theta)&=f_{XY}(h_1(r,\theta),h_2(r,\theta)) |J|\\ \nonumber &=f_{XY}(r \cos \theta,r \sin \theta) |J|. \end{align} where \begin{align} \nonumber J= \det \begin{bmatrix} \frac{\partial h_1}{\partial r} & \frac{\partial h_1}{\partial \theta} \\ & \\ \frac{\partial h_2}{\partial r} & \frac{\partial h_2}{\partial \theta} \\ \end{bmatrix} =\det \begin{bmatrix} \cos \theta & -r \sin \theta \\ & \\ \sin \theta & r \cos \theta \\ \end{bmatrix} =r \cos^2 \theta+r \sin^2 \theta=r. \end{align} We conclude that \begin{align} \nonumber f_{R\Theta}(r,\theta)&=f_{XY}(r \cos \theta,r \sin \theta) |J|\\ \nonumber &=\left\{ \begin{array}{l l} \frac{r}{\pi} & \quad r \in [0,1], \theta \in (-\pi,\pi] \\ & \quad \\ 0 & \quad \text{otherwise} \end{array} \right. \end{align} Note that from above we can write \begin{align} \nonumber f_{R\Theta}(r,\theta)=f_R(r)f_{\Theta}(\theta), \end{align} where \begin{equation} \nonumber f_R(r) = \left\{ \begin{array}{l l} 2r & \quad r \in [0,1] \\ & \quad \\ 0 & \quad \text{otherwise} \end{array} \right. \end{equation} \begin{equation} \nonumber f_\Theta(\theta) = \left\{ \begin{array}{l l} \frac{1}{2\pi} & \quad \theta \in (-\pi,\pi] \\ & \quad \\ 0 & \quad \text{otherwise} \end{array} \right. \end{equation} Thus, we conclude that $R$ and $\Theta$ are independent.
