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5.3.3 Solved Problems

Problem
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} 2 & \quad y+x \leq 1, x>0, y>0 \\ & \quad \\ 0 & \quad \text{otherwise} \end{array} \right. \end{equation} Find $\textrm{Cov}(X,Y)$ and $\rho(X,Y)$.
  • Solution
    • For $0 \leq x \leq 1$, we have \begin{align}%\label{} \nonumber f_X(x)&=\int_{-\infty}^{\infty} f_{XY}(x,y)dy \\ \nonumber &=\int_{0}^{1-x}2dy\\ \nonumber &=2(1-x). \end{align} Thus, \begin{equation} \nonumber f_{X}(x) = \left\{ \begin{array}{l l} 2(1-x) & \quad 0 \leq x \leq 1 \\ & \quad \\ 0 & \quad \text{otherwise} \end{array} \right. \end{equation} Similarly, we obtain \begin{equation} \nonumber f_{Y}(y) = \left\{ \begin{array}{l l} 2(1-y) & \quad 0 \leq y \leq 1 \\ & \quad \\ 0 & \quad \text{otherwise} \end{array} \right. \end{equation} Thus, we have \begin{align}%\label{} \nonumber EX&=\int_{0}^{1}2x(1-x)dx\\ \nonumber &=\frac{1}{3}=EY, \end{align} \begin{align}%\label{} \nonumber EX^2&=\int_{0}^{1}2x^2(1-x)dx\\ \nonumber &=\frac{1}{6}=EY^2. \end{align} Thus, \begin{align}%\label{} \nonumber Var (X)=Var(Y)=\frac{1}{18}. \end{align} We also have \begin{align}%\label{} \nonumber EXY&=\int_{0}^{1} \int_{0}^{1-x}2xydydx\\ \nonumber &=\int_{0}^{1}x(1-x)^2dx\\ \nonumber &=\frac{1}{12}. \end{align} Now, we can find $\textrm{Cov}(X,Y)$ and $\rho(X,Y)$: \begin{align}%\label{} \nonumber \textrm{Cov}(X,Y)&=EXY-EXEY\\ \nonumber &=\frac{1}{12}-\left(\frac{1}{3}\right)^2\\ \nonumber &=-\frac{1}{36}, \end{align} \begin{align}%\label{} \nonumber \rho(X,Y)&=\frac{\textrm{Cov}(X,Y)}{\sqrt{Var(X) Var(Y)}}\\ \nonumber &=-\frac{1}{2}. \end{align}

Problem
I roll a fair die $n$ times. Let $X$ be the number of $1$'s that I observe and let $Y$ be the number of $2$'s that I observe. Find $\textrm{Cov}(X,Y)$ and $\rho(X,Y)$. Hint: One way to solve this problem is to look at $Var(X+Y)$.
  • Solution
    • Note that you can look at this as a binomial experiment. In particular, we can say that $X$ and $Y$ are $Binomial(n,\frac{1}{6})$. Also, $X+Y$ is $Binomial(n,\frac{2}{6})$. Remember the variance of a $Binomial(n,p)$ random variable is $np(1-p)$. Thus, we can write \begin{align}%\label{} \nonumber n\frac{2}{6}.\frac{4}{6}&=Var(X+Y)\\ \nonumber &=Var(X)+Var(Y)+2\textrm{Cov}(X,Y)\\ \nonumber &=n\frac{1}{6}.\frac{5}{6}+n\frac{1}{6}.\frac{5}{6}+2\textrm{Cov}(X,Y). \end{align} Thus, \begin{align}%\label{} \nonumber \textrm{Cov}(X,Y)=-\frac{n}{36}. \end{align} And, \begin{align}%\label{} \nonumber \rho(X,Y)=\frac{\textrm{Cov}(X,Y)}{\sqrt{Var(X) Var(Y)}}=-\frac{1}{5}. \end{align}

Problem
In this problem, you will provide another proof for the fact that $|\rho(X,Y)| \leq 1$. By definition $\rho_{XY}=\textrm{Cov}(U,V)$, where $ U$ and $V$ are the normalized versions of $X$ and $Y$ as defined in Equation 5.22: \begin{align} U=\frac{X-EX}{\sigma_X}, \hspace{10pt} V=\frac{Y-EY}{\sigma_Y}. \end{align} Use the fact that $\textrm{Var}(U+V) \geq 0$ to show that $|\rho(X,Y)| \leq 1$.
  • Solution
    • We have \begin{align}%\label{} \nonumber \textrm{Var}(U+V)&=\textrm{Var}(U)+\textrm{Var}(V)+2 \textrm{Cov}(U,V)\\ \nonumber &=1+1+2\rho_{XY}. \end{align} Since $\textrm{Var}(U+V) \geq 0$, we conclude $\rho(X,Y) \geq -1$. Also, from this we conclude that \begin{align}%\label{} \nonumber \rho(-X,Y) \geq -1. \end{align} But $\rho(-X,Y)=-\rho(X,Y)$, so we conclude $\rho(X,Y) \leq 1$.

Problem
Let $X$ and $Y$ be two independent $Uniform(0,1)$ random variables. Let also $Z=\max(X,Y)$ and $W=\min(X,Y)$. Find $\textrm{Cov}(Z,W)$.
  • Solution
    • It is useful to find the distributions of $Z$ and $W$. To find the CDF of $Z$, we can write \begin{align}%\label{} \nonumber F_Z(z)&=P(Z \leq z) \\ \nonumber &=P(\max(X,Y) \leq z)\\ \nonumber &=P\bigg((X \leq z) \textrm{ and } (Y \leq z)\bigg)\\ \nonumber &=P(X \leq z) P(Y \leq z) &(\textrm{since }X \textrm{ and }Y \textrm{ are independent})\\ \nonumber &=F_X(z)F_Y(z). \end{align} Thus, we conclude \begin{equation} \nonumber F_Z(z) = \left\{ \begin{array}{l l} 0 & \quad z<0 \\ z^2 & \quad 0 \leq z \leq 1 \\ 1 & \quad z>1 \end{array} \right. \end{equation} Therefore, \begin{equation} \nonumber f_Z(z) = \left\{ \begin{array}{l l} 2z & \quad 0 \leq z \leq 1 \\ 0 & \quad \text{otherwise} \end{array} \right. \end{equation} From this we obtain $EZ=\frac{2}{3}$. Note that we can find $EW$ as follows \begin{align}%\label{} \nonumber 1=E[X+Y]&=E[Z+W] \\ \nonumber &=EZ+EW\\ \nonumber &=\frac{2}{3}+EW. \end{align} Thus, $EW=\frac{1}{3}$. Nevertheless, it is a good exercise to find the CDF and PDF of $W$, too. To find the CDF of $W$, we can write \begin{align}%\label{} \nonumber F_W(w)&=P(W \leq w) \\ \nonumber &=P(\min(X,Y) \leq w)\\ \nonumber &=1-P(\min(X,Y) > w)\\ \nonumber &=1-P\bigg((X > w) \textrm{ and } (Y > w)\bigg)\\ \nonumber &=1-P(X > w) P(Y > w) &(\textrm{since }X \textrm{ and }Y \textrm{ are independent})\\ \nonumber &=1-(1-F_X(w))(1-F_Y(w))\\ \nonumber &=F_X(w)+F_Y(w)-F_X(w)F_Y(w). \end{align} Thus, \begin{equation} \nonumber F_W(w) = \left\{ \begin{array}{l l} 0 & \quad w<0 \\ 2w-w^2 & \quad 0 \leq w \leq 1 \\ 1 & \quad w>1 \end{array} \right. \end{equation} Therefore, \begin{equation} \nonumber f_W(w) = \left\{ \begin{array}{l l} 2-2w & \quad 0 \leq w \leq 1 \\ 0 & \text{otherwise} \end{array} \right. \end{equation} From the above PDF we can verify that $EW=\frac{1}{3}$. Now, to find $\textrm{Cov}(Z,W)$, we can write \begin{align}%\label{} \nonumber \textrm{Cov}(Z,W)&=E[ZW]-EZEW \\ \nonumber &=E[XY]-EZEW\\ \nonumber &=E[X]E[Y]-E[Z]E[W] &(\textrm{since }X \textrm{ and }Y \textrm{ are independent})\\ \nonumber &=\frac{1}{2}.\frac{1}{2}-\frac{2}{3}.\frac{1}{3}\\ \nonumber &=\frac{1}{36}. \end{align} Note that $\textrm{Cov}(Z,W)>0$ as we expect intuitively.

Problem
Let $X$ and $Y$ be jointly (bivariate) normal, with $Var(X)=Var(Y)$. Show that the two random variables $X+Y$ and $X-Y$ are independent.
  • Solution
    • Note that since $X$ and $Y$ are jointly normal, we conclude that the random variables $X+Y$ and $X-Y$ are also jointly normal. We have \begin{align}%\label{} \nonumber \textrm{Cov}(X+Y,X-Y)&=\textrm{Cov}(X,X)-\textrm{Cov}(X,Y)+\textrm{Cov}(Y,X)-\textrm{Cov}(Y,Y)\\ \nonumber &=Var(X)-Var(Y)\\ \nonumber &=0. \end{align} Since $X+Y$ and $X-Y$ are jointly normal and uncorrelated, they are independent.

Problem
Let $X$ and $Y$ be jointly normal random variables with parameters $\mu_X=0$, $\sigma^2_X=1$, $\mu_Y=-1$, $\sigma^2_Y=4$, and $\rho=-\frac{1}{2}$.
  1. Find $P(X+Y>0)$.
  2. Find the constant $a$ if we know $aX+Y$ and $X+2Y$ are independent.
  3. Find $P(X+Y>0|2X-Y=0)$.
  • Solution
      1. Since $X$ and $Y$ are jointly normal, the random variable $U=X+Y$ is normal. We have \begin{align}%\label{} \nonumber &EU=EX+EY=-1, \end{align} \begin{align} \nonumber Var(U)&=Var(X)+Var(Y)+2 \textrm{Cov}(X,Y) \\ \nonumber &=1+4+2 \sigma_X \sigma_Y \rho(X,Y)\\ \nonumber &=5-2 \times 1\times2\times\frac{1}{2}\\ \nonumber &=3. \end{align} Thus, $U \sim N(-1,3)$. Therefore, \begin{align}%\label{} \nonumber P(U>0)=1-\Phi\left(\frac{0-(-1)}{\sqrt{3}}\right)=1-\Phi\left(\frac{1}{\sqrt{3}}\right)=0.2819 \end{align}
      2. Note that $aX+Y$ and $X+2Y$ are jointly normal. Thus, for them, independence is equivalent to having $\textrm{Cov}(aX+Y,X+2Y)=0$. Also, note that $\textrm{Cov}(X,Y)=\sigma_X \sigma_Y \rho(X,Y)=-1$. We have \begin{align} \nonumber \textrm{Cov}(aX+Y,X+2Y)&=a\textrm{Cov}(X,X)+2a\textrm{Cov}(X,Y)+\textrm{Cov}(Y,X)+2\textrm{Cov}(Y,Y) \\ \nonumber &=a-(2a+1)+8\\ \nonumber &=-a+7. \end{align} Thus, $a=7$.
      3. If we define $U=X+Y$ and $V=2X-Y$, then note that $U$ and $V$ are jointly normal. We have \begin{align} \nonumber &EU=-1, Var(U)=3,\\ \nonumber &EV=1, Var(V)=12, \end{align} and \begin{align}%\label{} \nonumber \textrm{Cov}(U,V)&=\textrm{Cov}(X+Y,2X-Y)\\ &=2\textrm{Cov}(X,X)-\textrm{Cov}(X,Y)+2\textrm{Cov}(Y,X)-\textrm{Cov}(Y,Y)\\ \nonumber &=2Var(X)+\textrm{Cov}(X,Y)-Var(Y)\\ \nonumber &=2-1-4\\ \nonumber &=-3. \end{align} Thus, \begin{align}%\label{} \nonumber \rho(U,V)&=\frac{\textrm{Cov}(U,V)}{\sqrt{Var(U) Var(V)}}\\ \nonumber &=-\frac{1}{2}. \end{align} Using Theorem 5.4, we conclude that given $V=0$, $U$ is normally distributed with \begin{align}%\label{} \nonumber &E[U|V=0]=\mu_U+ \rho(U,V) \sigma_U \frac{0-\mu_V}{\sigma_V}=-\frac{3}{4}, \\ \nonumber &Var(U|V=0)=(1-\rho_{UV}^2)\sigma^2_U=\frac{9}{4}. \end{align} Thus \begin{align}%\label{} \nonumber P(X+Y>0|2X-Y=0)&=P(U>0|V=0)\\ \nonumber &=1-\Phi\left(\frac{0-(-\frac{3}{4})}{\frac{3}{2}}\right)\\ &= 1-\Phi\left(\frac{1}{2}\right)=0.3085. \end{align}

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