W10 Homework A

Due date: Thursday 3/19, 11:59pm

Covariance and correlation

01

01

Covariance and correlation

The joint PMF of $X$ and $Y$ is given by the table:

$Y\downarrow X\to$	0	1	2	3
1	$\frac{1}{15}$	$\frac{1}{15}$	$\frac{2}{15}$	$\frac{1}{15}$
2	$\frac{1}{10}$	$\frac{1}{10}$	$\frac{1}{5}$	$\frac{1}{10}$
3	$\frac{1}{30}$	$\frac{1}{30}$	0	$\frac{1}{10}$

Compute:

(a) $E[X+Y]$ $$ (b) $E[{(X-Y)}^2]$ $$ (c) $\mathrm{Cov}[X,Y]$ $$ (d) $\rho [X,Y]$

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Solution

Solutions - 5240-01

(a)

$$\begin{array}{c}E[X+Y]=\sum _x\sum _y(x+y)p(x,y)\\ \\ \gg \gg 1\left(\frac{1}{15}\right)+2(\frac{1}{10}+\frac{1}{15})+3(\frac{1}{30}+\frac{2}{15}+\frac{1}{10})\\ \\ \gg \gg +4(\frac{1}{15}+\frac{1}{5}+\frac{1}{30})+5\left(\frac{1}{10}\right)+6\left(\frac{1}{10}\right)\gg \gg \frac{7}{2}\end{array}$$

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(b)

$$\begin{array}{cc}E\left[{(X-Y)}^2\right]& =\sum _x\sum _y{(x-y)}^{2}p(x,y)\\ & =1(\frac{1}{15}+\frac{1}{10}\frac{2}{15}+\frac{1}{10})+2(\frac{1}{10}+\frac{1}{30}+\frac{1}{15})+9\left(\frac{1}{30}\right)\\ & =1.5\end{array}$$

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(c)

(1) Recall the formula for $\text{Cov}[X,Y]$:

$$\text{Cov}[X,Y]=E[XY]-E[X]E[Y]$$

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(2) Compute $E[X]$ and $E[Y]$:

$$\begin{array}{cc}E[X]& =1(\frac{1}{15}+\frac{1}{10}+\frac{1}{30})+2(\frac{2}{15}+\frac{1}{5})+3(\frac{1}{15}+\frac{1}{10}+\frac{1}{10})\\ & =\frac{5}{3}\\ E[Y]& =1(\frac{1}{15}+\frac{1}{15}+\frac{2}{15}+\frac{1}{15})+2(\frac{1}{10}+\frac{1}{10}+\frac{1}{5}+\frac{1}{10})+3(\frac{1}{30}+\frac{1}{30}+\frac{1}{10})\\ & =\frac{11}{6}\end{array}$$

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(3) Compute $E[XY]$:

$$E[XY]=1\left(\frac{1}{15}\right)+2(\frac{1}{10}+\frac{2}{15})+3(\frac{1}{30}+\frac{1}{15})+4\left(\frac{1}{5}\right)+6\left(\frac{1}{10}\right)+9\left(\frac{1}{10}\right)=\frac{47}{15}$$

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(4) Compute $\text{Cov}[X,Y]$:

$$\begin{array}{c}\text{Cov}[X,Y]=\frac{47}{15}-\frac{5}{3}\cdot \frac{11}{6}\gg \gg \frac{7}{90}\end{array}$$

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(d)

(1) Recall the formula for $\rho [X,Y]$:

$$\rho [X,Y]=\frac{\text{Cov}[X,Y]}{\sqrt{\text{Var}[X]\text{Var}[Y]}}$$

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(2) Compute $E\left[X^2\right]$ and $E\left[Y^2\right]$:

$$\begin{array}{cc}E\left[X^2\right]& =1(\frac{1}{15}+\frac{1}{10}+\frac{1}{30})+4(\frac{2}{15}+\frac{1}{5})+9(\frac{1}{15}+\frac{1}{10}+\frac{1}{10})\\ & =\frac{59}{15}\\ E\left[Y^2\right]& =1(\frac{1}{15}+\frac{1}{15}+\frac{2}{15}+\frac{1}{15})+4(\frac{1}{10}+\frac{1}{10}+\frac{1}{5}+\frac{1}{10})+9(\frac{1}{30}+\frac{1}{30}+\frac{1}{10})\\ & =\frac{23}{6}\end{array}$$

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(3) Compute $\text{Var}[X]$ and $\text{Var}[Y]$:

$$\begin{array}{cc}\text{Var}[X]& =E\left[X^2\right]-{\left(E[X]\right)}^2=\frac{59}{15}-{\left(\frac{5}{3}\right)}^2=\frac{52}{45}\\ \text{Var}[Y]& =E\left[Y^2\right]-{\left(E[Y]\right)}^2=\frac{23}{6}-{\left(\frac{11}{6}\right)}^2=\frac{17}{36}\end{array}$$

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(4) Compute $\rho [X,Y]$:

$$\begin{array}{c}\rho [X,Y]=\frac{\frac{7}{90}}{\sqrt{\frac{52}{45}\cdot \frac{17}{36}}}\gg \gg \approx 0.1053\end{array}$$Link to original

02

02

Covariance etc. from independent densities

Suppose $X$ and $Y$ are independent variables with the following densities:

$$f_X(x)=\{\begin{array}{cc}\frac{1}{3}{e}^{-x/3}& x>0\\ 0& \text{ otherwise }\end{array}{f}_Y(y)=\{\begin{array}{cc}\frac{1}{8}{e}^{-y/8}& y>0\\ 0& \text{ otherwise }\end{array}$$

Compute:

(a) $P[X>Y]$ $$ (b) $E[XY]$ $$ (c) $\mathrm{Cov}[X,Y]$ $$ (d) $\rho [X,Y]$

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Solution

Solutions - 5240-02

(a)

$$\begin{array}{c}P[X>Y]={\int }_0^{\infty }{\int }_y^{\infty }{f}_X(x)f_Y(y)dxdy\\ \\ \gg \gg \frac{1}{24}{\int }_0^{\infty }{\int }_y^{\infty }{e}^{-\frac{x}{3}}{e}^{-\frac{y}{8}}dxdy\\ \\ \gg \gg \frac{1}{24}{\int }_0^{\infty }{e}^{-\frac{y}{8}}\underset{b\to \infty }{\lim }{[-3e^{-\frac{x}{3}}]|}_y^{b}dy\\ \\ \gg \gg \frac{1}{8}{\int }_0^{\infty }{e}^{-\frac{y}{8}}{e}^{-\frac{y}{3}}dy\gg \gg \frac{1}{8}{\int }_0^{\infty }{e}^{-\frac{11y}{24}}dy\\ \\ \gg \gg \frac{3}{11}\underset{b\to \infty }{\lim }{[-e^{-\frac{11}{24}y}]|}_0^b\gg \gg \frac{3}{11}\end{array}$$

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(b)

Since $X\sim \text{Exp}\left(\frac{1}{3}\right)$ and $Y\sim \text{Exp}\left(\frac{1}{8}\right)$, by independence:

$$E[XY]=E[X]E[Y]=3\cdot 8=24$$

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(c)

Since $X$ and $Y$ are independent:

$$\text{Cov}[X,Y]=0$$

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(d)

Since $\text{Cov}[X,Y]=0$:

$$\rho [X,Y]=0$$Link to original

03

03

Plumber completion time

A plumber is coming to fix the sink. He will arrive between 2:00 and 4:00 with uniform distribution in that range.

Sink fixes take an average of 45 minutes with completion times following an exponential distribution.

When do you expect the plumber to finish the job?

What is the variance for the finish time?

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Solution

Solutions - 5240-03

(1) Define random variables to describe the problem:

Let $X\sim \mathrm{Unif}[0,120]$ represent the arrival time of the plumber. Let $Y\sim \text{Exp}\left(\frac{1}{45}\right)$ represent the completion time of the sink fix.

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(2) Compute $E[X+Y]$:

This represents the expected time the plumber finishes the job.

$$\begin{array}{c}E[X+Y]=E[X]+E[Y]=60+45\gg \gg 105\end{array}$$

Thus, we expect the plumber to finish at $3:45$.

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(3) Compute the variance of the finish time:

$$\begin{array}{c}\text{Var}[X+Y]=\text{Var}[X]+\text{Var}[Y]+2\text{Cov}[X,Y]=\frac{{120}^2}{12}+\frac{1}{{\left(\frac{1}{45}\right)}^2}\gg \gg 3225{\text{ min}}^2\end{array}$$

As part of the problem interpretation, we assume that the time to fix a sink is independent of the starting time, so $\mathrm{Cov}[X,Y]=0$.

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