Unit 02 - Essential problems

Joint distributions

02

PMF calculations from a table

Suppose the joint PMF of $X$ and $Y$ has values given in this table:

$Y\downarrow X\to$	0	1	2	3
1	0.10	0.15	0	0.05
2	0.20	0.05	0.05	0.20
3	0.05	0	$x$	0.05

(a) Find $x$.

(b) Find the marginal PMF of $X$.

(c) Find the PMF of the random variable $Z=XY$.

(d) Find $P[X=Y]$ and $P[X>Y]$.

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Solution

Solutions - 5190-02

(a)

Find $x$ using the constraint ${\sum }_{x,y}{P}_{X,Y}(x,y)=1$:

$$\begin{array}{c}0.1+5(0.05)+0.15+2(0.2)+x=1\\ \\ \gg \gg x=1-0.1-0.25-0.15-0.4\gg \gg x=0.1\end{array}$$

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(b)

Find the marginal PMF of $X$ by summing rows:

$$P_X(x)=\{\begin{array}{cc}0.1+0.15+0.05=0.3& x=1\\ 2(0.2)+2(0.05)=0.5& x=2\\ 2(0.05)+0.1=0.2& x=3\end{array}$$

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(c)

(1) Define the possible values of $Z$:

Since $X\in \{1,2,3\}$ and $Y\in \{0,1,2,3\}$, we have that $Z\in \{0,1,2,3,4,6,9\}$.

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(2) Define PMF of $Z$:

Go through each possible value of $Z$ and see when it occurs.

$$P_Z(z)=\{\begin{array}{cc}P[Y=0]& z=0\\ P[X=1\cap Y=1]& z=1\\ P[X=1\cap Y=2]+P[X=2\cap Y=1]& z=2\\ P[X=1\cap Y=3]+P[X=3\cap Y=1]& z=3\\ P[X=2\cap Y=2]& z=4\\ P[X=2\cap Y=3]+P[X=3\cap Y=2]& z=6\\ P[X=3\cap Y=3]& z=9\end{array}$$

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(3) Substitute values for each probability:

$$P_Z(z)=\{\begin{array}{cc}0.35& z=0\\ 0.15& z=1\\ 0.05& z=2\\ 0.05& z=3\\ 0.05& z=4\\ 0.3& z=6\\ 0.05& z=9\end{array}$$

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(d)

(1) Find $P[X=Y]$:

$$\begin{array}{c}P[X=1\cap Y=1]+P[X=2\cap Y=2]+P[X=3\cap Y=3]\\ \\ \gg \gg 0.15+0.05+0.05\gg \gg 0.25\end{array}$$

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(2) Find $P[X>Y]$:

$$\begin{array}{c}P[Y=0]+P[X=2\cap Y=1]+P[X=3\cap Y=1]+P[X=3\cap Y=2]\\ \\ \gg \gg 0.35+0.05+0.05+0.05\gg \gg 0.5\end{array}$$Link to original

05

Marginals and probability from joint PDF

Suppose $X$ and $Y$ have joint PDF given by:

$$f_{X,Y}(x,y)=\{\begin{array}{cc}2e^{-(x+2y)}& \text{ if }x,y>0\\ 0& \text{ otherwise }\end{array}$$

(a) Find the marginal PDFs for $X$ and $Y$.

(b) Find $P[X>Y]$.

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Solution

Solutions - 5190-05

(a)

(1) Find the marginal PDF for $X$ by integrating the joint PDF with respect to $y$:

$$\begin{array}{c}{f}_X(x)={\int }_0^{\infty }2e^{-(x+2y)}dy\\ \\ \gg \gg \underset{b\to \infty }{\lim }{\int }_0^{b}2e^{-(x+2y)}dy\gg \gg -\underset{b\to \infty }{\lim }{\left[e^{-(x+2y)}\right]|}_0^b\gg \gg e^{-x}\end{array}$$

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(2) Find the marginal PDF for $Y$ by integrating the joint PDF with respect to $x$:

$$\begin{array}{c}{f}_Y(y)={\int }_0^{\infty }2e^{-(x+2y)}dx\\ \\ \gg \gg -\underset{b\to \infty }{\lim }{\left[2e^{-(x+2y)}\right]|}_0^b\gg \gg 2e^{-2y}\end{array}$$

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(b)

Integrate the joint PDF over the region above the line $y=x$:

$$\begin{array}{c}P[X>Y]={\int }_0^{\infty }{\int }_0^{x}2e^{-(x+2y)}dydx\\ \\ \gg \gg {\int }_0^{\infty }(e^{-x}-e^{-3x})dx\\ \\ \gg \gg {(-e^{-x}+\frac{1}{3}{e}^{-3x})|}_0^{\infty }\gg \gg 0-(-1+\frac{1}{3})\gg \gg \frac{2}{3}\end{array}$$Link to original

07

Grad student mentors

A university lab has an incoming group of researchers who need mentors. There are 4 graduate students, 2 undergraduate students, and the lab director who could serve as mentors. Suppose that exactly 3 of these 7 people will be selected as mentors.

(a) How many different groups of 3 people could be chosen to be the three mentors?

(b) Suppose exactly 2 graduate students and 1 undergraduate student are selected to be the three mentors. How many different groups of 3 people could be selected?

Let $X$ be the total number of graduate students chosen to be mentors, and $Y$ be the total number of undergraduate students chosen to be mentors.

(c) Construct the joint PMF of $X$ and $Y$: $P_{X,Y}(x,y)$

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Solution

Solutions - 5190-07

(a)

Count the total number of groups of 3 from 7:

$$\left(\genfrac{}{}{0px}{}{7}{3}\right)=\frac{7!}{4!3!}=\frac{7\cdot 6\cdot 5}{3\cdot 2}\gg \gg 35$$

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(b)

Choose 2 from 4 grad students and 1 from 2 undergrads:

$$\left(\genfrac{}{}{0px}{}{4}{2}\right)\left(\genfrac{}{}{0px}{}{2}{1}\right)=6\cdot 2\gg \gg 12$$

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(c)

Construct the joint PMF of $X$ and $Y$:

$Y\downarrow X\to$	0	1	2	3
0	0	0	$\frac{\left(\genfrac{}{}{0px}{}{4}{2}\right)}{35}=\frac{6}{35}$	$\frac{\left(\genfrac{}{}{0px}{}{4}{3}\right)}{35}=\frac{4}{35}$
1	0	$\frac{\left(\genfrac{}{}{0px}{}{4}{1}\right)\left(\genfrac{}{}{0px}{}{2}{1}\right)}{35}=\frac{8}{35}$	$\frac{\left(\genfrac{}{}{0px}{}{4}{2}\right)\left(\genfrac{}{}{0px}{}{2}{1}\right)}{35}=\frac{12}{35}$	0
2	$\frac{\left(\genfrac{}{}{0px}{}{2}{2}\right)}{35}=\frac{1}{35}$	$\frac{\left(\genfrac{}{}{0px}{}{4}{1}\right)\left(\genfrac{}{}{0px}{}{2}{2}\right)}{35}=\frac{4}{35}$	0	0

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12

Joint PMF with $y$-dependence

Suppose $X$ and $Y$ have the following joint PMF:

$$P_{X,Y}(x,y)=\{\begin{array}{cc}\frac{.05}{y}{(.95)}^{y-1}& x=\mathrm{1,2},\dots ,y;y=\mathrm{1,2},\dots \\ 0& \text{otherwise}\end{array}$$

(a) Find $P[Y<3]$.

(b) $P[X=1]=.1577$. Find $P[X=2]$.

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Solution

Solutions - 5190-12

(a)

(1) Tabulate the joint PMF for small values of $Y$:

$$\begin{array}{cccc}X\downarrow Y\to & 1& 2& 3\\ 1& \frac{.05}{1}{(.95)}^0& \frac{.05}{2}{(.95)}^1& \frac{.05}{3}{(.95)}^2\\ 2& 0& \frac{.05}{2}{(.95)}^1& \frac{.05}{3}{(.95)}^2\\ 3& 0& 0& \frac{.05}{3}{(.95)}^2\\ 4& 0& 0& 0\end{array}$$

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(2) Sum over $Y=1$ and $Y=2$:

$$\begin{array}{c}P[Y<3]=P[Y=1]+P[Y=2]\\ \\ \gg \gg \frac{.05}{1}{(.95)}^0+2\left(\frac{.05}{2}{(.95)}^1\right)\gg \gg .0975\end{array}$$

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(b)

$$\begin{array}{c}P[X=2]=P[X=1]-P_{X,Y}(\mathrm{1,1})\\ \\ \gg \gg .1577-.05\gg \gg .1077\end{array}$$Link to original

10

Soft-drink machine

A soft-drink machine has a random amount $Y$ in supply at the beginning of a given day and dispenses a random amount $X$ during the day (with measurements in gallons). It is not resupplied during the day, and therefore $X\le Y$. It has been observed that $X$ and $Y$ have a joint density given by:

$$f_{X,Y}(x,y)=\frac{1}{2},0\le x\le y\le 2$$

(a) Find the probability that the amount of soft-drink dispensed on a given day is greater than 1.5 gallons. (First compute the marginal PDF of $X$.)

(b) Find the probability that the amount of soda remaining in the machine at the end of the day is greater than $1/2$ gallon. That is, find $P[Y-X>1/2]$.

(c) Find the CDF of $W=Y-X$, the amount of soda remaining in the machine at the end of the day.

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Solution

Solutions - 5190-10

(a)

Marginal PDF of $X$:

$$f_X(x)=\{\begin{array}{cc}1-\frac{1}{2}x& 0\le x\le 2\\ 0& \text{otherwise}\end{array}$$

Then:

$$P[X>1.5]\gg \gg {\int }_{1.5}^{2.0}{f}_X(x)dx\gg \gg 0.0625$$

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(b)

$$\begin{array}{c}\gg \gg {\int }_0^{1.5}{\int }_{x+\frac{1}{2}}^2\frac{1}{2}dydx\\ \\ \gg \gg \frac{1}{2}(1.5)(1.5)\cdot \frac{1}{2}\gg \gg 9/16\end{array}$$

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(c)

$$\begin{array}{c}{F}_W(w)=P[W\le w]\gg \gg P[Y-X\le w]\\ \\ \gg \gg 1-{\int }_0^{2-w}{\int }_{w+x}^2\frac{1}{2}dydx\\ \\ \gg \gg w-\frac{w^2}{4}\end{array}$$ $$F_W(w)=\{\begin{array}{cc}0& w<0\\ w-\frac{w^2}{4}& 0\le w<2\\ 1& w\ge 2\end{array}$$Link to original

11

Alice and Bob meeting at a cafe

Alice and Bob plan to meet at a cafe to do homework together. Alice arrives at the cafe at a random time (uniform) between 12:00pm and 1:00pm; Bob independently arrives at a random time (uniform) between 12:00pm and 2:00pm on the same day. Let $X$ be the time, past 12:00pm, that Alice arrives (in hours) and $Y$ be the time, past 12:00pm, that Bob arrives (in hours). So $X=0$ and $Y=0$ represent 12:00pm.

(a) Consider $W=\mathrm{Max}(X,Y)$, the time that will pass (beyond 12:00pm) before Alice and Bob will begin working together. Find the CDF of W.

(b) Find the probability Alice will have to wait more than 75 minutes for Bob.

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Solution

Solutions - 5200-04

By independence, the joint PDF is the product of the marginal PDFs:

$$\begin{array}{c}{f}_X(x)=1,0<x<1f_Y(y)=\frac{1}{2},0<y<2\\ \\ f_{X,Y}(x,y)=\frac{1}{2},0<x<1,0<y<2\end{array}$$Link to original

Independent random variables

01

Random point in a triangle

Consider a joint distribution that is uniform over the triangle with vertices $(\mathrm{0,0}),(\mathrm{0,1})$, and $(\mathrm{1,0})$. Suppose a point $(X,Y)$ is chosen at random according to this distribution.

(a) Find the joint PDF $f_{X,Y}$.

(b) Find the marginal PDFs for $X$ and $Y$.

(c) Are $X$ and $Y$ independent?

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Solution

Solutions - 5200-01

(a)

Find the area of the triangle and write the PDF:

The area of the triangle is $\frac{1}{2}$. Therefore the PDF is

$$f_{X,Y}(x,y)=\{\begin{array}{cc}\frac{1}{1/2}=2& \begin{array}{c}{\displaystyle 0\le x\le 1}\\ {\displaystyle 0\le y\le 1-x}\end{array}\\ 0& \text{otherwise}\end{array}$$

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(b)

(1) Integrate with respect to $y$ to find the marginal PDF for $X$:

$$\begin{array}{c}{f}_X(x)={\int }_0^{1-x}2dy\gg \gg {2y|}_0^{1-x}\gg \gg 2(1-x)\end{array}$$ $$f_X(x)=\{\begin{array}{cc}2(1-x)& 0\le x\le 1\\ 0& \text{otherwise}\end{array}$$

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(2) Integrate with respect to $x$ to find the marginal PDF for $Y$:

$$\begin{array}{c}{f}_Y(y)={\int }_0^{1-y}2dx\gg \gg {2x|}_0^{1-y}\gg \gg 2(1-y)\end{array}$$ $$f_Y(y)=\{\begin{array}{cc}2(1-y)& 0\le y\le 1\\ 0& \text{otherwise}\end{array}$$

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(c)

(1) Compute the product $f_X(x)f_Y(y)$ and compare to $f_{X,Y}(x,y)$:

$$2\stackrel{?}{=}4(1-x)(1-y)$$

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(2) Consider the case $x=1,y=0$:

$$4(1-1)(1-0)=0\ne 2$$

Therefore, $X$ and $Y$ are not independent.

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Functions on two random variables

02

PDF of Min and Max

Suppose $X\sim \mathrm{Exp}(2)$ and $Y\sim \mathrm{Exp}(3)$ and these variables are independent. Find:

(a) The PDF of $W=\mathrm{Max}(X,Y)$

(b) The PDF of $W=\mathrm{Min}(X,Y)$

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Solution

Solutions - 5210-02

(a)

(1) State the PDFs and CDFs:

$$f_X(x)=\{\begin{array}{cc}2e^{-2x}& x\ge 0\\ 0& \text{else}\end{array}{F}_X(x)=\{\begin{array}{cc}1-e^{-2x}& x\ge 0\\ 0& \text{else}\end{array}$$ $$f_Y(y)=\{\begin{array}{cc}3e^{-3y}& y\ge 0\\ 0& \text{else}\end{array}{F}_Y(y)=\{\begin{array}{cc}1-e^{-3y}& y\ge 0\\ 0& \text{else}\end{array}$$

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(2) Find the CDF of $W=\text{Max}(X,Y)$ using independence:

Since $W=\text{Max}(X,Y)\ge X,Y$,

$$\begin{array}{c}{F}_W(w)=P[W\le w]=P[X\le w,Y\le w]\\ \\ \gg \gg P[X\le w]P[Y\le w]=F_X(w)F_Y(w)\end{array}$$ $$F_W(w)=\{\begin{array}{cc}1-e^{-2w}-e^{-3w}+e^{-5w}& w\ge 0\\ 0& \text{else}\end{array}$$

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(3) Differentiate to get the PDF:

$$f_W(w)=\{\begin{array}{cc}2e^{-2w}+3e^{-3w}-5e^{-5w}& w\ge 0\\ 0& \text{else}\end{array}$$

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(b)

(1) Find the CDF of $W=\text{Min}(X,Y)$ using independence:

$$\begin{array}{c}{F}_W(w)=P[W\le w]=1-P[W>w]\\ \\ \gg \gg 1-P[X>w,Y>w]=1-P[X>w]P[Y>w]\end{array}$$ $$F_W(w)=\{\begin{array}{cc}1-e^{-5w}& w\ge 0\\ 0& \text{else}\end{array}$$

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(2) Differentiate to get the PDF:

$$f_W(w)=\{\begin{array}{cc}5e^{-5w}& w\ge 0\\ 0& \text{else}\end{array}$$Link to original

01

PDF of sum from joint PDF

Suppose the joint PDF of $X$ and $Y$ is given by:

$$f_{X,Y}=\{\begin{array}{cc}{\displaystyle \frac{8}{81}}xy& 0\le y\le x\le 3\\ 0& \text{otherwise}\end{array}$$

Find the PDF of $X+Y$.

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Solution

Solutions - 5220-01

(1) Write the CDF of $W=X+Y$:

Since $0\le y\le x\le 3$, we have $W\in [0,6]$.

$$F_W(w)=P[X+Y\le w]=\underset{\begin{array}{c}0\le y\le x\le 3\\ x+y\le w\end{array}}{\iint }\frac{8}{81}xydydx$$

For fixed $x\in [\mathrm{0,3}]$, $y$ ranges from $0$ to $-x+w$ when $x<w/2$ and from $0$ to $3$ when $x>w/2$.

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(2) Evaluate $F_W$ for $0\le w\le 3$:

$$\begin{array}{c}{F}_W(w)={\int }_0^{w/2}{\int }_0^x\frac{8}{81}xydydx+{\int }_{w/2}^w{\int }_0^{-x+w}\frac{8}{81}xydydx\\ \\ \gg \gg \frac{4}{81}{\int }_0^{w/2}{x}^{3}dx+\frac{4}{81}{\int }_{w/2}^{w}x{(-x+w)}^{2}dx\\ \\ \gg \gg \frac{4}{81}\cdot \frac{w^4}{64}+\frac{4}{81}\cdot \frac{5w^4}{192}\gg \gg \frac{w^4}{486}\end{array}$$

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(3) Evaluate $F_W$ for $3<w\le 6$:

$$\begin{array}{c}{F}_W(w)={\int }_0^{w/2}{\int }_0^x\frac{8}{81}xydydx+{\int }_{w/2}^3{\int }_0^{-x+w}\frac{8}{81}xydydx\\ \\ \gg \gg \frac{4}{81}{\int }_0^{w/2}{x}^{3}dx+\frac{4}{81}{\int }_{w/2}^{3}x{(-x+w)}^{2}dx\\ \\ \gg \gg \frac{4}{81}[\frac{w^4}{64}+\frac{9w^2}{2}-18w+\frac{81}{4}-\frac{11w^4}{192}]\\ \\ \gg \gg -\frac{w^4}{486}+\frac{2w^2}{9}-\frac{8w}{9}+1\end{array}$$

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(4) Differentiate for the PDF:

$$f_W(w)=\{\begin{array}{cc}{\displaystyle \frac{2}{243}}{w}^3& 0\le w\le 3\\ -{\displaystyle \frac{8}{9}}+{\displaystyle \frac{4}{9}}w-{\displaystyle \frac{2}{243}}{w}^3& 3<w\le 6\\ 0& \text{else}\end{array}$$Link to original

06

Lights on

An electronic device is designed to switch house lights on and off at random times after it has been activated. Assume the device is designed in such a way that it will be switched on and off exactly once in a 1-hour period. Let $X$ denote the time at which the lights are turned on and Y the time at which they are turned off. Assume the joint density for $(X,Y)$ is given by:

$$f_{X,Y}(x,y)=8xy,0<x<y<1$$

Let $W=Y-X$, the time the lights remain on during the hour.

(a) Find the range of $W$.

(b) Compute a formula for the CDF of $W$, i.e. $F_W(w)$.

(c) Find the probability the lights remain on for at least 40 minutes in some given hour.

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Solution

Solutions - 5220-06

(a) Range of $W$ is $[\mathrm{0,1}]$.

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(b)

$$\begin{array}{ccc}& F_W(w)=P[Y-X\le w]\gg \gg P[Y\le w+X]& \\ & & \\ & \gg \gg 1-{\int }_0^{1-w}{\int }_{w+x}^{1}8xydydx& \\ & & \\ & \gg \gg \frac{8}{3}w-2w^2+\frac{1}{3}{w}^4& \text{(}w\in [\mathrm{0,1}]\text{)}\end{array}$$

Alternate:

$$F_W(w)={\int }_0^{1-w}{\int }_x^{w+x}8xydydx+{\int }_{1-w}^1{\int }_x^{1}8xydydx$$

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(c)

$$P[Y-X\ge 2/3]\gg \gg 1-F_W(2/3)\gg \gg 11/243$$Link to original

Covariance and correlation

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

04

Correlation between overlapping coin flip sequences

Suppose a coin is flipped 30 times.

Let $X$ count the number of heads among the first 20 flips, and $Y$ count the heads in the last 20.

Find $\rho [X,Y]$.

Hint: Partition the flips into three groups of 10. Create three variables, counting heads, and express $X$ and $Y$ using these. What is the variance of a binomial distribution?

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Solution

Solutions - 5240-04

(1) Define random variables for partitioning the 30 flips into groups of 10:

Let $A$ be the number of heads in the first 10 flips.

Let $B$ be the number of heads in the middle 10 flips.

Let $C$ be the number of heads in the last 10 flips.

Clearly, $A,B,C\sim \text{Bin}(10,\frac{1}{2})$ and are independent. Note that $X=A+B$ and $Y=B+C$.

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

$$\begin{array}{c}E[X]=E[A+B]=E[A]+E[B]=5+5=10\\ \\ E[Y]=E[B+C]=E[B]+E[C]=5+5=10\end{array}$$

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(3) Compute $E[XY]$:

$$\begin{array}{c}E[XY]=E\left[(A+B)(B+C)\right]\\ \\ \gg \gg E[AB]+E[AC]+E[BC]+E\left[B^2\right]\\ \\ \gg \gg E[A]E[B]+E[A]E[C]+E[B]E[C]+E\left[B^2\right]=75+E\left[B^2\right]\end{array}$$

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(4) Compute $E\left[B^2\right]$:

Since $E[B]=5$ and $\text{Var}[B]=\frac{5}{2}$, we have $E\left[B^2\right]=5^2+\frac{5}{2}=\frac{55}{2}$.

Thus, $E[XY]=\frac{205}{2}$.

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

$$\begin{array}{c}\text{Cov}[X,Y]=E[XY]-E[X]E[Y]=\frac{205}{2}-10\cdot 10=\frac{5}{2}\end{array}$$

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

$$\begin{array}{c}\rho [X,Y]=\frac{\text{Cov}[X,Y]}{\sqrt{\text{Var}[X]\text{Var}[Y]}}=\frac{\frac{5}{2}}{\sqrt{5\cdot 5}}\gg \gg \frac{1}{2}\end{array}$$Link to original

07

Covariance etc. from joint density

Suppose $X$ and $Y$ are random variables with the following joint density:

$$f_{X,Y}(x,y)=\{\begin{array}{cc}\frac{3}{2}(x^2+y^2)& x,y\in [\mathrm{0,1}]\\ 0& \text{ otherwise }\end{array}$$

Compute:

(a) $E[X]$ $$ (b) $E[Y]$ $$ (c) $E[XY]$ $$ (d) $\mathrm{Var}[X]$

(e) $\mathrm{Var}[Y]$ $$ (f) $\mathrm{Cov}[X,Y]$ $$ (g) $\rho [X,Y]$ $$ (h) Are $X$ and $Y$ independent?

(It is worth thinking through which of these can be computed in multiple ways.)

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Solution

Solutions - 5240-07

(a)

$$\begin{array}{c}E[X]=\frac{3}{2}{\int }_0^1{\int }_0^{1}x(x^2+y^2)dydx\gg \gg \frac{3}{2}{\int }_0^1{x}^3+\frac{x}{3}dx\\ \\ \gg \gg \frac{3}{2}{[\frac{x^4}{4}+\frac{x^2}{6}]}_0^1\gg \gg \frac{3}{2}[\frac{1}{4}+\frac{1}{6}]\gg \gg \frac{5}{8}\end{array}$$

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(b)

By symmetry, $E[Y]=E[X]$:

$$E[Y]=\frac{5}{8}$$

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(c)

$$\begin{array}{c}E[XY]=\frac{3}{2}{\int }_0^1{\int }_0^{1}xy(x^2+y^2)dydx\gg \gg \frac{3}{2}{\int }_0^1\frac{x^3}{2}+\frac{x}{4}dx\\ \\ \gg \gg \frac{3}{2}{[\frac{x^4}{8}+\frac{x^2}{8}]}_0^1\gg \gg \frac{3}{2}[\frac{1}{8}+\frac{1}{8}]\gg \gg \frac{3}{8}\end{array}$$

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(d)

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

$$\begin{array}{c}E\left[X^2\right]=\frac{3}{2}{\int }_0^1{\int }_0^1{x}^2(x^2+y^2)dydx\gg \gg \frac{3}{2}{\int }_0^1{x}^4+\frac{x^2}{3}dx\\ \\ \gg \gg \frac{3}{2}{[\frac{x^5}{5}+\frac{x^3}{9}]}_0^1\gg \gg \frac{3}{2}[\frac{1}{5}+\frac{1}{9}]\gg \gg \frac{7}{15}\end{array}$$

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(2) Compute $\text{Var}[X]$:

$$\begin{array}{c}\text{Var}[X]=E\left[X^2\right]-{\left(E[X]\right)}^2=\frac{7}{15}-{\left(\frac{5}{8}\right)}^2\gg \gg \frac{73}{960}\end{array}$$

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(e)

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

By symmetry, $E\left[Y^2\right]=E\left[X^2\right]=\frac{7}{15}$.

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(2) Compute $\text{Var}[Y]$:

$$\begin{array}{c}\text{Var}[Y]=\text{Var}[X]\gg \gg \frac{73}{960}\end{array}$$

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(f)

$$\begin{array}{c}\text{Cov}[X,Y]=E[XY]-E[X]E[Y]=\frac{3}{8}-\frac{25}{64}\gg \gg -\frac{1}{64}\end{array}$$

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(g)

$$\begin{array}{c}\rho [X,Y]=\frac{\text{Cov}[X,Y]}{\sqrt{\text{Var}[X]\text{Var}[Y]}}=\frac{-\frac{1}{64}}{\sqrt{{\left(\frac{73}{960}\right)}^2}}\gg \gg -\frac{15}{73}\end{array}$$

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(h)

Since $\text{Cov}[X,Y]\ne 0$, we can conclude that $X$ and $Y$ are not independent.

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Conditional distribution

01

Conditional density from joint density

Suppose that $X$ and $Y$ have joint probability density given by:

$$f_{X,Y}(x,y)=\{\begin{array}{cc}{\displaystyle \frac{12}{5}x(2-x-y)}& x,y\in [\mathrm{0,1}]\\ & \\ 0& \text{ otherwise }\end{array}$$

(a) Compute $f_{X|Y}(x|y)$, for $y\in [\mathrm{0,1}]$.

(b) Compute $P[X>1/2|Y=y]$.

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Solution

Solutions - 5250-01

(a)

(1) Compute the marginal $f_Y(y)$:

$$\begin{array}{c}{f}_Y(y)=\frac{12}{5}{\int }_0^1(2x-x^2-xy)dx\\ \\ \gg \gg \frac{12}{5}{[x^2-\frac{x^3}{3}-\frac{x^{2}y}{2}]|}_0^1\\ \\ \gg \gg \frac{12}{5}(\frac{2}{3}-\frac{y}{2})\end{array}$$

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(2) Apply the conditional density formula:

$$\begin{array}{c}{f}_{X|Y}(x|y)=\frac{f_{X,Y}(x,y)}{f_Y(y)}\\ \\ \gg \gg \frac{\frac{12}{5}x(2-x-y)}{\frac{12}{5}(\frac{2}{3}-\frac{y}{2})}\gg \gg \frac{6x(2-x-y)}{4-3y}\end{array}$$ $$f_{X|Y}(x|y)=\{\begin{array}{cc}{\displaystyle \frac{6x(2-x-y)}{4-3y}}& x,y\in [0,1]\\ 0& \text{otherwise}\end{array}$$

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(b)

Integrate the conditional density over $(\frac{1}{2},1)$:

$$\begin{array}{c}P[X>\frac{1}{2}|Y=y]={\int }_{1/2}^1\frac{6x(2-x-y)}{4-3y}dx\\ \\ \gg \gg \frac{6}{4-3y}{\int }_{1/2}^1(2x-x^2-xy)dx\\ \\ \gg \gg \frac{6}{4-3y}{[x^2-\frac{x^3}{3}-\frac{x^{2}y}{2}]|}_{1/2}^1\\ \\ \gg \gg {\displaystyle \frac{11-9y}{4(4-3y)}}\end{array}$$ $$P[X>\frac{1}{2}|Y=y]=\{\begin{array}{cc}{\displaystyle \frac{11-9y}{4(4-3y)}}& y\in [0,1]\\ 0& \text{otherwise}\end{array}$$Link to original

02

From conditional to joint, and back again

Suppose we have the following data about random variables $X$ and $Y$:

$$\begin{array}{c}{f}_X(x)=\{\begin{array}{cc}3x^2& 0\le x\le 1\\ 0& \text{ otherwise }\end{array}\\ \\ f_{Y|X}(y|x)=\{\begin{array}{cc}2y/x^2& 0\le y\le x\\ 0& \text{ otherwise }\end{array}\end{array}$$

(a) Find the joint distribution $f_{X,Y}(x,y)$.

(b) Find $f_{X|Y}(x|y)$.

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Solution

Solutions - 5250-02

(a)

Obtain the joint from the conditional and marginal:

$$\begin{array}{c}{f}_{X,Y}(x,y)=f_{Y|X}(y|x)\cdot f_X(x)\\ \\ \gg \gg 3x^2\cdot \frac{2y}{x^2}\gg \gg 6y\end{array}$$ $$f_{X,Y}(x,y)=\{\begin{array}{cc}6y& 0\le y\le x\le 1\\ 0& \text{otherwise}\end{array}$$

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(b)

(1) Compute the marginal $f_Y(y)$:

$$\begin{array}{c}{f}_Y(y)={\int }_y^{1}6ydx\\ \\ \gg \gg {6yx|}_y^1\gg \gg 6y(1-y)\end{array}$$

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(2) Apply the conditional density formula:

$$\begin{array}{c}{f}_{X|Y}(x|y)=\frac{f_{X,Y}(x,y)}{f_Y(y)}\\ \\ \gg \gg \frac{6y}{6y(1-y)}\gg \gg \frac{1}{1-y}\end{array}$$ $$f_{X|Y}(x|y)=\{\begin{array}{cc}{\displaystyle \frac{1}{1-y}}& 0\le y\le x\le 1\\ 0& \text{otherwise}\end{array}$$Link to original

03

Time till recharge

Let $X$ denote the amount of time (in hours) that a battery on a solar calculator will operate properly before needing to be recharged by exposure to light. The function below is the PDF of $X$.

$$f(x)=\{\begin{array}{cc}{\displaystyle \frac{50}{6}}{x}^{-3}& 2<x<10\\ & \\ 0& \text{ otherwise }\end{array}$$

Suppose that a calculator has already been in use for 5 hours. Find the probability it will operate properly for at least another 2 hours.

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Solution

Solutions - 5250-03

(1) Compute $P[B]=P[X>5]$:

$$\begin{array}{c}P[X>5]={\int }_5^{10}\frac{50}{6}{x}^{-3}dx\\ \\ \gg \gg {\frac{-25}{6}{x}^{-2}|}_5^{10}\\ \\ \gg \gg \frac{-25}{6}({10}^{-2}-5^{-2})\gg \gg 0.125\end{array}$$

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(2) Compute the conditional density $f_{X|B}$:

$$f_{X|B}(x)=\frac{\frac{50}{6}{x}^{-3}}{0.125},5<x<10$$

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(3) Compute $P[X>7|X>5]$:

$$\begin{array}{c}P[X>7|X>5]={\int }_7^{10}\frac{\frac{50}{6}{x}^{-3}}{0.125}dx\\ \\ \gg \gg \frac{1}{0.125}\cdot \frac{50}{6}{\left[\frac{-1}{2}{x}^{-2}\right]}_7^{10}\\ \\ \gg \gg \frac{-100}{3}({10}^{-2}-7^{-2})\gg \gg \approx 0.3469\end{array}$$Link to original

Conditional expectation

02

Conditional distribution and expectation from joint PDF

Suppose that $X$ and $Y$ have the following joint PDF:

$$f_{X,Y}(x,y)=\{\begin{array}{cc}cxy& 0<y<1,0<x<y\\ 0& \text{ otherwise }\end{array}$$

Notice that the range of possibilities for $x$ depends on the value of $y$.

First, show that $c=8$. Then compute:

(a) $f_X$ $$ (b) $f_{Y|X}$ $$ (c) $E[Y|X=0.5]$ $$ (d) $E[Y|X]$

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Solution

Solutions - 5260-02

Integrate the joint PDF and solve for $c$:

$$\begin{array}{c}{\int }_0^1{\int }_0^{y}cxydxdy=1\\ \\ \gg \gg {\int }_0^1\frac{c{y}^3}{2}dy\gg \gg \frac{c}{8}=1\gg \gg c=8\end{array}$$

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(a)

Integrate the joint PDF with respect to $y$ to obtain $f_X$:

$$\begin{array}{c}{f}_X(x)={\int }_x^{1}8xydy\\ \\ \gg \gg 8x{\left[\frac{y^2}{2}\right]|}_x^1\gg \gg 4x(1-x^2)\end{array}$$

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(b)

Apply the conditional density formula:

$$\begin{array}{c}{f}_{Y|X}(y|x)=\frac{f_{X,Y}(x,y)}{f_X(x)}\\ \\ \gg \gg \frac{8xy}{4x(1-x^2)}\gg \gg \frac{2y}{1-x^2},x<y\end{array}$$

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(c)

(1) Plug in $x=0.5$ into the conditional density from part (b):

$$\begin{array}{c}{f}_{Y|X}(y|0.5)=\frac{2y}{1-{0.5}^2}\gg \gg \frac{8y}{3}\end{array}$$

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(2) Compute the conditional expectation:

$$\begin{array}{c}E[Y|X=0.5]={\int }_{0.5}^1\frac{8y^2}{3}dy\\ \\ \gg \gg \frac{8}{3}{\left[\frac{y^3}{3}\right]|}_{0.5}^1\gg \gg {\displaystyle \frac{7}{9}}\end{array}$$

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(d)

Integrate the conditional density from part (b) to find $E[Y|X=x]$:

$$\begin{array}{c}E[Y|X=x]={\int }_x^1\frac{2y^2}{1-x^2}dy\\ \\ \gg \gg \frac{2}{1-x^2}{\left[\frac{y^3}{3}\right]|}_x^1\gg \gg {\displaystyle \frac{2(1-x^3)}{3(1-x^2)}}\end{array}$$Link to original