W09 Homework A

Due date: Thursday 3/12, 11:59pm

Functions on two random variables

01

03

PDF of Min

Let $X$ and $Y$ be independent copies of a $\mathrm{Unif}[\mathrm{0,1}]$ random variable.

Let $W=\mathrm{Min}(X,Y)$. Find the PDF of $W$.

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Solution

Solutions - 5210-03

(1) Write the PDFs and joint PDF:

PDFs of $X$ and $Y$:

$$f_X(x)=\{\begin{array}{cc}1& 0\le x\le 1\\ 0& \text{otherwise}\end{array}{f}_Y(y)=\{\begin{array}{cc}1& 0\le y\le 1\\ 0& \text{otherwise}\end{array}$$

Joint PDF using independence:

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

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

$$\begin{array}{c}{F}_W(w)=P[W\le w]=P[X\le w\text{OR}Y\le w]\\ \\ \gg \gg 1-P[X\ge w\text{AND}Y\ge w]\\ \\ \gg \gg 1-{\iint }_{x,y\ge w}{f}_{X,Y}dA\end{array}$$

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(3) Evaluate for $w\in [\mathrm{0,1}]$:

$$\begin{array}{c}{F}_W(w)=1-{\int }_w^1{\int }_w^{1}dxdy\\ \\ \gg \gg 1-{(1-w)}^2\gg \gg w(2-w)\end{array}$$

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(4) Differentiate to find $f_W(w)$:

$$f_W(w)=\{\begin{array}{cc}2-2w& w\in [\mathrm{0,1}]\\ 0& \text{otherwise}\end{array}$$Link to original

02

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

03

05

PDF of sum of uniforms

Let $X$ and $Y$ be independent copies of a $\mathrm{Unif}[\mathrm{0,1}]$ random variable. Let $W=X+Y$.

Find the PDF of $W$.

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Solution

Solutions - 5220-05

(1) Write PDFs:

PDFs of $X$ and $Y$:

$$f_X(x)=\{\begin{array}{cc}1& 0\le x\le 1\\ 0& \text{otherwise}\end{array}{f}_Y(y)=\{\begin{array}{cc}1& 0\le y\le 1\\ 0& \text{otherwise}\end{array}$$

Joint PDF using independence:

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

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Option 1: CDF first

(2) CDF of sum:

Write $W=X+Y$. Then:

$$\begin{array}{c}{F}_W(w)=P[X+Y\le w]\\ \\ \gg \gg {\iint }_{x+y\le w}{f}_{X,Y}dA\gg \gg \underset{\begin{array}{c}x+y\le w\\ x,y\in [\mathrm{0,1}]\end{array}}{\iint }1\cdot 1dydx\end{array}$$

When $0\le w<1$:

$$F_W(w)={\int }_{x=0}^w{\int }_{y=0}^{w-x}dydx\gg \gg \frac{w^2}{2}$$

When $1\le w<2$:

$$F_W(w)=1-{\int }_{x=w-1}^1{\int }_{y=w-x}^{1}dydx\gg \gg -w^2/2+2w-1$$

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(3) PDF from CDF:

$$f_W(w)=\{\begin{array}{cc}w& w\in [\mathrm{0,1}]\\ -w+2& w\in [\mathrm{1,2}]\\ 0& \text{otherwise}\end{array}$$

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Option 2: Convolution formula

(2) Convolution:

$$f_W(w)={\int }_{-\infty }^{+\infty }{f}_{X,Y}(u,w-u)du$$

For $w\in [\mathrm{0,1}]$:

$$f_W(w)={\int }_0^{w}du\gg \gg w$$

For $w\in [\mathrm{1,2}]$:

$$\begin{array}{c}{f}_W(w)={\int }_{w-1}^{1}du\gg \gg 1-(w-1)\\ \\ \gg \gg -w+2\end{array}$$Link to original

04

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

Review

Derived RV

05

01

Function on one variable

Suppose the PDF of $X$ is given by:

$$f_X(x)=\{\begin{array}{cc}\frac{2}{3}x& 1\le x\le 2\\ 0& \text{otherwise}\end{array}$$

Find the CDF and PDF of $W=\ln X$.

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Solution

Solutions - 5210-01

(1) Find the CDF of $W=\ln X$:

$$\begin{array}{c}{F}_W(w)=P[W\le w]=P[\ln X\le w]=P[X\le e^w]\end{array}$$

Compute $F_X(x)$:

$$\begin{array}{c}{F}_X(x)={\int }_1^x\frac{2}{3}tdt\gg \gg {\left[\frac{t^2}{3}\right]}_1^x\gg \gg \frac{x^2-1}{3}\end{array}$$

Substituting $x=e^w$:

$$F_W(w)=\{\begin{array}{cc}0& w<0\\ \frac{e^{2w}-1}{3}& 0\le w\le \ln 2\\ 1& w>\ln 2\end{array}$$

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(2) Differentiate to find $f_W(w)$:

$$f_W(w)=\{\begin{array}{cc}0& w<0\\ \frac{2e^{2w}}{3}& 0\le w\le \ln 2\\ 0& w>\ln 2\end{array}$$Link to original

Joint distributions

06

01

Finish a PMF table - Strange families

Suppose that 15 percent of the families in a strange community have no children, 20 percent have 1 child, 35 percent have 2 children, and 30 percent have 3 children. Assume the odds of a child being a boy or a girl are equal.

If a family is chosen at random from this community, then $B$, the number of boys, and $G$, the number of girls, in this family will have the joint PMF partially shown in this table:

$B\downarrow G\to$	0	1	2	3	$P_B(i):$
0	0.15	0.10	?	0.0375	0.3750
1	0.10	0.175	?	0.00	0.3875
2	?	?	0.00	0.00	0.2000
3	0.0375	0.00	0.00	0.00	0.0375
$P_G(j):$	0.3750	0.3875	0.2000	0.0375	[not used]

(a) Complete the table by finding the missing entries.

(b) What is the probability that “$B$ or $G$ is 1”?

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Solution

Solutions - 5190-01

(a)

Fill the cells using the respective column sum or row sum:

$P[B=2,G=0]$: We have $P[G=0]=0.375$, so

$$P[B=2,G=0]=0.375-0.15-0.1-0.0375\gg \gg 0.0875$$

$P[B=2,G=1]$: We have $P[G=1]=0.3875$, so

$$P[B=2,G=1]=0.3875-0.1-0.175\gg \gg 0.1125$$

$P[B=0,G=2]$: We have $P[B=0]=0.375$, so

$$P[B=0,G=2]=0.375-0.15-0.1-0.0375\gg \gg 0.0875$$

$P[B=1,G=2]$: We have $P[B=1]=0.3875$, so

$$P[B=1,G=2]=0.3875-0.1-0.175\gg \gg 0.1125$$

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(b)

(1) Add up the probabilities in which either $B=1$ or $G=1$:

$$\begin{array}{c}P[B=1\text{ or }G=1]\\ \\ \gg \gg 0.1+0.1+0.175+0.1125+0.1125\gg \gg 0.6\end{array}$$

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(2) Alternatively, use inclusion-exclusion with the marginal sums:

$$\begin{array}{c}P[B=1]+P[G=1]-P[B=1\text{ and }G=1]\\ \\ \gg \gg 0.3875+0.3875-0.175\gg \gg 0.6\end{array}$$Link to original

07

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

Independent random variables

08

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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