W09 Homework B

Due date: Tuesday 3/17, 11:59pm

Functions on two random variables

01

04

Coffee and danish

Joe visits a coffee shop every morning and orders a triple espresso and a cherry danish. Let $X$ represent the wait time for the espresso and $Y$ the wait time for the danish (both in minutes). The joint PDF of $X$ and $Y$ is:

$$f_{X,Y}(x,y)=\{\begin{array}{cc}\frac{1}{50}{e}^{-\frac{1}{5}x-\frac{1}{10}y}& x,y>0\\ 0& \text{ otherwise }\end{array}$$

Suppose that Joe will not sit down until he has both his espresso and his danish. Compute the probability that he will have to wait at least 5 minutes before sitting down, that is, find $P[\mathrm{Max}(X,Y)>5]$.

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Solution

Solutions - 5210-04

$$\begin{array}{c}P[\mathrm{Max}(X,Y)>5]=1-{\int }_0^5{\int }_0^5\frac{1}{50}{e}^{-\frac{x}{5}-\frac{y}{10}}dydx\\ \\ \gg \gg 0.7513\end{array}$$Link to original

02 $★$

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

Review

Derived RV

03

08

Derived random variable

Let $U=\mathrm{Unif}[\mathrm{0,1}]$ and suppose that $X=-\ln (1-U)$.

(a) Compute $F_X(x)$.

(b) Compute $f_X(x)$.

(c) Compute $E[X]$.

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Solution

Solutions - 5150-08

(a)

(1) Express $F_X(x)$ via the CDF of $U$:

For $x<0$: $F_X(x)=0$ since $X=-\ln (1-U)\ge 0$.

For $x\ge 0$:

$$\begin{array}{c}{F}_X(x)=P[-\ln (1-U)\le x]\gg \gg P[1-U\ge e^{-x}]\\ \\ \gg \gg P[U\le 1-e^{-x}]\end{array}$$

Since $0\le 1-e^{-x}\le 1$ for $x\ge 0$, this equals $1-e^{-x}$:

$$F_X(x)=\{\begin{array}{cc}0& x<0\\ 1-e^{-x}& x\ge 0\end{array}$$

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(b)

Differentiate $F_X(x)$:

$$f_X(x)=\{\begin{array}{cc}{e}^{-x}& x\ge 0\\ 0& \text{otherwise}\end{array}$$

This is the exponential PDF with parameter $a=1$.

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(c)

Since $X\sim \text{Exp}(1)$,

$$E[X]=1$$Link to original

Normal distribution

04

08

Blood sugar testing

A test for diabetes is a measurement $X$ of a person’s blood sugar level following an overnight fast. If a person has diabetes, $X$ is a Gaussian $(\mathrm{60,40})$ random variable.

Now suppose that the doctor has decided to use the following scores: positive for diabetes if $X\ge$ 140, negative if $X\le 110$, and the test is inconclusive if $110<X<140$.

Find the probability the test result will be inconclusive for a person who has diabetes.

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Solution

Solutions - 5180-08

(1) Write $X=\sigma Z+\mu$ and substitute into the probability:

$$\begin{array}{c}X=40Z+60\\ \\ P[110<X<140]\gg \gg P[110<40Z+60<140]\\ \\ \gg \gg P[\frac{110-60}{40}<Z<\frac{140-60}{40}]\gg \gg P[1.25<Z<2]\end{array}$$

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(2) Evaluate using the $\mathrm{\Phi }$ table:

$$\begin{array}{c}\mathrm{\Phi }(2)-\mathrm{\Phi }(1.25)\gg \gg 0.97725-0.8944\gg \gg 0.08285\end{array}$$Link to original

Joint distributions

05

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

06

08

Sales rep making calls

A sales representative is responsible for selling a particular item. On a given day, he has time to make a sales pitch to up to 3 customers. His goal is to sell the item to 2 customers; if he is successful with the first two, he will not try to sell to the $3^{\text{rd }}$ customer. The probability of any one customer purchasing the item is 0.6, independent of the others.

Let $X$ be the number of customers to which he tries to sell the item and $Y$ be the number of customers that purchase the item. Construct the joint PMF of $X$ and $Y$.

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Solution

Solutions - 5190-08

Construct the joint PMF table for $X$ (customers pitched) and $Y$ (customers who buy):

$X\backslash Y$	2	3
$0$	$0$	${(0.4)}^3$
$1$	$0$	$3(0.6)({0.4}^2)$
$2$	${(0.6)}^2$	$2({0.6}^2)(0.4)$

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07 $★$

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 - 5190-11

(a)

Express CDF of $W$:

$$F_W(w)=P[W\le w]=P[\mathrm{Max}(X,Y)\le w]$$

Case 1, $0<w<1$:

$$F_W(w)={\int }_0^w{\int }_0^w\frac{1}{2}dydx\gg \gg \frac{1}{2}{w}^2$$

Case 2, $1\le w<2$:

$$F_W(w)={\int }_0^1{\int }_0^w\frac{1}{2}dydx\gg \gg \frac{1}{2}w$$

Complete CDF:

$$F_W(w)=\{\begin{array}{cc}0& w<0\\ \frac{1}{2}{w}^2& 0\le w<1\\ \frac{1}{2}w& 1\le w<2\\ 1& w\ge 2\end{array}$$

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(b)

Compute $P[Y-X>1.25]=P[Y>X+1.25]$:

$$\begin{array}{c}\gg \gg {\int }_0^{\frac{3}{4}}{\int }_{x+1.25}^2\frac{1}{2}dydx\\ \\ \gg \gg \frac{3}{4}\cdot \frac{3}{4}\cdot \frac{1}{2}\cdot \frac{1}{2}\gg \gg \frac{9}{64}\gg \gg 0.1406\end{array}$$Link to original