W07 Homework A

Due date: Thursday 2/26, 11:59pm

Joint distributions

01

06

Stop lights

Let $X$ be the number of stop lights and $Y$ the number of red lights at which you must wait on your drive to grounds each day. The joint PMF of $X$ and $Y$ is given below.

$Y\downarrow X\to$	0	1	2
0	0.05	0.07	0.08
1	0.08	0.10	0.12
2	0.12	0.33	0.05

Find the marginal PMF of $X$ and compute $E[X]$.

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Solution

Solutions - 5190-06

(1) Find the marginal PMF of $X$ by summing rows:

$k$	0	1	2
$P_X(k)$	.25	.5	.25

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(2) Compute $E[X]$:

$$\begin{array}{c}E[X]=0(0.25)+1(0.5)+2(0.25)\gg \gg 1\end{array}$$Link to original

02

03

Marginals from joint PMF

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

$$P_{X,Y}(x,y)=\frac{x{y}^2}{30},x=\mathrm{1,2,3},y=\mathrm{1,2}$$

Compute the marginal PMFs $P_X(x)$ and $P_Y(y)$.

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Solution

Solutions - 5190-03

(1) Compute $P_X(x)$ by summing over $y$:

$$\begin{array}{c}{P}_X(x)=\sum _y{P}_{X,Y}(x,y)\\ \\ \gg \gg P_{X,Y}(x,1)+P_{X,Y}(x,2)\gg \gg \frac{x}{30}(1^2+2^2)\gg \gg \frac{x}{6}\end{array}$$

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(2) Compute $P_Y(y)$ by summing over $x$:

$$\begin{array}{c}{P}_Y(y)=\sum _x{P}_{X,Y}(x,y)\\ \\ \gg \gg \frac{y^2}{30}(1+2+3)\gg \gg \frac{y^2}{5}\end{array}$$Link to original

03

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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Independent random variables

04

04

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.

Find the joint PDF of $X$ and $Y$. (Hint: $X$ and $Y$ are independent.)

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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