W07 Homework B

Due date: Tuesday 3/10, 11:59pm

Joint distributions

01 $★$

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

02 $★$

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

03

09

Air pollution

In a certain community, levels of air pollution may exceed federal standards for ozone or for particulate matter on some days. In a particular summer week, let X be the number of days on which the ozone standard is exceeded, and let Y be the number of days on which the particulate matter is exceeded.

The following table represents the joint PMF for X and Y.

	$Y=1$	$Y=2$	$Y=3$
$X=1$	0.09	0.11	0.05
$X=2$	0.17	0.23	0.08
$X=3$	0.06	0.15	0.06

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

(b) Find $E[X+Y]$.

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Solution

Solutions - 5190-09

	$Y=1$	$Y=2$	$Y=3$	$P_X(k)$
$X=1$	0.09	0.11	0.05	0.25
$X=2$	0.17	0.23	0.08	0.48
$X=3$	0.06	0.15	0.06	0.27
$P_Y(\ell )$	0.32	0.49	0.19

(a)

Sum the joint probabilities where $X>Y$:

$$\begin{array}{c}P[X=2,Y=1]+P[X=3,Y=1]+P[X=3,Y=2]\\ \\ \gg \gg 0.17+0.06+0.15\gg \gg 0.38\end{array}$$

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(b)

Compute $E[X]$, $E[Y]$, and use linearity:

$$\begin{array}{c}E[X]=0.25+2(0.48)+3(0.27)\gg \gg 2.02\\ \\ E[Y]=0.32+2(0.49)+3(0.19)\gg \gg 1.87\\ \\ E[X+Y]=E[X]+E[Y]\gg \gg 3.89\end{array}$$Link to original

04 $★$

03

Composite PDF from joint PDF

The joint density of random variables $X$ and $Y$ is given by:

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

Compute the PDF of $Z=X/Y$.

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Solution

Solutions - 5200-03

(1) Define $Z=X/Y$ and find the CDF of $Z$:

$$\begin{array}{c}P[Z\le z]=P[X/Y\le z]=P[X\le zY]\\ \\ \gg \gg {\int }_0^{\infty }{\int }_0^{zy}{e}^{-x}{e}^{-y}dxdy\\ \\ \gg \gg -{\int }_0^{\infty }{e}^{-y}{\left[e^{-x}\right]|}_0^{zy}dy\\ \\ \gg \gg 1-{\int }_0^{\infty }{e}^{-y(z+1)}dy\gg \gg 1-\frac{1}{z+1},z>0\end{array}$$

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(2) Differentiate to find $f_Z(z)$:

$$f_Z(z)=\frac{1}{{(z+1)}^2},z>0$$Link to original

Independent random variables

05

5

One car outlasts the other

Suppose that $X,Y\sim \mathrm{Exp}(0.1)$ are two independent exponential random variables.

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

(b) Find the probability:

$$P[X-5>Y-15|X>5\text{AND}Y>15]$$

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Compare to W05-B Q02 “Vehicle Lifetimes.” Which method is easier? (For yourself, not this HW.)

Solution

Solutions - 5200-05

(a)

Since $X$ and $Y$ are independent, the joint PDF is the product of the marginals:

$$f_{X,Y}(x,y)=\{\begin{array}{cc}(0.1e^{-0.1x})(0.1e^{-0.1y})=0.01e^{-0.1(x+y)}& x,y>0\\ 0& \text{otherwise}\end{array}$$

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(b)

(1) Expand the conditional probability:

$$\begin{array}{c}P[X-5>Y-15|X>5\text{AND}Y>15]=\frac{P[X>Y-10\text{AND}X>5\text{AND}Y>15]}{P[X>5\text{AND}Y>15]}\end{array}$$

The three conditions in the numerator are partially redundant because $Y>15$ implies $Y-10>5$ and therefore $X>5$ follows automatically. So we may drop the middle term:

$$\gg \gg \frac{P[X>Y-10\text{AND}Y>15]}{P[X>5\text{AND}Y>15]}$$

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(2) Evaluate the denominator using independence:

$$\begin{array}{c}P[X>5\text{AND}Y>15]=P[X>5]\cdot P[Y>15]\\ \\ \gg \gg e^{-0.1(5)}\cdot e^{-0.1(15)}\gg \gg e^{-0.5}\cdot e^{-1.5}\gg \gg e^{-2}\end{array}$$

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(3) Evaluate the numerator:

The region is $x>y-10$ and $y>15$, i.e. above $y=15$ and below $y=x+10$.

For the outer variable $x$, the lower bound is $x=5$ (where the line $y=x+10$ meets $y=15$). For a given $x$ value, $y$ ranges from $15$ to $x+10$.

Inner integral over $y$:

$$\begin{array}{c}{\int }_5^{\infty }{\int }_{15}^{x+10}0.01e^{-0.1(x+y)}dydx\\ \\ \gg \gg {\int }_5^{\infty }0.01e^{-0.1x}{\left[\frac{e^{-0.1y}}{-0.1}\right]}_{y=15}^{x+10}dx\\ \\ \gg \gg {\int }_5^{\infty }0.1e^{-0.1x}(e^{-1.5}-e^{-0.1x-1})dx\end{array}$$

Outer integral over $x$:

$$\begin{array}{c}0.1e^{-1.5}{\int }_5^{\infty }{e}^{-0.1x}dx-0.1e^{-1}{\int }_5^{\infty }{e}^{-0.2x}dx\\ \\ \gg \gg 0.1e^{-1.5}\cdot \frac{e^{-0.5}}{0.1}-0.1e^{-1}\cdot \frac{e^{-1}}{0.2}\\ \\ \gg \gg e^{-2}-0.5e^{-2}\gg \gg 0.5e^{-2}\end{array}$$

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(4) Plug into original ratio:

$$\begin{array}{c}\frac{0.5e^{-2}}{e^{-2}}\gg \gg {\displaystyle \frac{1}{2}}\end{array}$$

This result is what we would expect to obtain by applying the memoryless property of exponential distributions.

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Review

06

05

Normal distribution - cars passing toll booth

The number of cars passing a toll booth on Wednesdays has a normal distribution $𝒩(1200,40000)$.

(a) What is the probability that on a randomly chosen Wednesday, more than 1,400 cars pass the toll booth?

(b) What is the probability that between 1,000 and 1,400 cars pass the toll booth on a random Wednesday?

(c) Suppose it is learned that at least 1200 cars passed the toll booth last Wednesday. What is the probability that at least 1300 cars passed the toll booth that day?

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Solution

Solutions - 5180-05

(a)

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

$$\begin{array}{c}X=200Z+1200\\ \\ P[X>1400]\gg \gg P[200Z+1200>1400]\\ \\ \gg \gg P[Z>\frac{1400-1200}{200}]\gg \gg P[Z>1]\end{array}$$

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

$$\begin{array}{c}1-\mathrm{\Phi }(1)\gg \gg 1-0.8413\gg \gg 0.1587\end{array}$$

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(b)

(1) Substitute $X=200Z+1200$ into the probability:

$$\begin{array}{c}P[1000<X<1400]\gg \gg P[1000<200Z+1200<1400]\\ \\ \gg \gg P[-1<Z<1]\end{array}$$

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(2) Express in terms of $\mathrm{\Phi }$ and evaluate:

$$\begin{array}{c}\mathrm{\Phi }(1)-\mathrm{\Phi }(-1)\gg \gg 2\mathrm{\Phi }(1)-1\\ \\ \gg \gg 2(0.8413)-1\gg \gg 0.6826\end{array}$$

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(c)

(1) Set up conditional probability expression:

$$P[X\ge 1300|X\ge 1200]=\frac{P[X\ge 1300\cap X\ge 1200]}{P[X\ge 1200]}=\frac{P[X\ge 1300]}{P[X\ge 1200]}$$

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(2) Substitute $X=200Z+1200$ in numerator and denominator:

$$\begin{array}{c}\frac{P[200Z+1200\ge 1300]}{P[200Z+1200\ge 1200]}\gg \gg \frac{P[Z\ge 0.5]}{P[Z\ge 0]}\\ \\ \gg \gg \frac{1-\mathrm{\Phi }\left(\frac{1}{2}\right)}{1-\mathrm{\Phi }(0)}\end{array}$$

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

$$\begin{array}{c}\frac{1-0.6915}{1-0.5}\gg \gg 0.617\end{array}$$Link to original