W05 Homework A

Due date: Thursday 2/12, 11:59pm

Poisson process

01

05

Potholes on the highway

On a certain terrible stretch of highway, the appearance of potholes can be modeled by a Poisson process. Let the RV $X$ denote the distance between successive potholes (measured in miles). The CDF of X is:

$$F_X(x)=\{\begin{array}{cc}0& x\le 0\\ 1-e^{-0.5x}& x>0\end{array}$$

(a) What is the mean number of potholes in a 2-mile stretch of the highway?

(b) What is the probability that there will be at least 2 potholes in a 2-mile stretch of the highway?

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Solution

Solutions - 5160-05

(a)

$X\sim \mathrm{Exp}(0.5)$, so the Poisson rate for a 2-mile stretch is $\alpha =0.5\cdot 2=1$, giving $Y\sim \mathrm{Poisson}(1)$.

The mean number of potholes in a 2-mile stretch is $E[Y]=\alpha =1$.

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(b)

$$\begin{array}{c}P[Y\ge 2]=1-P[Y=0]-P[Y=1]\\ \\ \gg \gg 1-e^{-1}\cdot \frac{1^0}{0!}-e^{-1}\cdot \frac{1^1}{1!}\gg \gg 0.2642\end{array}$$Link to original

02

04

Selling Christmas trees

An online company sells artificial Christmas trees. During the holiday season, the amount of time between sales, $T$, is an exponential random variable with an expected value of 2.5 hours.

(a) Find the probability that the store will sell more than 2 trees in a 1-hour period of time.

(b) Find the probability the time between the sales of two trees will be between 4-5 hours.

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Solution

Solutions - 5160-04

(a)

$X\sim \mathrm{Poisson}(2/5)$

$$\begin{array}{c}P[X>2]=1-P[X\le 2]\\ \\ \gg \gg 1-(\frac{e^{-2/5}{\left(\frac{2}{5}\right)}^0}{0!}+\frac{e^{-2/5}{\left(\frac{2}{5}\right)}^1}{1!}+\frac{e^{-2/5}{\left(\frac{2}{5}\right)}^2}{2!})\\ \\ \gg \gg 1-0.9921\gg \gg 0.0079\end{array}$$

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(b)

$$\begin{array}{c}P[4\le T\le 5]=F_T(5)-F_T(4)\\ \\ \gg \gg (1-e^{-\frac{2}{5}(5)})-(1-e^{-\frac{2}{5}(4)})\\ \\ \gg \gg e^{-8/5}-e^{-2}\gg \gg 0.0666\end{array}$$Link to original

Derived random variable

03

01

Constants in PDF from expectation

Suppose $X$ has PDF given by:

$$f_X(x)=\{\begin{array}{cc}a+b{x}^2& 0\le x\le 1\\ 0& \text{otherwise}\end{array}$$

Suppose $E[X]=\frac{7}{10}$. Find the only possible values for $a$ and $b$. Then find $\mathrm{Var}[X]$.

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Solution

Solutions - 5150-01

(1) Recall formula for expectation of a continuous random variable:

$$E[X]={\int }_{-\infty }^{\infty }x{f}_X(x)dx$$

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(2) Use formula to find an equation relating $a$ and $b$:

$$\begin{array}{cc}{\int }_0^{1}x(a+b{x}^2)dx& =\frac{7}{10}\\ {[\frac{a{x}^2}{2}+\frac{b{x}^4}{4}]|}_0^1& =\frac{7}{10}\\ \frac{a}{2}+\frac{b}{4}=\frac{7}{10}& \end{array}$$

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(3) Integrate the PDF to get a second equation:

Since integrating a PDF should yield $1$:

$$\begin{array}{cc}{\int }_0^{1}a+b{x}^{2}dx& =1\\ {[ax+\frac{b{x}^3}{3}]|}_0^1& =1\\ a+\frac{b}{3}& =1\end{array}$$

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(4) Solve system of equations for $a$ and $b$:

Isolating $a$ in the second equation yields $a=1-\frac{b}{3}$.

Plugging this into the first equation yields $\frac{1-\frac{b}{3}}{2}+\frac{b}{4}=\frac{7}{10}$.

Solving for $b$ yields $b=2.4$, and thus $a=0.2$.

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(5) Compute variance:

Using $\text{Var}[X]=E\left[X^2\right]-{\left(E[X]\right)}^2$, first compute $E\left[X^2\right]$:

$$\begin{array}{c}E\left[X^2\right]={\int }_0^1{x}^2(a+b{x}^2)dx\\ \\ \gg \gg {[\frac{a{x}^3}{3}+\frac{b{x}^5}{5}]|}_0^1\\ \\ \gg \gg \frac{a}{3}+\frac{b}{5}\gg \gg \frac{41}{75}\end{array}$$

So:

$$\text{Var}[X]\gg \gg \frac{41}{75}-{\left(\frac{7}{10}\right)}^2\gg \gg \frac{17}{300}$$Link to original

04

02

Variance: Direct integral formula

Suppose $X$ has PDF given by: $f_X(x)=\{\begin{array}{cc}3e^{-3x}& x\ge 0\\ 0& \text{otherwise}\end{array}$

Find $\mathrm{Var}[X]$ directly using the integral formula.

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Solution

Solutions - 5150-02

(1) Recall the integral formula for variance:

Use the fact that $\text{Var}[X]=E\left[{(X-\mu )}^2\right]$.

$$\text{Var}[X]={\int }_{-\infty }^{\infty }{(x-\mu )}^2{f}_X(x)dx$$

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(2) Compute $\mu =E[X]$:

$$E[X]={\int }_0^{\infty }3x{e}^{-3x}dx=\underset{b\to \infty }{\lim }{[-x{e}^{-3x}-\frac{e^{-3x}}{3}]|}_0^b=\frac{1}{3}$$

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

$$\text{Var}[X]={\int }_0^{\infty }{(x-\frac{1}{3})}^{2}3e^{-3x}dx=\underset{b\to \infty }{\lim }{\int }_0^{b}3x^2{e}^{-3x}-2x{e}^{-3x}+\frac{e^{-3x}}{3}dx$$ $$=\underset{b\to \infty }{\lim }{[-x^2{e}^{-3x}-\frac{e^{-3x}}{9}]|}_0^b=\frac{1}{9}$$Link to original

Review problems

05

05

Binomial - Repeated coin flips

A coin is flipped 7 times and the sequence of results recorded as an outcome.

(a) How many possible outcomes have exactly 3 heads?

(b) How many possible outcomes have at least 3 heads?

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Solution

Solutions - 5070-05

(a)

Out of $7$ trials, we choose $3$ of them to be heads.

$$\left(\genfrac{}{}{0px}{}{7}{3}\right)=35$$

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(b)

Out of 7 trials, we choose at least 3 of them to be heads.

$$\sum _{k=3}^7\left(\genfrac{}{}{0px}{}{7}{k}\right)=99$$Link to original