W05 Homework B

Due date: Tuesday 2/17, 11:59pm

Poisson process

01 $★$

03

Application of Poisson: meteor shower

The UVA astronomy club is watching a meteor shower. Meteors appear at an average rate of $4$ per hour.

(a) Write a short explanation to justify the use of a Poisson distribution to model the appearance of meteors. Why should appearances be Poisson distributed?

(b) What is the probability that the club sees more than 2 meteors in a single hour?

(c) Suppose we find out that over a four hour evening, 13 meteors were spotted. What is the probability that none of them happened in the first hour, conditioned on that information?

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Solution

Solutions - 5130-03

(a)

Poisson processes model events that occur randomly when you know the mean number of events within any given interval and all disjoint intervals (of any size) are independent.

Since meteors arrive independently of each other, a Poisson process is reasonable.

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(b)

Compute probability:

We have $P[X>2]=1-P[X\le 2]$.

$$\begin{array}{c}P[X>2]=1-P[X\le 2]\\ \\ \gg \gg 1-\frac{e^{-4}{4}^0}{0!}-\frac{e^{-4}{4}^1}{1!}-\frac{e^{-4}{4}^2}{2!}\gg \gg \approx 0.7619\end{array}$$

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(c)

Let Window A be the first hour, Window B be the second through fourth hours, and Window C be the complete four hours.

The background rate is $\lambda =4$, so we have ${\alpha }_A=4$, ${\alpha }_B=12$, and ${\alpha }_C=16$. Define three Poisson variables accordingly: $X_A\sim \mathrm{Pois}(4)$, $X_B\sim \mathrm{Pois}(12)$, $X_C\sim \mathrm{Pois}(16)$.

Now the desired probability is:

$$\begin{array}{c}P[X_A=0|X_C=13]\gg \gg \frac{P[X_A=0\text{AND}{X}_C=13]}{P[X_C=13]}\\ \\ \gg \gg \frac{P[X_A=0\text{AND}{X}_B=13]}{P[X_C=13]}\end{array}$$

The last line follows because $X_C=X_A+X_B$. By independence of Poisson process windows:

$$\begin{array}{c}\gg \gg \frac{P[X_A=0]\cdot P[X_B=13]}{P[X_C=13]}\gg \gg \frac{P_{X_A}(0)\cdot P_{X_B}(13)}{P_{X_C}(13)}\\ \\ \gg \gg \frac{\left(e^{-4}\frac{4^0}{0!}\right)\left(e^{-12}\frac{{12}^{13}}{13!}\right)}{e^{-16}\frac{{16}^{13}}{13!}}\gg \gg 0.0238\end{array}$$Link to original

02 $★$

02

Vehicle lifetimes

Suppose that vehicle lifetimes follow an exponential distribution with an expected lifetime of 10 years.

Suppose you have one car that is 5 years old, and one that is 15 years old, at the present moment.

What is the probability that the first car outlives the second? (I.e. that the second breaks at an earlier time than the first breaks, both starting now.)

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Solution

Solutions - 5160-02

By the memoryless property of exponential distributions, elapsed time has no effect on future events. The fact that one car is older than the other has no effect on the remaining lifetimes.

Since both cars have the same remaining lifetime distribution, the probability that either car outlives the other is 0.5.

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03

03

Wait time for 5 calls - two methods

Consider the Poisson process of phone calls coming to a call center at an average rate of 1 call every 6 minutes.

Let us model the wait time for 5 calls to come in. You may use Desmos or similar to perform the integration numerically.

(a) Method One: An arrival of ‘1-call’ comes in at an average rate of $\lambda =10$ calls per hour. So a Bundle of ‘5-calls’ comes in at an average rate of ${\lambda }_B=2$ Bundles per hour. Use an exponential variable with ${\lambda }_B=2$ to determine the probability that the wait time for a Bundle (of 5 calls) is at most $1\mathrm{hr}$.

(b) Method Two: Use $\lambda =10$ calls per hour with an Erlang distribution at $\ell =5$ to determine the probability that the wait time for 5 calls is at most $1\mathrm{hr}$.

(c) Compare the results of (a) and (b). Can you explain why they agree or disagree? Which is correct??

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Solution

Solutions - 5160-03

(a)

Compute probability the wait time for a Bundle is at most 1 hr:

$$\begin{array}{c}P[X\le 1]={\int }_0^{1}2e^{-2x}dx\\ \\ \gg \gg {[-e^{-2t}]|}_0^1\gg \gg \approx 0.8647\end{array}$$

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(b)

Erlang distribution:

$$f_Y(y)=\frac{{\lambda }^{\ell }}{(\ell -1)!}{y}^{\ell -1}{e}^{-\lambda y},y\ge 0$$

Desired probability:

$$\begin{array}{c}P[Y\le 1]={\int }_0^1\frac{{10}^5}{4!}{y}^4{e}^{-10y}dy\\ \\ \gg \gg \approx 0.9707\end{array}$$

This integral requires iterated IBP.

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(c)

The results disagree. Method 2 is the correct approach assuming individual calls arrive according to a Poisson process.

It turns out that if you have some Poisson process, bundles of arrivals do not themselves follow a Poisson process.

Think of a Poisson process as some dots scattered on a timeline. Place a heavy dot at the location of each $5^{\text{th}}$ dot in order, and erase all other dots. These heavy dots do not follow a Poisson process.

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04

01

Mean and variance of exponential

Show that $E[X]=\frac{1}{\lambda }$ and $\mathrm{Var}[X]=\frac{1}{{\lambda }^2}$ for $X\sim \mathrm{Exp}(\lambda )$.

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Solution

Solutions - 5160-01

(1) State the PDF of an exponential distribution:

$$f_X(x)=\lambda e^{-\lambda x},x>0$$

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(2) Compute $E[X]$ using the integral formula:

$$\begin{array}{c}E[X]={\int }_0^{\infty }\lambda x{e}^{-\lambda x}dx\\ \\ \gg \gg \underset{b\to \infty }{\lim }{[-x{e}^{-\lambda x}-\frac{e^{-\lambda x}}{\lambda }]|}_0^b\gg \gg \frac{1}{\lambda }\end{array}$$

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(3) Compute $E\left[X^2\right]$ using the integral formula:

$$\begin{array}{c}E\left[X^2\right]={\int }_0^{\infty }\lambda x^2{e}^{-\lambda x}dx\\ \\ \gg \gg \underset{b\to \infty }{\lim }{[-x^2{e}^{-\lambda x}-\frac{2x{e}^{-\lambda x}}{\lambda }-\frac{2e^{-\lambda x}}{{\lambda }^2}]|}_0^b\gg \gg \frac{2}{{\lambda }^2}\end{array}$$

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

$$\begin{array}{c}\text{Var}[X]=E\left[X^2\right]-{\left(E[X]\right)}^2\\ \\ \gg \gg \frac{2}{{\lambda }^2}-\frac{1}{{\lambda }^2}\gg \gg \frac{1}{{\lambda }^2}\end{array}$$Link to original

Derived random variables

05 $★$

05

Expectation from CDF

The CDF of random variable X is given by:

$$F_X(x)=\{\begin{array}{cc}0& x<-1\\ {\displaystyle \frac{x+1}{2}}& -1\le x<0\\ 1/2& 0\le x<2\\ {\displaystyle \frac{x-1}{2}}& 2\le x<3\\ 1& 3\le x\end{array}$$

Compute $E[X]$ and $\mathrm{Var}[X]$ (use the shorter formula).

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Solution

Solutions - 5120-05

(a)

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

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(b)

First find expectation of $X$:

$$\begin{array}{c}E[X]={\int }_{-\infty }^{\infty }xf(x)dx\\ \\ \gg \gg {\int }_{-1}^0\frac{1}{2}xdx+{\int }_2^3\frac{1}{2}xdx\\ \\ \gg \gg {\frac{1}{4}{x}^2|}_{-1}^0+{\frac{1}{4}{x}^2|}_2^3\\ \\ \gg \gg 0-\frac{1}{4}+\frac{1}{4}(9-4)\gg \gg 1\end{array}$$

To use the variance formula $\mathrm{Var}[X]=E[X^2]-E{[X]}^2$, we also need to find $E[X^2]$. For this we use $g(X)=X^2$ and the formula for $E[g(X)]$:

$$\begin{array}{c}E[X^2]={\int }_{-\infty }^{+\infty }{x}^{2}f(x)dx\\ \\ \gg \gg {\int }_{-1}^0\frac{1}{2}{x}^{2}dx+{\int }_2^3\frac{1}{2}{x}^{2}dx\\ \\ \gg \gg \frac{1}{6}{x}^3{|}_{-1}^0+\frac{1}{6}{x}^3{|}_2^3\\ \\ \gg \gg (0-\frac{1}{6}(-1{)}^3)+(\frac{1}{6}(3{)}^3-\frac{1}{6}(2{)}^3)\\ \\ \gg \gg \frac{1}{6}+\frac{19}{6}\gg \gg \frac{20}{6}\end{array}$$

Therefore:

$$\begin{array}{c}\mathrm{Var}[X]=E[X^2]-E{[X]}^2\\ \\ \gg \gg \frac{20}{6}-{(1)}^2\gg \gg \frac{7}{3}\end{array}$$Link to original

06

03

PDF of derived variable for $E[X]$ and $\mathrm{Var}[X]$

Suppose the PDF of an RV is given by:

$$f_X(x)=\{\begin{array}{cc}\frac{3}{4}x(2-x)& 0\le x\le 2\\ 0& \text{otherwise}\end{array}$$

(a) Find $E[X]$ using the integral formula.

(b) Find $f_{X^2}(x)$, the PDF of $X^2$ (by calculating the CDF first).

(c) Find $E[X^2]$ using $f_{X^2}(x)$.

(d) Find $\mathrm{Var}[X]$ using results of (a) and (c).

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Solution

Solutions - 5150-03

(a)

Compute $E[X]$:

$$\begin{array}{c}E[X]={\int }_0^2\frac{3}{4}{x}^2(2-x)dx\\ \\ \gg \gg {[\frac{x^3}{2}-\frac{3}{16}{x}^4]|}_0^2\gg \gg 1\end{array}$$

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(b)

(1) Find the CDF of $X$:

$$\begin{array}{c}{F}_X(x)={\int }_{-\infty }^x{f}_X(t)dt\\ \\ \gg \gg {\int }_0^x\frac{3}{4}t(2-t)dt\gg \gg {[\frac{3}{4}{t}^2-\frac{t^3}{4}]|}_0^x\\ \\ \gg \gg \frac{3}{4}{x}^2-\frac{1}{4}{x}^3\end{array}$$

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(2) Find the CDF of $X^2$:

Since $X^2$ is monotone increasing, $F_{X^2}(x)=F_X\left(\sqrt{x}\right)$.

$$F_X\left(\sqrt{x}\right)=\{\begin{array}{cc}0& x<0\\ \frac{3}{4}x-\frac{1}{4}{x}^{\frac{3}{2}}& 0\le x<4\\ 1& 4\le x\end{array}$$

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(3) Find the PDF of $X^2$ by differentiating:

$$f_{X^2}(x)=\{\begin{array}{cc}0& x<0\\ \frac{3}{4}-\frac{3}{8}{x}^{\frac{1}{2}}& 0\le x<4\\ 0& 4\le x\end{array}$$

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(c)

Find $E\left[X^2\right]$:

$$\begin{array}{c}E\left[X^2\right]={\int }_0^{4}x(\frac{3}{4}-\frac{3}{8}{x}^{\frac{1}{2}})dx\\ \\ \gg \gg {[\frac{3}{8}{x}^2-\frac{3}{20}{x}^{\frac{5}{2}}]|}_0^4\gg \gg \frac{6}{5}\end{array}$$

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(d)

Compute $\text{Var}[X]$:

$$\begin{array}{c}\text{Var}[X]=E\left[X^2\right]-{\left(E[X]\right)}^2\\ \\ \gg \gg \frac{6}{5}-1\gg \gg \frac{1}{5}\end{array}$$Link to original

Review problems

07

03

Rolling until a six

A fair die is rolled until a six comes up.

What are the odds that it takes at least 10 rolls? (Use a geometric random variable.)

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Solution

Solutions - 5110-03

(1) Define random variable:

Let $X\sim \text{Geom}\left(\frac{1}{6}\right)$.

We wish to find $P[X\ge 10]$.

For all $X=k$, $k\ge 10$, the first $k-1$ trials result in failure, and the $k^{\text{th}}$ trial is a success.

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(2) Compute probability:

$$\begin{array}{c}P[X\ge 10]\gg \gg \sum _{k=10}^{\infty }\frac{1}{6}{(1-\frac{1}{6})}^{k-1}\\ \\ \gg \gg \frac{1}{6}\sum _{k=10}^{\infty }{\left(\frac{5}{6}\right)}^{k-1}\gg \gg \frac{1}{6}\left(\frac{{\left(\frac{5}{6}\right)}^9}{1-\frac{5}{6}}\right)\\ \\ \gg \gg {\left(\frac{5}{6}\right)}^9\gg \gg \approx 0.1938\end{array}$$

Note that the summation is simplified using the formula for a geometric series.

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