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W05 - HW Solutions

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Poisson process

01 - Poisson satisfies

Show that a Poisson variable satisfies the total probability rule for a CDF, namely that .

Solution

  1. & State CDF of a Poisson distribution.
    • We know that
    • We know that .
  2. & Compute limit as .
    • Note that

02 - Expectation of Poisson

Derive the formula for a Poisson variable .

Solution

  1. & State PMF of a Poisson distribution.
  2. & Find expectation.
    • We know that .

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 that over a four hour evening, 13 meteors were spotted. What is the probability that none of them happened in the first hour?

Solution (a)

  1. & Write explanation.
    • You use Poisson distribution if events occur randomly and you know the mean number of events that occur within a given interval of time.
    • In addition, Poisson distributions are advantageous when describing rare events. Since meteors are a rare occurrence, it makes sense to use a Poisson distribution.

(b)

  1. & Compute probability.
    • Since , it’s easier to compute the latter.

(c)

  1. & Compute probability.
    • We know that there are 13 meteors in 4 hours, so we see an average of meteors per hour. Let
    • We wish to find the probability .

04 - Silver dimes

Suppose 1 our of 350 dimes in circulation is made of silver. Consider a tub of dimes worth 40$.

  • (a) Find a formula for the exact probability that this collection contains at least 2 silver dimes. Can your calculator evaluate this formula?
  • (b) Estimate the probability in question using a Poisson approximation.

Solution (a)

  1. & Identify distribution.
    • Clearly, this scenario follows a binary distribution.
    • We have a chance that the dime is made of silver.
    • Since we have 40$ worth of dimes, there are 400 dimes.
    • Thus, .
  2. & Find formula for probability.

(b)

  1. & Find corresponding Poisson distribution.
    • Let .
    • .
  2. & Compute probability.
    • We have that .

05 - Applications of Poisson approximation of binomial

Let and consider the Poisson approximation to .

  • (a) Estimate the possible error of the approximation (for an arbitrary probability).
  • (b) Compute the exact error of the approximation for the specific value .

Solution (a)

  1. & Define random variable that is the Poisson approximation to .
    • .
  2. & Estimate error.

(b)

  1. & Compute using the binomial distribution.
    • We have that .
  2. & Compute using the Poisson distribution.
  3. & Compute error.

06 - Constants in PDF from expectation.

Suppose has PDF given by Suppose . Find the only possible values for and . Then, find .

Solution

  1. & Recall formula for expectation of a continuous random variable.
  2. & Use formula to find an equation relating and .
Misplaced &\int_{0}^{1}x\left(a+bx^{2}\right)dx &= \frac{7}{10} \\ \left.\left[\frac{ax^{2}}{2} + \frac{bx^{4}}{4}\right]\right|_{0}^{1} &= \frac{7}{10} \\ \frac{a}{2} + \frac{b}{4} = \frac{7}{10} \end{align*} $$ 3. & Recall that integrating a PDF should yield $1$. Integrate the PDF and find a second equation relating $a$ and $b$. $$\begin{align*}\int_{0}^{1}a+bx^{2}dx &= 1 \\ \left.\left[ax + \frac{bx^{3}}{3}\right]\right|_{0}^{1} &= 1 \\ a + \frac{b}{3} &= 1\end{align*} $$ 3. & Solve system of equations for $a$ and $b$. - Isolating $a$ in the second equation yields $a = 1 - \frac{b}{3}$. - Plugging this expression into the first equation yields $\frac{1- \frac{b}{3}}{2} + \frac{b}{4} = \frac{7}{10}$ . - Solving for $b$ yields 2.4, and thus $a = 0.2$. 4. & Compute variance using the formula $\text{Var}[X] = E\left[X^{2}\right] - \left(E[X]\right)^{2}$ - Compute $E\left[X^{2}\right]$. $$E\left[X^{2}\right] = \int_{0}^{1}x^{2}\left(a+bx^{2}\right)dx = \left.\left[\frac{ax^{3}}{3} + \frac{bx^{5}}{5}\right]\right|_{0}^{1} = \frac{a}{3} + \frac{b}{5} = \frac{41}{75}$$ - $\text{Var}[X] = \frac{41}{75} - \left(\frac{7}{10}\right)^{2} = \frac{17}{300}$. #### 07 - Variance: Direct integral formula Suppose $X$ has PDF given by: $$f_{X}(x) = \begin{cases} 3e^{-3x} & x \geq 0 \\ 0 & \text{otherwise} \end{cases} $$ Find $\text{Var}[X]$ using the integral formula. **Solution** 1. & Recall the integral formula for variance. - Use the fact that $\text{Var}[X] = E\left[(X - \mu)^{2}\right]$ $$\text{Var}[X] = \int_{-\infty}^{\infty} (x - \mu)^{2}f_{X}(x)dx$$ 2. & Compute $\mu = E[X]$. $$E[X] = \int_{0}^{\infty}3xe^{-3x}dx = \lim_{b \to \infty}\left.\left[-xe^{-3x}-\frac{e^{-3x}}{3}\right]\right|_{0}^{b} = \frac{1}{3}$$ 1. && Compute $\text{Var}[X]$ $$\text{Var}[X] = \int_{0}^{\infty}\left(x - \frac{1}{3}\right)^{2}3e^{-3x}dx = \lim_{b \to \infty}\int_{0}^{b}3x^{2}e^{-3x}-2xe^{-3x}+\frac{e^{-3x}}{3}dx$$ $$= \lim_{b\to\infty}\left.\left[-x^{2}e^{-3x}-\frac{e^{-3x}}{9}\right]\right|_{0}^{b} = \frac{1}{9}$$ #### 08 - PDF of derived variable for $E[X]$ and $\text{Var}[X]$ Suppose the PDF of an RV is given by $$f_{X}(x) = \begin{cases} \frac{3}{4}x(2-x) & 0 \leq x \leq 2 \\ 0 & \text{otherwise} \end{cases} $$ - (a) Find $E[X]$ using the integral formula. - (b) Find $f_{X^{2}}(x)$, the PDF of $X^{2}$ (by calculating the CDF first). - (c) Find $E\left[X^{2}\right]$ using $f_{X^{2}}(x)$. - (d) Find $\text{Var}[X]$ using results of (a) and (c). **Solution** **(a)** 1. & Compute $E[X]$. $$E[X] = \int_{0}^{2}\frac{3}{4}x^{2}(2-x)dx = \left.\left[\frac{x^{3}}{2}-\frac{3}{16}x^{4}\right]\right|_{0}^{2} = 1$$ **(b)** 1. & Find the CDF of $X$. $$F_{X}(x) = \int_{-\infty}^{x}f_{X}(t)dt = \int_{0}^{x}\frac{3}{4}t(2-t)dt = \left.\left[\frac{3}{4}t^{2}-\frac{t^{3}}{4}\right]\right|_{0}^{x} = \frac{3}{4}x^{2}-\frac{1}{4}x^{3}$$ 2. Find the CDF of $X^{2}$. - Since $X^{2}$ is monotone increasing, $F_{X^{2}}(x) = F_{X}\left(\sqrt{x}\right)$. $$F_{X}\left(\sqrt{x}\right) = \begin{cases} 0 & x < 0 \\ \frac{3}{4}x - \frac{1}{4}x^{\frac{3}{2}} & 0 \leq x < 4 \\ 1 &4 \leq x \end{cases} $$ 3. & Find the PDF of $X^{2}$ by differentiating. $$f_{X^{2}}(x) = \begin{cases} 0 & x < 0 \\ \frac{3}{4} - \frac{3}{8}x^{\frac{1}{2}} & 0 \leq x < 4 \\ 0 & 4 \leq x \end{cases} $$ **(c)** 1. & Find $E\left[X^{2}\right]$. $$E\left[X^{2}\right] = \int_{0}^{4}x\left(\frac{3}{4}-\frac{3}{8}x^{\frac{1}{2}}\right)dx = \left.\left[\frac{3}{8}x^{2}-\frac{3}{20}x^{\frac{5}{2}}\right]\right|_{0}^{4} = \frac{6}{5}$$ **(d)** 1. & Compute $\text{Var}[X]$. $$\text{Var}[X] = E\left[X^{2}\right] - \left(E[X]\right)^{2} = \frac{6}{5} - 1 = \frac{1}{5}$$ #### 09 - Mean and variance of exponential Show that $E[X] = \frac{1}{\lambda}$ and $\text{Var}[X] = \frac{1}{\lambda^{2}}$ for $X \sim \text{Exp}(\lambda)$. **Solution** 1. & State the PDF of an exponential distribution. $$f_{X}(x) = \lambda e^{-\lambda x}, \qquad x > 0 $$ 2. && Compute $E[X]$ using the integral formula. $$E[X] = \int_{0}^{\infty}\lambda x e^{-\lambda x}dx = \lim_{b \to \infty}\left.\left[- xe^{-\lambda x} - \frac{e^{-\lambda x}}{\lambda}\right]\right|_{0}^{b} = \frac{1}{\lambda}$$ 3. && Compute $E\left[X^{2}\right]$ using the integral formula. $$E\left[X^{2}\right] = \int_{0}^{\infty}\lambda x^{2}e^{-\lambda x}dx = \lim_{b \to \infty}\left.\left[- x^{2}e^{-\lambda x} - \frac{2 xe^{-\lambda x}}{\lambda} - \frac{2e^{-\lambda x}}{\lambda^{2}} \right]\right|_{0}^{b} = \frac{2}{\lambda^{2}}$$ 4. & Compute $\text{Var}[X]$ $$\text{Var}[X] = E\left[X^{2}\right]- \left(E[X]\right)^{2} = \frac{2}{\lambda^{2}} - \frac{1}{\lambda^{2}} = \frac{1}{\lambda^{2}}$$ #### 10 - Vehicle lifetimes Suppose that vehicle lifetimes follow an exponential distribution with an expected lifetime of 10 years. Suppose you have one car that is 5 years old, and one that is 15 years old. What is the probability that the first car outlives the second? **Solution** 1. & Recall the memoryless property of exponential distributions. - Elapsed time has no effect on future events. - Therefore, the fact that one car is older than the other has no effect on the remaining lifetimes. 2. & Derive conclusions. - Since both cars have the same remaining lifetime distribution, the probability that either car outlives the other is 0.5. #### 11 - Wait time for 5 calls - two methods Consider the Poisson process of phone calls coming to a call center at an average rate of 1 call every 6 minutes. Let us model the wait time for 5 calls to come in. - (a) Method One: An arrival of '1-call' comes in at an average rate of $\lambda = 10$ calls per hour. So, a Bundle of '5-calls' comes in at an average wait of $\lambda_{B} = 2$ Bundles per hour. Use an exponential variable with $\lambda_{B} = 2$ to determine the probability that the wait time for a Bundle (of 5 calls) is at most 1 hr. - (b) Method Two: Use $\lambda = 10$ calls per hour with an Erlang distribution at $\ell = 5$ to determine the probability that the wait time for 5 calls is at most 1 hr. - (c) Compare the results of (a) and (b). Can you explain why they agree or disagree? Which is correct? **Solution** **(a)** 1. & Compute probability the wait time for a Bundle is at most 1 hr. - Our bounds will be from 0 to 1 since we are only concerned about 1 hour. $$P[X \leq 1] = \int_{0}^{1}2e^{-2x}dx = \left.\left[-e^{-2t}\right]\right|_{0}^{1} \approx 0.8647$$ **(b)** 1. & State the Erlang distribution. $$f_{Y}(y) = \frac{\lambda^{\ell}}{(\ell - 1)!}y^{\ell - 1}e^{-\lambda y}, \qquad y \geq 0$$ 2. & Compute desired probability. $$P[Y \leq 1] = \int_{0}^{1}\frac{{10}^{5}}{4!}y^{4}e^{-10y}dy \approx 0.9707$$ **(c)** 1. State conclusions. - Clearly, the results disagree. This is because method 1 considers calls coming in at bundles at a time instead of considering 5 discrete calls. Method 2 is more accurate since it considers the rates of individual calls.