Solved Problems | Special Continuous Distributions

4.2.6 Solved Problems:
Special Continuous Distributions

Problem

Suppose the number of customers arriving at a store obeys a Poisson distribution with an average of $\lambda$ customers per unit time. That is, if $Y$ is the number of customers arriving in an interval of length $t$ , then $Y \sim Poisson (\lambda t)$ . Suppose that the store opens at time $t=0$ . Let $X$ be the arrival time of the first customer. Show that $X \sim Exponential(\lambda)$ .

Solution

We first find $P(X > t)$ :

$P(X > t)$	$=P(\textrm{No arrival in }[0,t])$
	$=e^{-\lambda t}\frac{(\lambda t)^0}{0!}$
	$=e^{-\lambda t}$ .

Thus, the CDF of

$X$ for

$x>0$ is given by

$F_X(x)=1-P(X>x)=1-e^{-\lambda x},$ which is the CDF of

$Exponential(\lambda)$ . Note that by the same argument, the time between the first and second customer also has

$Exponential(\lambda)$ distribution. In general, the time between the

$k$ 'th and

$k+1$ 'th customer is

$Exponential(\lambda)$ .

Problem (Exponential as the limit of Geometric)

Let $Y \sim Geometric(p)$ , where $p=\lambda \Delta$ . Define $X=Y \Delta$ , where $\lambda, \Delta>0$ . Prove that for any $x \in (0,\infty)$ , we have $\lim_{\Delta \rightarrow 0} F_X(x)=1-e^{-\lambda x}.$

Solution

If $Y \sim Geometric(p)$ and $q=1-p$ , then

$P(Y \leq n)$	$=\sum_{k=1}^{n} pq^{k-1}$
	$=p. \frac{1-q^{\large{n}}}{1-q}=1-(1-p)^n$ .

Then for any

$y \in (0,\infty)$ , we can write

$P(Y \leq y) =1-(1-p)^{\lfloor y\rfloor},$ where

$\lfloor y\rfloor$ is the largest integer less than or equal to

$y$ . Now, since

$X=Y \Delta$ , we have

$F_X(x)$	$=P(X \leq x)$
	$=P\left(Y \leq \frac{x}{\Delta}\right)$
	$=1-(1-p)^{\lfloor \frac{\Large{x}}{\Delta} \rfloor}=1-(1-\lambda \Delta)^{\lfloor \frac{\Large{x}}{\Delta} \rfloor}$ .

Now, we have

$\lim_{\Delta \rightarrow 0} F_X(x)$	$=\lim_{\Delta \rightarrow 0} 1-(1-\lambda \Delta)^{\lfloor \frac{\Large{x}}{\Delta} \rfloor}$
	$=1-\lim_{\Delta \rightarrow 0} (1-\lambda \Delta)^{\lfloor \frac{\Large{x}}{\Delta} \rfloor}$
	$=1-e^{-\lambda x}$ .

The last equality holds because

$\frac{x}{\Delta}-1 \leq \lfloor \frac{x}{\Delta} \rfloor \leq \frac{x}{\Delta}$ , and we know

$\lim_{\Delta \rightarrow 0^{+}} (1-\lambda \Delta)^{\frac{1}{\Delta}}=e^{-\lambda}.$

Problem

Let $U \sim Uniform(0,1)$ and $X=-\ln (1-U)$ . Show that $X \sim Exponential(1)$ .

Solution

First note that since $R_U=(0,1)$ , $R_X=(0,\infty)$ . We will find the CDF of $X$ . For $x \in(0,\infty)$ , we have

$F_X(x)$	$=P(X \leq x)$
	$=P(-\ln (1-U) \leq x)$
	$=P\left(\frac{1}{1-U} \leq e^x\right)$
	$=P(U \leq 1-e^{-x})=1-e^{-x}$ ,

which is the CDF of an

$Exponential(1)$ random variable.

Problem

Let $X \sim N(2,4)$ and $Y=3-2X$ .

Find $P(X > 1)$ .
Find $P(-2 < Y < 1)$ .
Find $P(X > 2 |Y < 1)$ .

Solution

Find

$P(X > 1)$ : We have

$\mu_X=2$ and

$\sigma_X=2$ . Thus,

$P(X > 1)$	$=1-\Phi\left(\frac{1-2}{2}\right)$
	$=1-\Phi(-0.5)=\Phi(0.5)=0.6915$

Find

$P(-2 < Y < 1)$ : Since

$Y=3-2X$ , using Theorem 4.3, we have

$Y \sim N(-1,16)$ . Therefore,

$P(-2 < Y < 1)$	$=\Phi\left(\frac{1-(-1)}{4}\right)-\Phi\left(\frac{(-2)-(-1)}{4}\right)$
	$=\Phi(0.5)-\Phi(-0.25)=0.29$

Find

$P(X > 2 |Y < 1)$ :

$P(X > 2 \|Y < 1)$	$=P(X > 2 \|3-2X < 1)$
	$=P(X > 2 \|X > 1)$
	$=\frac{P(X > 2,X > 1)}{P(X > 1)}$
	$=\frac{P(X > 2)}{P(X > 1)}$
	$=\frac{1-\Phi(\frac{2-2}{2})}{1-\Phi(\frac{1-2}{2})}$
	$=\frac{1-\Phi(0)}{1-\Phi(-0.5)}$
	$\approx 0.72$

Problem

Let $X \sim N(0,\sigma^2)$ . Find $E|X|$

Solution

We can write $X=\sigma Z$ , where $Z \sim N(0,1)$ . Thus, $E|X|=\sigma E|Z|$ . We have

$E\|Z\|$	$=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} \|t\| e^{-\frac{t^2}{2}}dt$
	$=\frac{2}{\sqrt{2\pi}}\int_{0}^{\infty} \|t\| e^{-\frac{t^2}{2}}dt \hspace{20pt}(\textrm{integral of an even function})$
	$=\sqrt{\frac{2}{\pi}}\int_{0}^{\infty} t e^{-\frac{t^2}{2}}dt$
	$=\sqrt{\frac{2}{\pi}}\bigg[-e^{-\frac{t^2}{2}} \bigg]_{0}^{\infty}=\sqrt{\frac{2}{\pi}}$

Thus, we conclude

$E|X|=\sigma E|Z|=\sigma\sqrt{\frac{2}{\pi}}$ .

Problem

Show that the constant in the normal distribution must be $\frac{1}{\sqrt{2 \pi}}$ . That is, show that $I=\int_{-\infty}^{\infty} e^{-\frac{x^2}{2}}dx=\sqrt{2 \pi}.$ Hint: Write $I^2$ as a double integral in polar coordinates.

Solution

Let $I=\int_{-\infty}^{\infty} e^{-\frac{x^2}{2}}dx$ . We show that $I^2=2\pi$ . To see this, note

$I^2$	$= \int_{-\infty}^{\infty} e^{-\frac{x^2}{2}}dx \int_{-\infty}^{\infty} e^{-\frac{y^2}{2}}dy$
	$=\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-\frac{x^2+y^2}{2}}dxdy$ .

To evaluate this double integral we can switch to polar coordinates. This can be done by change of variables

$x=r \cos \theta, y=r \sin \theta$ , and

$dx dy=rdrd\theta$ . In particular, we have

$I^2$	$=\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-\frac{x^2+y^2}{2}}dxdy$
	$=\int_{0}^{\infty} \int_{0}^{2\pi} e^{-\frac{r^2}{2}}r d\theta dr$
	$=2 \pi \int_{0}^{\infty} r e^{-\frac{r^2}{2}} dr$
	$=2 \pi \bigg[-e^{-\frac{r^2}{2}}\bigg]_{0}^{\infty}=2 \pi$ .

Problem

Let $Z \sim N(0,1)$ . Prove for all $x \geq 0$ , $\frac{1}{\sqrt{2\pi}} \frac{x}{x^2+1} e^{-\frac{x^2}{2}} \leq P(Z \geq x) \leq \frac{1}{\sqrt{2\pi}} \frac{1}{x} e^{-\frac{x^2}{2}}.$

Solution

To show the upper bound, we can write

$P(Z \geq x)$	$= \frac{1}{\sqrt{2 \pi}} \int_{x}^{\infty}e^{-\frac{u^2}{2}} du$
	$\leq \frac{1}{\sqrt{2 \pi}} \int_{x}^{\infty} \frac{u}{x} e^{-\frac{u^2}{2}} du \hspace{10pt} (\textrm{since $u \geq x > 0$})$
	$= \frac{1}{\sqrt{2 \pi}}\frac{1}{x} \left[-e^{-\frac{u^2}{2}}\right]^{\infty}_{x}$
	$=\frac{1}{\sqrt{2\pi}} \frac{1}{x} e^{-\frac{x^2}{2}}$ .

To show the lower bound, let

$Q(x)=P(Z \geq x)$ , and

$h(x)=Q(x)-\frac{1}{\sqrt{2\pi}} \frac{x}{x^2+1} e^{-\frac{x^2}{2}}, \hspace{10pt} \textrm{ for all }x \geq 0.$ It suffices to show that

$h(x)\geq 0, \hspace{10pt} \textrm{ for all }x \geq 0.$ To see this, note that the function

$h$ has the following properties

$h(0)=\frac{1}{2}$ ;
$\lim \limits_{x \rightarrow \infty} h(x)=0$ ;
$h'(x)=-\frac{2}{\sqrt{2\pi}}\left( \frac{e^{-\frac{x^2}{2}}}{(x^2+1)^2}\right) < 0$ , for all $x \geq 0$ .

Therefore,

$h(x)$ is a strictly decreasing function that starts at

$h(0)=\frac{1}{2}$ and decreases as

$x$ increases. It approaches

$0$ as

$x$ goes to infinity. We conclude that

$h(x) \geq 0$ , for all

$x \geq 0$ .

Problem

Let $X \sim Gamma(\alpha,\lambda)$ , where $\alpha, \lambda \gt 0$ . Find $EX$ , and $Var(X)$ .

Solution
- To find $EX$ we can write $\begin{align*} EX &= \int_0^\infty x f_X(x) dx \\ &= \int_0^\infty x \cdot \frac{\lambda^{\alpha}}{\Gamma{\alpha}} x^{\alpha - 1} e^{-\lambda x} {\rm d}x \\ &= \frac{\lambda^{\alpha}}{\Gamma(\alpha)} \int_0^{\infty} x \cdot x^{\alpha - 1} e^{-\lambda x} {\rm d}x \\ &= \frac{\lambda^{\alpha}}{\Gamma(\alpha)} \int_0^{\infty} x^{\alpha} e^{-\lambda x} {\rm d}x \\ &= \frac{\lambda^{\alpha}}{\Gamma(\alpha)} \frac{\Gamma(\alpha + 1)}{\lambda^{\alpha + 1}} &\textrm{(using Property 2 of the gamma function)} \\ &= \frac{\alpha\Gamma(\alpha)}{\lambda\Gamma(\alpha)} &\textrm{(using Property 3 of the gamma function)} \\ &= \frac{\alpha}{\lambda}. \end{align*}$
  
  Similarly, we can find $EX^2$ : $\begin{align*} EX^2 &= \int_0^{\infty} x^2 {\rm d}x \\ &= \int_0^{\infty} x^2 \cdot \frac{\lambda^{\alpha}}{\Gamma(\alpha)} x^{\alpha - 1} e^{-\lambda x} {\rm d}x \\ &= \frac{\lambda^{\alpha}}{\Gamma(\alpha)} \int_0^{\infty} x^2 \cdot x^{\alpha - 1} e^{-\lambda x} {\rm d}x \\ &= \frac{\lambda^{\alpha}}{\Gamma(\alpha)} \int_0^{\infty} x^{\alpha + 1} e^{-\lambda x} {\rm d}x \\ &= \frac{\lambda^{\alpha}}{\Gamma(\alpha)} \frac{\Gamma(\alpha + 2)}{\lambda^{\alpha + 2}} &\textrm{(using Property 2 of the gamma function)} \\ &= \frac{(\alpha + 1)\Gamma(\alpha + 1)}{\lambda^2 \Gamma(\alpha)} &\textrm{(using Property 3 of the gamma function)} \\ &= \frac{(\alpha + 1) \alpha \Gamma(\alpha)}{\lambda^2 \Gamma(\alpha)} &\textrm{(using Property 3 of the gamma function)} \\ &= \frac{\alpha (\alpha + 1)}{\lambda^2}. \end{align*}$
  
  So, we conclude $\begin{align*} Var(X) &= EX^2 - (EX)^2 \\ &= \frac{\alpha (\alpha + 1)}{\lambda^2} - \frac{\alpha^2}{\lambda^2} \\ &= \frac{\alpha}{\lambda^2}. \end{align*}$

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$E\|Z\|$	$=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} \|t\| e^{-\frac{t^2}{2}}dt$
	$=\frac{2}{\sqrt{2\pi}}\int_{0}^{\infty} \|t\| e^{-\frac{t^2}{2}}dt \hspace{20pt}(\textrm{integral of an even function})$
	$=\sqrt{\frac{2}{\pi}}\int_{0}^{\infty} t e^{-\frac{t^2}{2}}dt$
	$=\sqrt{\frac{2}{\pi}}\bigg[-e^{-\frac{t^2}{2}} \bigg]_{0}^{\infty}=\sqrt{\frac{2}{\pi}}$

4.2.6 Solved Problems: Special Continuous Distributions

4.2.6 Solved Problems:
Special Continuous Distributions