Q. 8.23 - A First Course in Probability

Question

Let $X$be a Poisson random variable with a mean of$20$.

$\left(a\right)$Use the Markov inequality to obtain an upper bound$p=P(X\ge 26)$.

$\left(b\right)$Use the one-sided Chebyshev inequality to obtain an upper bound$p$.

$\left(c\right)$Use the Chernoff bound to obtain an upper bound$p$.

$\left(d\right)$ Approximate $p$by making use of the central limit theorem.

$\left(e\right)$Determine $p$by running an appropriate program.

Step-by-Step Solution

Answer

$\left(a\right)$ $p=P\{X\ge 26\}\le .7692$

 $\left(b\right)$  $p=P\{X\ge =26\}\le .3571$

$\left(c\right)$  $p=P\{X\ge 26\}\le \mathbf{.}\mathbf{4398}$

 $\left(d\right)$  $p=P\{X\ge 26\}\approx .\mathbf{1093}$

 $\left(e\right)$ $p=P\{X\ge 26\}=\mathbf{0}\mathbf{.}\mathbf{112185}$

Step 1 Given information.

Let $X$be a Poisson random variable with a mean of$20$.

Step 2 Explanation.

Suppose that $X$is a Poisson random variable with a parameter$\lambda$. It is given that $X$is a random variable with a mean of$20$. Since the expected value and variance of a Poisson random variable are both equal to its parameter$\lambda ,X$ has mean $\mu =\lambda =20$and variance${\sigma }^2=\lambda =20$.

Step 3 Part (a) Explanation.

By Markov's inequality,

$p=P\{X\ge 26\}\le \frac{E\left[X\right]}{26}=\frac{20}{26}=.7692$

Step 4 Part (b) Explanation.

Using Corollary$5.1$, $\underset{¯}{a=6}$we get:

$p=P\{X\ge \underset{=26}{\underbrace{20+6}}\}\le \frac{{\sigma }^2}{{\sigma }^2+6^2}=\frac{20}{20+36}=.\mathbf{3571}$

Step 5 Part (c) Explanation.

 See Example$5d$. Since $X$is a Poisson random variable with parameter$\lambda =20$ using the result

$P\{X\ge i\}\le \frac{e^{-\lambda }(e\lambda {)}^i}{i^i}$

We get,

$p=P\{X\ge 26\}\le \frac{e^{-20}(20e{)}^{26}}{{26}^{26}}=.4398$

Step 6 Part (d) Explanation.

Using the central limit theorem we get:

$p=P\{X\ge 26\}\stackrel{\text{ the continuity correction }}{=}P(X\ge 25.5)$

$=1-P\{X<25.5\}=1-P\left\{\frac{X-20}{\sqrt{20}}<\frac{25.5-20}{\sqrt{20}}\right\}\approx 1-\Phi (1.23)$

$\text{ Table 5.1 (textbook, Chapter 5) }1-.8907=.\mathbf{1093}\text{. }$

Step 7 Part (e) Explanation.

Using software package by personal choice, we obtain:

$p=P\{X\ge 26\}=1-P\{X<26\}=1-\sum _{i=0}^{25}{e}^{-20}\frac{{20}^i}{i!}$

$=1-.887815=\mathbf{0}\mathbf{.}\mathbf{112185}$