Q.7.61 - A First Course in Probability

Question

Let $X1, ...$be independent random variables with the common distribution function$F$, and suppose they are independent of $N$, a geometric random variable with a parameter $p$. Let $M = max(X1, ... , XN).$

(a) Find$P\{M\dots x\}$by conditioning on$N$. 

(b) Find$P\{M\dots x|N=1\}$.

(c) Find$P\{M\dots x|N>1\}$ 

(d) Use (b) and (c) to rederive the probability you found in (a) 

Step-by-Step Solution

Answer

a)$P\{M\le x\}=\frac{NF\left(e\right)}{1-(\mid P)F\left(x\right)}$

b)$P\{M\le x\mid N=1\}=F\left(x\right)$

c)$P\{M\le x\mid N>1\}=F\left(x\right)P\{M\le x\}$

d) The probability of part (a) is rederived by using (b) and (c) as $P\{M\le x\}=\frac{F\left(x\right)p}{1-(1-p)F\left(x\right)}$

Step 1:Given Information(part a)

Given that the distribution function$F$, and suppose they are independent of $N$, a geometric random variable with parameter$p$. Let$M = max(X1, ... , XN).$

Step 2:Explanation(part a)

Discover by $p\{M...x\}$conditioning on $N$

$P\{M\le x\}=\sum _{n=1}^{\mathrm{\infty }} P\{M\le x\mid N=n\}P\{N=n\}$

$Sn \backslash displaystyle\backslash \{sum\_\backslash \{n=1\backslash \}*\backslash \{\backslash infty\backslash \} \$F^\{\backslash wedge\}\backslash \{n\backslash \left\}\right(x) p(1-p)^\{*\}\backslash \{n-1\backslash \} s\$$

$\mathrm{\$}=\mathrm{\setminus }\mathrm{frac}\left\{pF\right(x\left)\right\}\{1-(1-p\left)F\right(x\left)\right\}\mathrm{\$}$

Step 3:Final Answer(part a)

$p\{M...x\}$by conditioning on $N$ is

$P\{M\le x\}=\sum _{n=1}^{\mathrm{\infty }} P\{M\le x\mid N=n\}P\{N=n\}$

Step 4:Given Information(part b)

Given that $M=max(X1,...,XN)$ and the parameter $p$.

Step 5:Explanation(part b)

Discover $P\{M\le x\mid N=1\}$

We have

$P\{M\le x\mid N=1\}=F\left(x\right)$

Step 6:Final Answer(part b)

$P\{M\le x\mid N=1\}=F\left(x\right)$

Step 7:Given Information(part c)

Given that $M\le x$ and $N>1$.

Step 8:Explanation(part c)

Discover $P\{M\le x\mid N>1\}$

We have

$P\{M\le x\mid N>1\}=F\left(x\right)P\{M\le x\}$

Step 9:Final Answer(part c)

$P\{M\le x\mid N>1\}=F\left(x\right)P\{M\le x\}$

Step 10:Given Information(part d)

Given that $P\{M\le x\mid N=1\}$ and $P\{M\le x\mid N>1\}$.

Step 11:Explanation(part d)

$P\{M\le x\}\cdot P\{M\le x\mid N=1\}P\{N=1\}\cdot P\{M\le x\mid N>1\}P\{N>1\}$

$=F\left(x\right)p+F\left(x\right)P\{M\le x\}(1-p)$

Hence,

$P\{M\le x\}=\frac{F\left(x\right)p}{1-(1-p)F\left(x\right)}$

Step 12:Final Answer(part d)

The probability of part (a) is rederived by using (b) and (c) as $P\{M\le x\}=\frac{F\left(x\right)p}{1-(1-p)F\left(x\right)}$