91Ó°ÊÓ

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

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

$P\{M≤x\}=\underset{n=1}{\overset{\mathrm{∞}}{∑}} P\{M≤xâˆ\pounds N=n\}P\{N=n\}$

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

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

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

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

Discover$P\{M≤xâˆ\pounds N=1\}$

We have

$P\{M≤xâˆ\pounds N=1\}=F\left(x\right)$

Given that$M≤x$and$N>1$.

Discover $P\{M≤xâˆ\pounds N>1\}$

We have

$P\{M≤xâˆ\pounds N>1\}=F\left(x\right)P\{M≤x\}$

Given that$P\{M≤xâˆ\pounds N=1\}$and$P\{M≤xâˆ\pounds N>1\}$.

$P\{M≤x\}â‹\dots P\{M≤xâˆ\pounds N=1\}P\{N=1\}â‹\dots P\{M≤xâˆ\pounds N>1\}P\{N>1\}$

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

Hence,

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

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