If $X_1,X_2,X_3,X_4,X_5$are independent and identically distributed exponential random variables with the parameter $\lambda$, compute

(a) role="math" localid="1647168400394" $P\{min(X_1,...,X_5)\le a\}$;

(b) role="math" localid="1647168413468" $P\{max(X_1,...,X_5)\le a.$

The smallest or least value in a set of data is called the minimum.

Each of $X_1,X_2,X_3,X_4,X_5~Exp\left(\lambda \right)$is identically and independent distributed.

Formula :

$F(T\le x)={\int }_0^{x}f\left(t\right)dt$

If the distribution are independent and identical, then CDF will be

$$\begin{gathered} T~Exp\left(\lambda \right) \\ F(T\le x)={\int }_0^x\lambda \times e^{-\lambda t}dt \\ F(T\le x)={\left[-e^{-\lambda t}\right]}_0^x \\ F(T\le x)=1-e^{-\lambda x} \end{gathered}$$

Finding,

$$\begin{gathered} P\left\{min\right(X_1,X_2,X_3,X_4,X_5)\le a\} \\ =1-P\{min(X_1,X_2,X_3,X_4,X_5)>a\} \end{gathered}$$

If the minimum of X is greater than a, then all of them will be greater than a.

$F(X\le a)=1-e^{-\lambda a}$

Hence,

$$\begin{gathered} 1-P\{min(X_1,X_2,X_3,X_4,X_5)>a\} \\ =1-\left(P\right(X_1>a)\times P(X_2>a)\times P(X_3>a)\times P(X_4>a)\times P(X_5>a\left)\right) \\ =1-(P{(X_1>a))}^5 \\ =1-(1-P{(X_1\le a))}^5 \end{gathered}$$

So,

localid="1647167229630" $$\begin{gathered} =1-(1-{(1-e^{-\lambda a}))}^5 \\ =1-{(1-1+e^{-\lambda a})}^5 \\ =1-{\left(e^{-\lambda a}\right)}^5 \\ =1-e^{-5\lambda a} \end{gathered}$$

The highest or greatest value in a set of data is called the maximum.

Using result from subpart (a), the CDF of $X_1$will be $1-e^{-\lambda x}$.

So, $P(X_1\le a)=1-e^{-\lambda a}$.

Finding $P\left\{max\right(X_1,X_2,X_3,X_4,X_5\le a\left)\right\}$

If maximum $X_i$is less than a, then all $X_{i}s$will be less than a.

Hence,

localid="1647168323267" $$\begin{gathered} P\left\{max\right(X_1,X_2,X_3,X_4,X_5\le a\left)\right\} \\ =P(X_1\le a)\times P(X_2\le a)\times P(X_3\le a)\times P(X_4\le a)\times P(X_5\le a) \\ =(P{(X_1\le a))}^5 \\ {(1-e^{-\lambda a})}^5 \end{gathered}$$