Suppose that $X_1$, $X_2$, ..., $X_n$ are independent, identical, exponential random variables with mean $\mu ^{-1}$, and let $X_{(1)}$ be the smallest among all $X_i$. Why is it P{$X_{(1)}$ x} = P{$X_1$x}P{$X_2$x}...P{$X_n$x}?

Question

Suppose that $X_1$, $X_2$, ..., $X_n$ are independent, identical, exponential random variables with mean $\mu ^{-1}$, and let $X_{(1)}$ be the smallest among all $X_i$. Why is it P{$X_{(1)}$ > x} = P{$X_1$>x}P{$X_2$>x}...P{$X_n$>x}?

zarkon · Answer

if the smallest is larger than x then all are larger than x

zarkon · Answer

ie if $X_{(1)}>x$ then $X_i>x$ for all $i$

anonymous · Answer

But I don't understand how to get this step $P(X_{(1)} > x) = P(X_1>x)P(X_2>x)...P(X_n>x)$

zarkon · Answer

$$\{X_{(1)}>x\}=\{X_1>x,X_2>x,\ldots, X_n>x\}$$

then 

$$P(\{X_{(1)}>x\})=P(\{X_1>x,X_2>x,\ldots, X_n>x\})$$

and by independence we get the product

$$P(\{X_{(1)}>x\})=P(\{X_1>x\})P(\{X_2>x\})\cdots P(\{X_n>x\})$$

anonymous · Answer

I see! Thank you!!!