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

(a) $P{(min(X_1,...,X_5) \le a});$

(b) $P{(max(X_1,...,X_5) \le a}).$

I am real lost on what the question is even asking here. I know how to work with exponential random variables in terms of $f(x)=λe^{-λx}$ and $F(x)=1-e^{-λx}$. But how would I use those ideas here?


2 Answers 2



For part (a), you can express the probability as $$P{(min(X_1,...,X_5) \le a}) = 1-P((X_1>a) \cap (X_2>a) \cap (X_3>a) \cap (X_4>a) \cap (X_5>a))$$

For part (b), you have $$P{(max(X_1,...,X_5) \le a}) = P((X_1 \leq a) \cap (X_2 \leq a) \cap (X_3 \leq a) \cap (X_4 \leq a) \cap (X_5 \leq a))$$

Based on the fact that the random variables are i.i.d. exponential distributed, can you take it from here?

  • 1
    $\begingroup$ would you then solve using the inclusion exclusion principle for each one. Then break down each piece with the exponential? In short no. $\endgroup$ Mar 24, 2015 at 19:48
  • $\begingroup$ right they are independent. So they will all just be multiplied out which makes it much simpler. Silly me. $\endgroup$ Mar 24, 2015 at 19:56
  • $\begingroup$ Yes, Jack, you are correct - and also as they are identically distributed, it simplifies further. $\endgroup$ Mar 24, 2015 at 19:56

Let $X$ be exponentially distributed with parameter $\lambda$ with distribution function F.

Using the fact that. $$ \max(X_{1},\dotsc, X_{5}) \leq a \iff X_{1} \leq a, X_{2} \leq a,\dotsc, X_{5} \leq a $$ $$ P(\max(X_{1},\dotsc, X_{5})\leq a ) = \prod_{i=1}^5P(X_{i}\leq a) = F(a)^5 $$ where in the first equality we use independence. In the second, the fact that they are identically distributed.

For the second one use the fact that $$ \min(X_{1}, \dotsc, X_{5}) \ge a \iff X_{1} \geq a, X_{2} \geq a,\dotsc, X_{5} \geq a $$ to compute $P(\min(X_{1}, \dotsc, X{5}) \ge a )$ using similar reasoning as above.


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