Original Problem

This is problem from MIT6.041 course.

pdf: enter image description here

They ask :

"Suppose $X_1$, $X_2$, and $X_3$ are independent exponential random variables, each with param­eter $\lambda$. Find the PDF of $Z= \max\{X_1, X_2, X_3 \}$."

The answer is

The maximum of a set is upper bounded by $z$ when each element of the set is upper bounded by $z$. Thus for any positive $z$, $P(Z \le z)= P(\max\{X_1,X_2,X_3\}\le z) = \dots$

My problem

I don't know why it gives $Z \leq z$, if $Z=\max$ of something it should be larger than... Why is it smaller than, I can't understand this kind of inequality. How can I interpret this $Z$?

  • 1
    $\begingroup$ If $Z$ is defined to be equal to $\max \{X_1, X_2, X_3\}$ and we know that $\max \{X_1, X_2, X_3\}\leq z$, then by definition, we must also have $Z\leq z$. $\endgroup$
    – Bajas
    Aug 9, 2023 at 11:31
  • 1
    $\begingroup$ The answer starts by calculating the pdf of the random variable $Z=max{X_1,X_2,X_2}$ which is per definition for a continuous random variable given by $F_Z(z) = \mathbb{P}(Z \leq z)$ . From there, you can obtain the pdf of $Z$ by differentiating $F$. $\endgroup$ Aug 9, 2023 at 11:32
  • $\begingroup$ @Bajas, why $\max\{X_1,X_2,X_3\} \leq z$, I know $z$ is >0, $X_n$ is probability which should $\leq 1$ $\endgroup$
    – Yiffany
    Aug 13, 2023 at 5:38

1 Answer 1


The problem asks us to find the PDF of $Z=\max\{X_1,X_2,X_3\}$ ($Z$ is just notation for the maximum here). The probability $P(Z\le z)$ is the cumulative distribution function (CDF) of $Z$. If we are able to derive the CDF of $Z$, then we just need to calculate the derivative of the CDF to find the PDF of $Z$ (or $\max\{X_1,X_2,X_3\}$). So that is why the solution starts with deriving the CDF of $Z$.

I hope this helps.

  • 2
    $\begingroup$ Derive here of course means obtain, but unfortunately often means differentiate, which is the next step. $\endgroup$
    – J.G.
    Aug 9, 2023 at 11:37

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