# Original Problem

This is problem from MIT6.041 course.

pdf:

"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 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 $$Z$$?

• 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 $Z\leq z$. Aug 9, 2023 at 11:31
• 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 $F$. Aug 9, 2023 at 11:32
• @Bajas, why $\max\{X_1,X_2,X_3\} \leq z$, I know $z$ is >0, $X_n$ is probability which should $\leq 1$ Aug 13, 2023 at 5:38

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 $$Z$$.