The inequality of PDF of $Z = \max\{X_1,X_2,X_3\}$

probabilityprobability distributionsprobability theoryself-learning

Original Problem

This is problem from MIT6.041 course.

pdf:

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 $Z$?

Best Answer

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.

Related Solutions

[Math] CDF of $Y=\max\{X_1, X_2, X_3\},$ given $f(x)=2,$ for $0

Comment: Maximum of three uniform random variables.

You begin with independent $X_i \sim Unif(0, 1/2),$ for $i = 1, 2, 3,$ and $Y = \max(X_1, X_2, X_3).$

@AndreNicolas has told you how to obtain $F_Y(y) = P(Y \le y) = (2y)^3,\,$ for $0 < y \le 1/2.$

And you have differentiated to get $f_Y(y) = 24y^2,\,$ for $0 <y \le 1/2.$

Below is R code for a simulation of 100,000 performances of this experiment. That is 100,000 observations of the random variable $Y$. I have made a histogram of them and superimposed your density curve on the histogram. Perhaps you know how to use your density function to find $E(Y)$ and $SD(Y)$ to see how close my approximations are to the exact values (within 2 or 3 places, I'd suppose).

 x1 = runif(10^5, 0, 1/2)
 x2 = runif(10^5, 0, 1/2)
 x3 = runif(10^5, 0, 1/2)
 y = pmax(x1, x2, x3)
 mean(y);  sd(y)
 ## 0.3748052   # approx. of E(Y)
 ## 0.09656252  # approx. of SD(Y)
 hist(y, prob=T, col="wheat")
 curve(24*x^2, lwd=2, col="blue", add=T) # (syntax of 'curve' mandates argument x)

[Math] Joint PDF of $Y=\min\{{X_1,X_2,X_3}\}$ and $Z=\max\{{X_1,X_2,X_3}\}$

The following may get you started, by finding the density functions of $Y$ and $Z.$

For $Z = \max(X_1, X_2, X_3),$ we have the CDF $$F_Z(t) = P(Z \le t) = P(X_1 \le t)P(X_2 \le t)P(X_3 \le t) = t^3,$$ for $0 <t <1.$ [If the maximum is less than $t,$ then so must each of the three independent $X_i$'s be less than $t.$] So $f_Z(t) = 3t^2,$ which is the density function of the distribution $\mathsf{Beta}(3,1).$

Similarly, for the minimum, $F_Y(t) = (1-t)^3,$ for $0 < t < 1,$ and $Y \sim \mathsf{Beta}(1,3).$

For illustration, the plot below shows histograms of simulations of 100,000 realizations of $Y$ and of $Z,$ along with the appropriate Beta densities.