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

Question

The continuous random variable $X$ has pdf $f(x) = 2,$ for $0 < x < 1/2;\;$ $0$ otherwise.

Let $X_1, X_2, X_3$ be independent random variables, all with the same distribution as $X.$ Now, suppose we define $Y = \max\{X_1, X_2, X_3\}.$

Why is $F(y) = Pr(X_1 ≤ y) \cdot Pr(X_2 ≤ y) \cdot Pr(X_3 ≤ y) = (2y) \cdot (2y) \cdot (2y)$?

I made an error when typing the pdf. The limits are 0<x<1/2, which should make it a proper density function. My instructor gave the solution for F(y), but I am confused as to how that is the solution. How did he get Pr(X1≤ y) = 2y? — m10islife, Oct 11 '15 at 02:26
I evaluated the CDF as F(y) = 8y^3, and the pdf as f(y)=F'(y)=24y^2 — m10islife, Oct 11 '15 at 02:36
The max $Y$ is $\le y$ iff all the $X_i$ are $\le y$. The probability $X_1\le y$ is $2y$, same for $X_2$, $X_3$, so by independence $\Pr(Y\le y)=(2y)^3$. — André Nicolas, Oct 11 '15 at 05:04

BruceET · Accepted Answer · 2015-10-12T21:20:34.557

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)

If $W \sim Beta(3, 1),$ then your 'max' random variable can be expressed as $Y = W/2.$ You can read about Beta random variables on Wikipedia (or perhaps your textbook). — BruceET, Oct 11 '15 at 05:40

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

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