Let $X_1$,…,$X_n$ be $n$ independent random variables each with the identical pdf $f(x)$. Let $Z = max(X_1,…,X_n).$ Find the pdf of $Z$.
My approach:
Step 1: Find the CDF
$$F_Z(z) = n\int\limits_{z}^{z + dz}f(X_{max})dX_{max}*\int\limits_{-\infty}^{z}(f(x))^{n-1} dx$$
My reasoning here is to find the probability of one variable being at the max value and multiply that with the probability of all the other variables being at that value or less. Then I multiply all of that by $n$ because there are n ways this situation can occur.
Step2: Find the PDF
$$f_Z(z) = \frac{d}{dz}F_Z(z)$$
I'm stuck on step 2. Which integral do I take the derivative of? Or is it unclear because I messed up step 1 somehow?
Best Answer
You can use $P(X_{(n)} \leq x) = [P(X_i \leq x)]^n$
Since they $X_i$ are all independent and identically distributed,
$P(X_{(n)} \leq x) = P(X_1\leq x $ and $ X_2 \leq x $ and $... $ and $ X_n \leq x) = P(X_1 \leq x)^n = \big(F_{X_1}(x)\big)^n$
Then the derivative of that is $$f_{X_{(n)}}(x) = n[F_{X_i}(x)]^{n-1}\cdot f_{X_i}(x)$$ by the chain rule.