Fourth order statistic on a continuous pdf $f(x) = 3x^2$ when $0<x<1$

Question

The random variable $X$ has pdf $f(x) = 3x^2$ when $0<x<1$ and $0$ otherwise. A sample size $n = 6$ is to be taken. Determine the density of $X_{(4)}$, the fourth order statistic.

I tried$$\int \frac{6!}{2!3!} f(x)F(x)^4(1-F(x))^2$$ where $F(x)$ is the integral of $f(x)$ from $0$ to $x$, but was marked wrong

Answer 1

You're almost right. There was no need to take the integral. The general formula for the density of the $k$th order statistic of $n$ iid random variables with PDF $f$ and CDF $F$ is $$f_{(k)}(x) = \frac{n!}{(k-1)!(n-k)!} f(x) [F(x)]^{k-1}[1-F(x)]^{n-k}$$ for $-\infty< x<\infty$. You can find the proof in any standard textbook.

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You were given that the common density is $$ f(x) = 3x^2$$ when $0<x<1$. For $0<x<1$, we have $$F(x) = \int_0^x f(t)\,dt = \int_0^x3t^2\,dt = x^3.$$

Now you just plug into the formula for $f_{(k)}(x)$.

Answer 2

Given

$$f(x) = 3x^2$$ $$F(x) = x^3 $$

when $0<x<1$.

The pdf of the $r$th order statistic is given by

$$f_{r:n}(x)_=\frac{n !}{(r - 1)! (n - r)!} F(x)^{r-1} (1-F(x))^{n-r} f(x)$$

We just plug $f(x)$ and $F(x)$ into the above for $n=6$ and $r=4$, and we get

$$g_{4:6}(x)=180 x^{11} \left(x^3-1\right)^2$$

with cdf

$$ G_{4:6}(x) =\begin{array}{cc} \left\{ \begin{array}{cc} x^{12} \left(10 x^6-24 x^3+15\right) & 0<x<1 \\ 1 & x\geq 1\\ \end{array} \\ \right. \end{array} $$

${\bf EDIT}:$

For the "fun of it", let's find a formula for the expected values of the rth order statistic given 6 samples...

It turns out that

$$E[X_{r:6}] = \frac{720 \Gamma \left(r+\frac{1}{3}\right)}{\Gamma \left(\frac{22}{3}\right) \Gamma (r)}$$