Let $X_i$ be iid random variables with common density $f$ and
distribution $F$. Let $Y_k = X_{(k)}$ be the k-th order statistics
(that is, $Y_1 = X_{(1)} = \min(X_i)$ etc.). Show that the joint
density of the order statistics is given by $$f_{X_{(1)}, X_{(2)},
\dots, X_{(n)}}(y_1, \dots, y_n) = \begin{cases} n! \prod_{i=1}^n
f(y_i) & \text{for $y_1 < y_2 < \dots < y_n$} \\ 0& \text{otherwise}
\end{cases}$$ Then show that $$f_{(k)}(y) = k {n \choose k} f(y)
(1-F(y))^{n-k}F(y)^{k-1}$$
I am having trouble with this exercise. I can't seem solve it.
Thoughts on the first part
Well I thought let's start with $P(Y_1 \le y_1, \dots, Y_n \le y_n)$. This can be rewritten as $P(\exists i: X_i \le y_1, \exists i, j: X_i \le y_2, X_j \le y_2, \dots, \forall i: X_i \le y_n )$.
This is ugly and the events are not even independent. I thought about using $P(A_1 \cap A_2 \cap \dots \cap A_n) = P(A_1) P(A_2 \mid A_1) \cdots P(A_n \mid A_1 \cap A_2 \cap \dots \cap A_{n-1})$ but I can't seem to get nowhere.
Thought on the second part
Given the first result, I thought about finding the marginal density, but this seems to be more complicated than I thought.
I mean, $$f_{(k)}(y) = \int_{-\infty}^y \int_{y_1}^y\int_{y_2}^y \int_{y_{k-1}}^y \int_y^\infty \dots \int_{y_{n-1}}^\infty n! \prod_{i=1}^n f(y_i) dy_n \ \dots \ dy_1$$
which I doubt is correct and anyhow I am not able to solve all those integrals.
Can somebody offer some help? I don't think this way of doing it is feasible.
Best Answer
The probability that the $k$th order statistic is less than $x$ is the probability of the union of mutually exclusive events where at least $j \geq k$ of the samples are less than $x$ but the remaining $n-j$ samples are greater than $x$ -- where there are ${n}\choose{j}$ different combinations satisfying this condition. The probability of this union is a sum of probabilities over $j=k, \ldots,n$.
Assuming an iid sample of size $n$, the distribution function for $X_{(k)}$ is
$$\mathbb{P} (X_{(k)}\leq x) = \sum_{j=k}^{n} {{n}\choose{j}}[F(x)]^j[1-F(x)]^{n-j}$$
To get the density function, take the derivative of the distribution function with respect to $x$
$$f_{(k)}(x)\\=\sum^n_{j=k} {n \choose j}\cdot \left(jF(x)^{j-1}(1-F(x))^{n-j}f(x)-F(x)^{j}(n-j)(1-F(x))^{n-j-1}f(x) \right)\\=f(x)\left[\sum^n_{j=k} j{n \choose j} F(x)^{j-1}(1-F(x))^{n-j}-\sum^{n-1}_{j=k} (n-j){n \choose j} F(x)^{j}(1-F(x))^{n-j-1}\right].$$
Note that
$$j{n \choose j}=n{n-1 \choose j-1},\\ (n-j){n \choose j}=n{n-1 \choose j}.$$
Then
$$f_{(k)}(x)\\=nf(x)\left[\sum^n_{j=k} {n-1 \choose j-1} F(x)^{j-1}(1-F(x))^{(n-1)-(j-1)}-\sum^{n-1}_{j=k} {n-1 \choose j} F(x)^{j}(1-F(x))^{(n-1)-j}\right].$$
The first sum is
$$\sum^n_{j=k} {n-1 \choose j-1} F(x)^{j-1}(1-F(x))^{(n-1)-(j-1)} \\={n-1 \choose k-1}F(x)^{k-1}(1-F(x))^{n-k} + \sum^n_{j=k+1} {n-1 \choose j-1} F(x)^{j-1}(1-F(x))^{(n-1)-(j-1)}$$
The second sum is
$$\sum^{n-1}_{j=k} {n-1 \choose j} F(x)^{j}(1-F(x))^{(n-1)-j)} \\=\sum^n_{j=k+1} {n-1 \choose j-1} F(x)^{j-1}(1-F(x))^{(n-1)-(j-1)}.$$
The remaining sums cancel leaving
$$f_{(k)}(x) = n{n-1 \choose k-1}F(x)^{k-1}(1-F(x))^{n-k}f(x)$$