Joint density of first k order statistics

Question

I need to compute the joint density of the first $k$($<n$) order statistics

I was trying to do it by getting the marginal density function:

$$\int \cdots \int f(x_1)\cdots f(x_n)n!dx_{k+1}\cdots dx_{n}$$ but I dont know my interval of integration; is it $(-\infty,\infty)$ for each integral?

I would really appreciate if you can help me with this problem

Answer 1

The joint density is, for $x_1\lt x_2\lt \cdots\lt x_k,$

\begin{eqnarray*} && g(x_1,\ldots,x_k) \;=\; \int_{x_n\;=\;x_k}^{\infty} \int_{x_{n-1}\;=\;x_k}^{x_n} \cdots \int_{x_{k+1}\;=\;x_k}^{x_{k+2}} f(x_1)\cdots f(x_n)n!\;dx_{k+1}\cdots dx_{n} \\ && \\ &&\; =\; n! f(x_1)\cdots f(x_k) \int_{x_k}^{\infty} \int_{x_k}^{x_n} \cdots \int_{x_k}^{x_{k+3}} \left(F(x_{k+2})-F(x_k)\right) f(x_{k+2})\cdots f(x_n)\;dx_{k+2}\cdots dx_{n} \\ && \\ &&\; =\; n! f(x_1)\cdots f(x_k) \int_{x_k}^{\infty} \int_{x_k}^{x_n} \cdots \int_{x_k}^{x_{k+4}} \left[\dfrac{1}{2!} \left(F(x_{k+2})-F(x_k)\right)^2 \right]_{x_k}^{x_{k+3}} f(x_{k+3})\cdots f(x_n)\;dx_{k+3}\cdots dx_{n} \\ && \\ &&\; =\; \dfrac{n!}{2!} f(x_1)\cdots f(x_k) \int_{x_k}^{\infty} \int_{x_k}^{x_n} \cdots \int_{x_k}^{x_{k+4}} \left(F(x_{k+3})-F(x_k)\right)^2 f(x_{k+3})\cdots f(x_n)\;dx_{k+3}\cdots dx_{n} \\ && \\ &&\;=\;\ldots \\ && \\ &&\;=\; \dfrac{n!}{(n-k)!} f(x_1)\cdots f(x_k) \left(1-F(x_k)\right)^{n-k}. \end{eqnarray*}

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You can also see this result intuitively, since, for $x_1\lt x_2\lt \cdots\lt x_k$ to be the first $k$ order statistics, we need $k$ of the $n$ r.v.'s to take one of these $k$ values. Also we need each of the remaining $n-k$ r.v.'s to take any value in the interval $(x_k,\infty)$. There are $\dfrac{n!}{(n-k)!}$ such arrangements of r.v.'s. Hence the result.