Joint density of first k order statistics

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

• Try to answer the following questions first: why do you integrate $n! f(x_1)\cdots f(x_n)$, what is this function and where we need to assume it is zero. – zhoraster Sep 29 '15 at 5:56

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*}

$$\\$$

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.

• I have one question: why are the lower limits of the integrals $x_k$? I don´t understand that part, we have order values $x_1<\cdots <x_k<x_{k+1}<x_{k+2}<\cdots x_n$ so I think that $f(x_{k+1})$ just takes values from $x_{k+1}$ to $x_{k+2}$ – user128422 Oct 22 '15 at 21:39
• @user128422 The order $x_{k+1}\lt x_{k+2}\lt\cdots\lt x_n$ is retained using the upper limits of integration. Our outermost integral is over $x_n$ so each successive integration variable ($x_{n-1},\ldots)$ must be less than the previous variable but still greater than $x_k$. If we reversed the order of integration (outermost integral over $x_{k+1}$) we would then have the integration limits that I think you refer to: $x_k\lt x_{k+1}\lt\infty$ then $x_{k+1}\lt x_{k+2}\lt\infty$ then $x_{k+2}\lt x_{k+3}\lt\infty$ and so on. – Mick A Oct 22 '15 at 22:25
• I´m really sorry but I still don´t undersand it. I think the domain of each $f(x_{j})$ is from $x_{j-1}$ to $x_{j+1}$ so why do we take the domain from $x_{j-1}$ to $\infty$? – user128422 Oct 22 '15 at 23:51
• @user128422 I'll try to explain one integral at a time. Start with the outermost $\int_{x_n=x_k}^\infty$. For this we are given just the values (parameters) $x_1,\ldots,x_k$ but not the values of $x_{k+1},\ldots,x_{n-1}$ - because they are in the inner integrals. Given this, it is possible for $x_n$ to take any value from $x_k$ through to $\infty$. We cannot have a lower limit of $x_{n-1}$ because $x_{n-1}$ is not in an outer integral or a parameter of the required pdf - for the purpose of this particular integral we have no information on the value of $x_{n-1}$ and cannot use it here. ... – Mick A Oct 23 '15 at 0:31
• @user128422 ... Next integral is $\int_{x_{n-1}=x_k}^{x_n}$. Here we are given the values of $x_1,\ldots,x_k$ and also $x_n$ but not $x_{k+1},\ldots,x_{n-1}$. Given this, it is possible for $x_{n-1}$ to take any value from $x_k$ through to $x_n$. I hope this make sense. – Mick A Oct 23 '15 at 0:32