# Obtain order statistics using uniform order statistics

This is a homework questions. Can you guys give me some hints?

Let $U_{(1)}<\cdots<U_{(n)}$ be the order statistics of a sample of size $n$ from a Uniform$(0,1)$ population. Show that $F^{-1}(U_{(1)})<\cdots<F^{-1}(U_{(n)})$ are distributed as the order statistics of a sample of size $n$ from a population with density $f$.

Attempt:

Let $U=(U_{(1)},\ldots,U_{(n)})$, and $V=(F^{-1}(U_{(1)}),\ldots,F^{-1}(U_{(n)}))=F^{-1}(U)$. I know that the joint pdf of the order statistics is: $f_{X_{(1)},\ldots,X_{(n)}}(x_1,\ldots,x_n)=n!\prod_{i=1}^n f_X(x_i)$. So I thought I could use the jacobian method or something:

\begin{align*}f_V(\mathbf{v})&=f_U(F(\mathbf{v}))|J_{F^{-1}}(F(\mathbf{v}))|\\ &= n!\prod_{i=1}^n F(v_i)|J_{F^{-1}}(F(\mathbf{v}))|\end{align*}

But I have no idea what the jacobian could be, and the $F(v_i)$ doesn't seem right either. Any ideas?

• Think about the properties of $F^{-1}$ as applied to a random sample of uniforms, to start. Does the ordering change? Do they have the right marginal distributions? Are they still iid? Now, conclude. – cardinal Sep 7 '13 at 19:18
• I know that if $U\sim Uniform(0,1)$ then $F^{-1}(U)$ has density $f$. Hmm..if$F^{-1}$ is a strictly increasing function then i guess it would map the smallest uniform to the smallest from density f. – bdeonovic Sep 7 '13 at 19:30
• Yes! (+1) You're essentially done. – cardinal Sep 7 '13 at 20:08