Let $X_1,\dots,X_n$ be $iid$ with distribution function F and $Y_1,\dots,Y_n$ $iid$ with distribution function G. I've proved for some function $g$ that $X_1$ is equal in distribution to $g(Y_1)$, therefore $$g(Y_i)\overset d= X_j,~\text{ for $i,j=1,\dots,n$}.\tag{1}$$ I also proved for the order statistics $X_{1,n}\le\dots\le X_{n,n}$: $$X_{i,n}\overset d= g(Y_{i,n}),~\text{ for $i=1,\dots,n$}\tag{2}$$

Now I want to prove that their common distribution function is equal in distribution, too. If the $X_i$ have an absolutely continuous density function $f$, we can use the formula for their common density function $$ f_{X_{1,n},\dots,X_{n,n}}(t_1,\dots,t_n)= \begin{cases} n!\cdot f(t_1)\cdots f(t_n),&\text{ for $t_1\le~\dots~\le t_n,$}\\ 0,&\text{ otherwise.} \end{cases} $$ and (1) to show equality in distribution.

So basically my question is: How can I show that the following condition regarding the common distribution $$(X_{1,n},\dots,X_{n,n})\overset d= (g(Y_{1,n}),\dots,g(Y_{n,n}))$$ holds in general - not just for $X_i$ with continuous density function?


Hint: $P(g(Y_1) \in A_1, g(Y_2) \in A_2, \ldots, g(Y_n) \in A_n) = \prod_i P(g(Y_i) \in A_i) = \prod_i P(X_i \in A_i) = P(X_1 \in A_1, X_2 \in A_2, \ldots, X_n \in A_n)$

EDIT: So the joint distribution of $(X_1, \ldots, X_n)$ is the same as the joint distribution of $(g(Y_1), \ldots, g(Y_n)$. The joint distribution of the order statistics of $(X_1,\ldots,X_n)$ is then the same as the joint distribution of the order statistics of $(g(Y_1),\ldots,g(Y_n))$, because $(x_1, \ldots, x_n ) \to (x_{1,1}, \ldots, x_{1,n})$ is a continuous function.

But unless $g$ is a nondecreasing function, the distributions of the order statistics $X_{i,n}$ might not be the same as the distributions of $g(Y_{i,n})$, because the $g(Y_i)$ will not be in the same order as the $Y_i$.

  • $\begingroup$ I don't see how that helps with the common distribution of the order statistics, as they aren't independent and we therefore can't use the same trick - am I missing the point? $\endgroup$ – Scooby Oct 8 '14 at 16:00
  • $\begingroup$ So, to sum it up: Because $$(X_1,\dots,X_n)\overset d= (g(Y_1),\dots,g(Y_n))$$ we can use the continuous transformation $$h(x_1,\dots,x_n)=(x_{1,n},\dots,x_{n,n})$$ that permutes the random variables with $x_{1,n}\le\dots\le x_{n,n}$. So we get $$h(X_1,\dots,X_n)\overset d= h(g(Y_1),\dots,g(Y_n)),$$ which we wanted to prove. Thanks for your help! $\endgroup$ – Scooby Oct 8 '14 at 17:32

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