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Let $X_1$ and $X_2$ be i.i.d. normal.

The question is: Can we find the joint distribution of a pair \begin{align}(U_1,U_2)=(\max(X_1,X_2), \min(X_1,X_2)). \end{align}

What I did

Note that this is just the ordering of $X_1$ and $X_2$. In other words, If $X_1>X_2$ then $(U_1,U_2)=(X_1,X_2)$ and if $X_2<X_1$ then $(U_1,U_2)=(X_2,X_1)$.

Then for every set $A$ we have that \begin{align} P( (U_1,U_2) \in A)&= P( (U_1,U_2) \in A \mid X_1>X_2) P( X_1>X_2)+ P( (U_1,U_2) \in A \mid X_1\le X_2) P( X_1 \le X_2)\\ &= \frac{1}{2} P( (U_1,U_2) \in A \mid X_1>X_2)+ \frac{1}{2} P( (U_1,U_2) \in A \mid X_1 \le X_2)\\ &=P( (X_1,X_2) \in A \mid X_1>X_2)+ \frac{1}{2} P( (X_2,X_1) \in A \mid X_1 \le X_2) \end{align}

Can somehow finish this?

Lisa
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2 Answers2

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The density of this distribution is the same as for $X_1$ and $X_2$, just multiplied by $2$ and restricted to $U_1\gt U_2$. That's true for arbitrary i.i.d. variables, not necessarily normally distributed.

joriki
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    So, just to be clear; $f_{U_1,U_2}(u_1,u_2)= 2 f_{X_1,X_2}(u_1,u_2), u_1 \ge u_2$ and $f_{U_1,U_2}(u_1,u_2)=0, u_1<u_2$. Is this right? – Lisa Jun 22 '18 at 22:20
  • Also, could please put a proof of steps. I am very much interested in all the details. – Lisa Jun 22 '18 at 23:02
  • @Lisa: Yes, that's right. I'm not good at proving things formally that seem intuitively obvious to me; others on this site are better at that. It's true simply because the half of the distribution where $X_1\ge X_2$ stays the same and the other half is reflected and added onto that: There are exactly two ways to get $U_1=u_1\land U_2=u_2$, namely $X_1=u_1\land X_2=u_2$ and $X_1=u_2\land X_2=u_1$. If it's not clear to you why that leads to this result, I'll be happy to delete my answer so the question can attract answers with more formal proofs. – joriki Jun 23 '18 at 02:29
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    @joriki I seems fairly clear that $f_{U_1,U_2}(u,v)=f_{X_1,X_2}(u,v)~\mathbf 1_{u\leqslant v}+f_{X_1,X_2}(v,u)~\mathbf 1_{u<v}$ because you are folding the support. The fact that $X_1,X_2$ are independent and identically distributed takes care of the rest. – Graham Kemp Jun 23 '18 at 08:43
  • @GrahamKemp: Thanks. I think you meant "$u\gt v$" in the second term. – joriki Jun 23 '18 at 08:49
  • @joriki No. $u<v$, Notice the order of the arguments in the function. – Graham Kemp Jun 23 '18 at 09:10
  • @GrahamKemp: Ah, I see, thanks. – joriki Jun 23 '18 at 09:11
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Suppose $X_{(1)}=\min(X_1,X_2)$ and $X_{(2)}=\max(X_1,X_2)$. Let $f$ and $F$ denote the density and distribution function of the (normal) population.

Then for $x<y$, the distribution function of $(X_{(1)},X_{(2)})$ is

\begin{align}F(x,y)&=\Pr(X_{(1)}\le x,X_{(2)}\le y)\\&=\Pr(X_{(2)}\le y)-\Pr(X_{(1)}>x,X_{(2)}\le y)\\&=\Pr(X_1,X_2\le y)-\Pr(x<X_1,X_2\le y)\\&=(F(y))^2-\left(F(y)-F(x)\right)^2\end{align}

The second equality follows from the fact that $\Pr(A\cap B^c)=\Pr(A)-\Pr(A\cap B)$.

So the pdf of $(X_{(1)},X_{(2)})$ is \begin{align}f(x,y)&=\frac{\partial^2}{\partial x\partial y}F(x,y)\\&=\frac{\partial}{\partial x}\left(2F(y)f(y)-2(F(y)-F(x))f(y)\right)\mathbf1_{x<y}\\&=2f(x)f(y)\mathbf1_{x<y}\end{align}

This method is easily generalised to find the distribution function and hence the density of $(\min (X_1,\cdots,X_n),\max (X_1,\cdots,X_n))$ for any continuous population when your sample size is $n$, say. Here of course $n=2$.

StubbornAtom
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