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Suppose that $X_1$ and $X_2$ are independent with common pdf $f(x)$ and cdf $F(x)$. $Y_1=\max\{X_1, X_2\}$ and $Y=\min\{X_1, X_2\}$. Find the joint pdf of $(Y_1, Y_2)$.

My solution: I have no idea about the pdf of $(Y_1, Y_2)$ but I can find the cdf and pdf of $Y_1$, $Y_2$:

$$ P(Y_1\le y_1)=F(y_1)^2 $$ and then $f_{Y_1}(y_1)=2F(y_1)f(y_1)$ for $y_1\in A$.

Similarly, $$ P(Y_2\le y_2)=1-(1-F(y_2))^2 $$ and then $f_{Y_2}(y_2)=2(1-F(y_2))f(y_2)$ for $y_2\in A$.

How to go the next step?

Update:

$$ P(\max\{X_1, X_2\}\le x_1, \min\{X_1, X_2\}\le x_2)=1-P(\max\{X_1, X_2\}\le x_1, \min\{X_1, X_2\}> x_2) $$ $$ =1-P(X_1\le x_1, X_2\le x_1, X_1>x_2, X_2> x_2) $$

When $x_1>x_2$, we get $$ =1-(F(x_1)-F(x_2))^2 $$

Hermi
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1 Answers1

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Let $X_1$ and $X_2$ iid continuous random variables. I will write $Y_1:=\min(X_1,X_2)$ and $Y_2:=\max(X_1,X_2)$ since this notation it's more classic and preserves the idea of an order statistic (we could be more precise and write $(Y_{(1)},Y_{(2)})$ instead $(Y_1,Y_2)$). Recall that $Y_1\leqslant Y_2$ by definition. We want $P(Y_1\leqslant x_1,Y_2\leqslant x_2)$.

Casework:

  • If $x_1\leqslant x_2$, then $$[Y_1\leqslant x_1,Y_2\leqslant x_2]=[(X_1\leqslant x_1 , X_2\leqslant x_2)]\cup [(X_2\leqslant x_1,X_1\leqslant x_2)]$$ Thus, additive law of the probability give $$P(Y_1\leqslant x_1,Y_2\leqslant x_2)=P(X_1\leqslant x_1,X_2\leqslant x_2)+P(X_2\leqslant x_1,X_1\leqslant x_2)-P(X_1\leqslant x_1,X_2\leqslant x_1)$$ Independence give $$P(Y_1\leqslant y_1,Y_2\leqslant y_2)=F(x_1)F(x_2)+F(x_1)F(x_2)-F(x_1)F(x_1)$$

  • If $x_1>x_2$, then since $Y_1\leqslant Y_2$ we have $$P(Y_1\leqslant x_1,Y_2\leqslant x_2)=P(Y_1\leqslant x_2,Y_2\leqslant x_2)=F(x_2)F(x_2)$$

End casework.

Thus, the cdf of $(Y_1,Y_2)$ is just $$F_{Y_1,Y_2}(x_1,x_2)=\begin{cases}2F(x_1)F(x_2)-[F(x_1)]^2,\quad x_1\leqslant x_2,\\ [F(x_2)]^2,\quad x_1>x_2\end{cases}$$

Therefore, the pdf of $(Y_1,Y_2)$ is the derivative then $$f_{Y_1,Y_2}(x_1,x_2)=\begin{cases}2f(x_1)f(x_2),\quad x_1\leqslant x_2\\ 0,\quad \text{otherwise}\end{cases}$$

A. P.
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    +1 Yes, though it may be worth adding that with continuous iid random variables your final conclusion on the joint density is intuitively obvious – Henry Jan 23 '23 at 10:14
  • Thank you Henry, It's nice to read your comment. I will follow your recommendation, of course. – A. P. Jan 23 '23 at 14:36
  • Thanks. Can you please see my update solution? Does it seem that I got a different result from yours: $=1-(F(x_1)-F(x_2))^2$ – Hermi Jan 24 '23 at 04:57
  • @Henry indeed. So is there a way to get at it with a shorter argument? – Benjamin Wang Jan 24 '23 at 21:49
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    @BenjaminWang Clearly $f_{Y_1,Y_2}(x_1,x_2) =0$ when $x_1 > x_2$ by construction and with continuous iid random variables the probability of them being equal is $0$, while when $x_1 \lt x_2$ you have $f_{Y_1,Y_2}(x_1,x_2)$ $=f_{X_1,X_2}(x_1,x_2)+f_{X_1,X_2}(x_2,x_1)$ $=f_{X_1}(x_1)f_{X_2}(x_2) +f_{X_1}(x_2)f_{X_2}(x_1)$ $=2f(x_1)f(x_2)$ – Henry Jan 24 '23 at 22:20
  • @BenjaminWang Can you please see my update solution? It seems that I got a different result. – Hermi Jan 25 '23 at 23:41
  • @Hermi $(A\cap B)^c \color{red}{\neq} A \cap B^c$ – Benjamin Wang Jan 26 '23 at 06:31