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The Problem: Suppose that $X_1,\dots,X_n$ are independent random variables with the same absolutely continuous distribution. Let $f$ denote their common marginal PDF. Set $Y=\min(X_1,\dots,X_n)$ and $Z=\max(X_1,\dots,X_n)$. Find the joint PDF of $(Y,Z)$.

My Attempt: First we find the joint CDF. For any $y,z\in\mathbb R$ we have that \begin{equation*}\begin{split} F_{YZ}((y,z))&=P(Y\leq y,Z\leq z)\\ &=P(\min(X_1,\dots,X_n)\leq y,\max(X_1,\dots,X_n)\leq z)\\ &=P(\min(X_1,\dots,X_n)\leq y,X_1\leq z,\dots,X_n\leq z)\\ &=[1-P(X_1>y)\cdots P(X_n>y)]F(z)^n\\ &=[1-(1-F(y))^n]F(z)^n, \end{split}\end{equation*} where in the fourth step we used the independence of the random variables $X_1,\dots,X_n$ and the fact that $P(A\cap B)=[1-P(A^\complement)]P(B)$ for any independent events $A,B$. By the absolute continuity of the random variables $X_1,\dots,X_n$ we may differentiate the expression for $F_{YZ}$ to obtain the joint PDF of $(Y,Z)$, \begin{equation*}\begin{split} f_{YZ}(y,z)&=\frac{\partial^2}{\partial y\partial z}[1-(1-F(y))^n]F(z)^n\\ &=\frac{\partial}{\partial y}nf(z)[1-(1-F(y))^n]F(z)^{n-1}\\ &=n^2f(z)f(y)F(z)^{n-1}(1-F(y))^{n-1}\\ \end{split}\end{equation*}

Do you agree with my solution above? Any feedback is much appreciated. Thank you for your time.

StubbornAtom
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Stackman
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1 Answers1

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Mind that $Y,Z$ are NOT independent. $$ P(Y<y,Z<z) = P(Z<z)- P(Y>y, Z<z) = P\bigg( \bigcap_i \{X_i<z\}\bigg)- P\bigg(\bigcap_i \{X_i\in [y,z]\} \bigg) = \ F(z)^n- (F(z)-F(y))^n $$ Taking the derivative w.r.t to $y$ and $z$ yields $$ f_{Y,Z}(y,z)= \partial_y (nF(z)^{n-1}f(z) -n(F(z)-F(y))^{n-1} f(z)) = n(n-1)(F(z)-F(y))^{n-2}f(z)f(y) $$

Mick
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  • Does the first step follow because ${Z\leq z}={Y\leq y,Z\leq z}\cup{Y>y,Z\leq z}$ is a union of disjoint events? – Stackman Mar 13 '20 at 15:36
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    yes, exactly. And due to the absolute continuity it doesn't matter if you write $\le$ or $<$. – Mick Mar 13 '20 at 15:44