[Math] Convolution of maximum and minimum of uniform random variables

convolutionprobabilityrandomrandom variablesuniform distribution

Let $X_1,\ldots, X_n$ be $n$ independent random variables uniformly distributed on $[0,1]$.
Let be $Y=\min(X_i)$ and $Z=\max(X_i) $. Calculate the cdf of $(Y,Z)$ and verify $(Y,Z)$ has independent components.

Please help me out

Best Answer

Even without computing anything, one can guess at the onset that $(Y,Z)$ is probably not independent since $Y\leqslant Z$ almost surely. Now, the most direct way to compute the distribution of $(Y,Z)$ might be to note that, for every $(y,z)$, $$ [y\lt Y,Z\leqslant z]=\bigcap_{i=1}^n[y\lt X_i\leqslant z], $$ hence, for every $0\lt y\lt z\lt1$, $$ P(y\lt Y,Z\leqslant z)=P(y\lt X_1\leqslant z)^n=(z-y)^n. $$ In particular, using this for $y=0$ yields, for every $z$ in $(0,1)$, $$ F_Z(z)=P(Z\leqslant z)=z^n. $$ Thus, for every $(y,z)$ in $(0,1)$, $$ F_{Y,Z}(y,z)=P(Y\leqslant y,Z\leqslant z) $$ is also $$ F_{Y,Z}(y,z)=P(Z\leqslant z)-P(y\lt Y,Z\leqslant z)=z^n-(z-y)^n\cdot\mathbf 1_{y\lt z}. $$ One may find more convenient to describe the distribution of $(Y,Z)$ by its PDF $f_{Y,Z}$, obtained as $$ f_{Y,Z}=\frac{\partial^2F_{Y,Z}}{\partial y\partial z}. $$ In the present case, for every $n\geqslant2$, $$ f_{Y,Z}(y,z)=n(n-1)(z-y)^{n-2}\mathbf 1_{0\lt y\lt z\lt1}. $$