# Solved – Independence of \$min(X,Y)\$ and \$max(X,Y)\$ for independent \$X\$, \$Y\$

What's the reasoning for checking the independence of

\$\$min(X,Y)\$\$ and \$\$max(X,Y)\$\$ for independent r.v.s \$X,Y\$?

Is it possible that \$min\$ and \$max\$ both select the same r.v. in which case they would be dependent? No, because that would mean that \$X=Y\$, i.e. \$X\$ and \$Y\$ would not be independent.

Does the independence regarding functions of independent r.v.s apply here? Yes? In that case \$X,Y\$ independent \$implies\$ \$min(X,Y),space max(X,Y)\$ independent.

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If \$X\$ and \$Y\$ are independent continuous random variables, then \$max(X,Y)\$ and \$min(X,Y)\$ are independent random variables if and only if one of the following two conditions holds:

• \$P(X > Y) = 1\$

• \$P(X < Y) = 1\$

Note that the above conditions mean that \$P(X=Y) = 0\$ but this does not mean that \$(X=Y)\$ is the same as the impossible event, that is, there is no outcome \$omega\$ in the sample space for which \$X(omega) = Y(omega)\$. Those thoroughly confused by this notion should recall that they might have been told that for a continuous random variable \$V\$, \$P(V = a) = 0\$ for all real numbers \$a\$ even though it is manifestly true that \$V\$ can take on value \$a\$ for some particular \$a\$, and if they have swallowed that whopper, then accepting that \$P(X=Y)=0\$ does not mean that the event \$(X=Y)\$ will never occur is just a small additional stretch of their credulity.

When \$X\$ and \$Y\$ are independent discrete random variables, then the above condition needs to be relaxed slightly, and it is possible to have \$P(X=Y) > 0\$. For example, if \$(X,Y)\$ takes on values \$(1,0), (2,0), (1,1), (2,1)\$ with equal probability \$frac 14\$, then \$(min(X,Y), max(X,Y))\$ takes on values \$(0,1), (0,2), (1,1), (1,2)\$ with equal probability \$frac 14\$ and thus \$min(X,Y)\$ and \$max(X,Y))\$ are independent. A little thought will show that \$(min(X,Y), max(X,Y))\$ is the same as \$(Y,X)\$ in this case. A little further thought will show that if \$P(X=Y)>0\$, then it must be that there is a unique \$a\$ such that \$P(X=a, Y= a) >0\$ and that for all other real numbers \$b\$, \$P(X=b, Y= b) =0\$. For independent discrete random variables \$X\$ and \$Y\$, the probability mass function has nonzero values at all points on a rectangular grid, and this grid must be strictly below or strictly above the line \$x=y\$ or must have only one point (the upper left corner or the lower right corner) on the line \$x=y\$; the point \$(1,1)\$ in the example above.

An interesting follow-up question is:

When \$X\$ and \$Y\$ are dependent random variables, is it possible for \$max(X,Y)\$ and \$min(X,Y)\$ to be independent random variables?

to which the answer is Yes, it is possible. Consider the case when \$X\$ and \$Y\$ are jointly continuous random variables uniformly distributed on the set

\$\$left{(x,y)colon frac 12 leq x leq 1, 0 leq y leq x-frac 12right} bigcup left{(x,y)colon 0 leq x leq frac 12, frac 12 leq y < x + frac 12right}\$\$ The joint density of the minimum and maximum can be worked out as described here where it is shown that if \$Z = min(X,Y)\$ and \$W = max(X,Y)\$, then \$\$f_{Z,W}(z,w) = begin{cases} f_{X,Y}(z,w) + f_{X,Y}(w,z), & text{if}~w > z,\ \ 0, & text{if}~w < z. end{cases} \$\$ Applying this, it can be shown that the joint density of \$Z\$ and \$W\$ is uniform on interior of the square with vertices \$(0,frac 12), (frac 12, frac 12), (frac 12, 1), (0,1)\$, and so \$Z sim U[0,frac 12]\$ and \$W sim U[frac 12,1]\$ are independent random variables.

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# Solved – Independence of \$min(X,Y)\$ and \$max(X,Y)\$ for independent \$X\$, \$Y\$

What's the reasoning for checking the independence of

\$\$min(X,Y)\$\$ and \$\$max(X,Y)\$\$ for independent r.v.s \$X,Y\$?

Is it possible that \$min\$ and \$max\$ both select the same r.v. in which case they would be dependent? No, because that would mean that \$X=Y\$, i.e. \$X\$ and \$Y\$ would not be independent.

Does the independence regarding functions of independent r.v.s apply here? Yes? In that case \$X,Y\$ independent \$implies\$ \$min(X,Y),space max(X,Y)\$ independent.

If \$X\$ and \$Y\$ are independent continuous random variables, then \$max(X,Y)\$ and \$min(X,Y)\$ are independent random variables if and only if one of the following two conditions holds:

• \$P(X > Y) = 1\$

• \$P(X < Y) = 1\$

Note that the above conditions mean that \$P(X=Y) = 0\$ but this does not mean that \$(X=Y)\$ is the same as the impossible event, that is, there is no outcome \$omega\$ in the sample space for which \$X(omega) = Y(omega)\$. Those thoroughly confused by this notion should recall that they might have been told that for a continuous random variable \$V\$, \$P(V = a) = 0\$ for all real numbers \$a\$ even though it is manifestly true that \$V\$ can take on value \$a\$ for some particular \$a\$, and if they have swallowed that whopper, then accepting that \$P(X=Y)=0\$ does not mean that the event \$(X=Y)\$ will never occur is just a small additional stretch of their credulity.

When \$X\$ and \$Y\$ are independent discrete random variables, then the above condition needs to be relaxed slightly, and it is possible to have \$P(X=Y) > 0\$. For example, if \$(X,Y)\$ takes on values \$(1,0), (2,0), (1,1), (2,1)\$ with equal probability \$frac 14\$, then \$(min(X,Y), max(X,Y))\$ takes on values \$(0,1), (0,2), (1,1), (1,2)\$ with equal probability \$frac 14\$ and thus \$min(X,Y)\$ and \$max(X,Y))\$ are independent. A little thought will show that \$(min(X,Y), max(X,Y))\$ is the same as \$(Y,X)\$ in this case. A little further thought will show that if \$P(X=Y)>0\$, then it must be that there is a unique \$a\$ such that \$P(X=a, Y= a) >0\$ and that for all other real numbers \$b\$, \$P(X=b, Y= b) =0\$. For independent discrete random variables \$X\$ and \$Y\$, the probability mass function has nonzero values at all points on a rectangular grid, and this grid must be strictly below or strictly above the line \$x=y\$ or must have only one point (the upper left corner or the lower right corner) on the line \$x=y\$; the point \$(1,1)\$ in the example above.

An interesting follow-up question is:

When \$X\$ and \$Y\$ are dependent random variables, is it possible for \$max(X,Y)\$ and \$min(X,Y)\$ to be independent random variables?

to which the answer is Yes, it is possible. Consider the case when \$X\$ and \$Y\$ are jointly continuous random variables uniformly distributed on the set

\$\$left{(x,y)colon frac 12 leq x leq 1, 0 leq y leq x-frac 12right} bigcup left{(x,y)colon 0 leq x leq frac 12, frac 12 leq y < x + frac 12right}\$\$ The joint density of the minimum and maximum can be worked out as described here where it is shown that if \$Z = min(X,Y)\$ and \$W = max(X,Y)\$, then \$\$f_{Z,W}(z,w) = begin{cases} f_{X,Y}(z,w) + f_{X,Y}(w,z), & text{if}~w > z,\ \ 0, & text{if}~w < z. end{cases} \$\$ Applying this, it can be shown that the joint density of \$Z\$ and \$W\$ is uniform on interior of the square with vertices \$(0,frac 12), (frac 12, frac 12), (frac 12, 1), (0,1)\$, and so \$Z sim U[0,frac 12]\$ and \$W sim U[frac 12,1]\$ are independent random variables.

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