Let X∼Dist(θX) and Y∼Dist(θY) be independent continuous random variables generated from the same unspecified distributional form but with allowance for different parameter values. I am interested in finding a parametric distribution form for which the following sampling probability holds for all allowable parameter values:

P(X>Y|θX,θY)=θ2Xθ2X+θ2Y.

My question:Can anyone tell me a continuous distributional form for which this holds? Are there any (non-trivial) general conditions that lead to this?

My preliminary thoughts:If you multiply both parameters by any non-zero constant then the probability remains unchanged, so it makes sense for θ to be some kind of scale parameter.

**Answer**

If we take two Exponential random variablesX∼E(θX)X∼E(θY) we get thatP(X>Y|Y=y)=exp{−θXy}and

EY[exp{−θXY}]=∫∞0exp{−θXy}θYexp{−θYy}dy=θYθX+θYNow, ifX∼E(θ−2X)X∼E(θ−2Y)thenP(X>Y)=θ2Xθ2X+θ2Y

A more interesting question is whether or not this is the only possible case of distribution for which it works. (For instance, this is the only element of the Gamma family for which it works.) Assuming a scale family structure, a necessary and sufficient on the underlying density f of X and Y is that

∫∞0zf(z)f(τz)dz=1(1+τ)2

But the generic answer is no: as noted in the answer by @soakley, this also works for Weibulls which is not a surprise sinceP(X>Y)=P(Xα>Yα)for all α>0 (and Weibulls are powers of exponentials). A more general class of examples is thus provided byX′=ϕ(X)Y′=ϕ(Y)**for all strictly increasing functions ϕ**, where X,Y are exponentials as above, since then we have P(X′>Y′)=P(ϕ(X)>ϕ(Y))=P(X>Y)=θ2Xθ2X+θ2Y.

**Attribution***Source : Link , Question Author : Ben , Answer Author : D.W.*