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Let $X \sim \text{Dist}(\theta_X)$ and $Y \sim \text{Dist}(\theta_Y)$ 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:

$$\mathbb{P}(X > Y| \theta_X, \theta_Y) = \frac{\theta_X^2}{\theta_X^2 + \theta_Y^2}.$$

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 $\theta$ to be some kind of scale parameter.

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    $\begingroup$ Maybe this will help: en.wikipedia.org/wiki/… $\endgroup$ – John Coleman Feb 5 '18 at 11:35
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    $\begingroup$ Can you provide a context or references for this question? $\endgroup$ – Xi'an Feb 5 '18 at 17:42
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If we take two Exponential random variables$$X\sim\mathcal{E}(\theta_X)\qquad X\sim\mathcal{E}(\theta_Y)$$ we get that$$\mathbb{P}(X>Y|Y=y)=\exp\{-\theta_X y\}$$and $$\mathbb{E}^Y[\exp\{-\theta_X Y\}]=\int_0^\infty \exp\{-\theta_X y\}\,\theta_Y\exp\{-\theta_Y y\}\text{d}y=\dfrac{\theta_Y}{\theta_X+\theta_Y}$$Now, if$$X\sim\mathcal{E}(\theta_X^{-2})\qquad X\sim\mathcal{E}(\theta_Y^{-2})$$then$$\mathbb{P}(X>Y)=\dfrac{\theta_X^2}{\theta_X^2+\theta_Y^2}$$

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 $$\int_0^\infty z\, f(z)\, f(\tau z) \,\text{d}z = \frac{1}{(1+\tau)^2}$$

But the generic answer is no: as noted in the answer by @soakley, this also works for Weibulls which is not a surprise since$$\mathbb{P}(X>Y)=\mathbb{P}(X^\alpha>Y^\alpha)$$for all $\alpha>0$ (and Weibulls are powers of exponentials). A more general class of examples is thus provided by$$X' = \phi(X) \qquad Y' = \phi(Y)$$for all strictly increasing functions $\phi$, where $X,Y$ are exponentials as above, since then we have $$\mathbb{P}(X'>Y') =\mathbb{P}(\phi(X)>\phi(Y))=\mathbb{P}(X>Y)=\dfrac{\theta_X^2}{\theta_X^2+\theta_Y^2}.$$

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If $X$ is Weibull $\left( \alpha,\beta_1 \right)$ and $Y$ is an independent Weibull $\left( \alpha, \beta_2 \right) $, where alpha is the shape parameter and the betas are scale parameters, then it is known that $$P \left[ X > Y \right]= \frac{\beta_1^\alpha}{\beta_1^\alpha + \beta_2^\alpha} $$

This can be derived following the same approach given in Xi'an's answer.

Now let $\alpha=2$ for both $X$ and $Y$. If $X$ has scale parameter $\theta_X$ and $Y$ has scale parameter $\theta_Y,$ we have $$P \left[ X > Y \right]= \frac{\theta_X^2}{\theta_X^2 + \theta_Y^2} $$

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  • $\begingroup$ (+1): Given the vague notion of parameterisation adopted in the question, you can parametrise the Weibulls by $\theta_X$ and $\theta_Y$ for all $\alpha$'s. So the result holds for all $\alpha$'s. $\endgroup$ – Xi'an Feb 5 '18 at 17:48
  • $\begingroup$ Indeed, just as you have shown. I assumed the OP wanted something more direct with the parameters. $\endgroup$ – soakley Feb 5 '18 at 17:50

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