Let $x$ and $y$ be two independent random variables and assume that both are exponentially distributed, \begin{align*} x & \sim\exp\left( \lambda\right) \\ y & \sim\exp\left( \mu\right) \end{align*} then we have that $$ \Pr\left( x<y\right) =\frac{\lambda}{\lambda+\mu}. $$ Suppose I need to know the probability of the scaled random variables $\Pr\left( x^{\alpha}<y^{\alpha}\right) $ for $\alpha>0$. Is it true that $$ \Pr\left( x<y\right) =\Pr\left( x^{\alpha}<y^{\alpha}\right) ? $$ If yes, how can I show that? Does this hold for other distributions as well?
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1$\begingroup$ Since both are non-negative random variables, x<y <=> x^a < y^a. It may not hold if x,y can take negative values. $\endgroup$– LiiCommented Dec 14, 2017 at 15:23
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3$\begingroup$ $X^a$ and $Y^a$ are not scaled versions of $X$ and $Y$. Scaled versions are $aX$ and $aY$. $\endgroup$– Dilip SarwateCommented Dec 14, 2017 at 16:03
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2$\begingroup$ This isn't a question about distributions: it's about events. That's what the notations like "$x\lt y$" and "$x^\alpha\lt y^\alpha$" are referring to. $\endgroup$– whuber ♦Commented Dec 14, 2017 at 19:33
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1 Answer
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Supposing $x$, $y$ positive:
$$ \Pr\left( x^{\alpha}<y^{\alpha}\right) \\ = \Pr\left( \ln (x^{\alpha})<\ln(y^{\alpha})\right) \\ = \Pr\left( \alpha \ln (x)<\alpha\ln(y)\right) \\ = \Pr\left( \ln (x)< \ln(y)\right) \\ = \Pr\left( x< y\right)$$
