# Scaling the random variables x and y in P(x>y) by some positive parameter

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?

• Since both are non-negative random variables, x<y <=> x^a < y^a. It may not hold if x,y can take negative values. – Lii Dec 14 '17 at 15:23
• $X^a$ and $Y^a$ are not scaled versions of $X$ and $Y$. Scaled versions are $aX$ and $aY$. – Dilip Sarwate Dec 14 '17 at 16:03
• 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. – whuber Dec 14 '17 at 19:33

Supposing $x$, $y$ positive:
$$\Pr\left( x^{\alpha}<y^{\alpha}\right) \\ = \Pr\left( \ln (x^{\alpha})<\ln(y^{\alpha})\right) \\ = \Pr\left( \alpha \ln (x)<\alpha\ln(y)\right) \\ = \Pr\left( \ln (x)< \ln(y)\right) \\ = \Pr\left( x< y\right)$$