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Consider two random variables following a common distribution F. Is $P(X_{1} > X_{2}) = P(f(X_{1})>f(X_{2}))$ for any function f?

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    $\begingroup$ If $f $ was a constant function then the right hand side would be zero. $\endgroup$ – copper.hat Oct 31 '19 at 14:59
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    $\begingroup$ Take f(x) = -x for an obvious counterexample. $\endgroup$ – Brady Gilg Oct 31 '19 at 16:40
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    $\begingroup$ You need to make assumptions about f in order for this to be true. Assuming that f is strictly monotonic and increasing should do it $\endgroup$ – Adrian Oct 31 '19 at 21:45
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Let $X_1, X_2$ be two independent random variables each with pmf

$$ P_{X}(x)= \begin{cases} \frac{1}{2} & x =1 \\ \frac{1}{2} & x =-1 \\ \end{cases} $$

Then $\mathsf P(X_1>X_2)=0.25$ but $\mathsf P\big(X_1^2>X_2^2\big)=0$

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No. That equation does not hold in general. Although that equation does not hold in general, it does hold for monotonically increasing functions. If $f$ is monotone increasing (e.g., exponential or logarithmic functions) then you have the event equivalence:

$$\{ X > Y \} \quad \quad \iff \quad \quad \{ f(X) > f(Y) \}.$$

In this case the underlying events are equivalent and so the probability equation you give holds regardless of the distribution of $X_1$ and $X_2$. Note also that there are some special combinations of a distribution $F$ and a function $f$ for which your probability equation holds, but it will not hold as a general property unless $f$ is a monotonically increasing function.

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    $\begingroup$ Might want to add the qualifier "strictly" to "monotonic". "monotonic" is often used to refer to functions with $x > y \rightarrow f(x) \ge f(y)$ rather than $x > y \rightarrow f(x) > f(y)$ $\endgroup$ – Acccumulation Oct 31 '19 at 20:56
  • $\begingroup$ Maybe. But for non-strict version I would usually say "monotone non-decreasing" rather than "monotone increasing". I would generally take "increasing" to imply strictness, unless the contrary is stipulated. Rather than change it, I think I'll just leave these comments here for the OP to see. $\endgroup$ – Reinstate Monica Oct 31 '19 at 21:02
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  • $\begingroup$ That's all very well, but I've seen the alternative terminology elsewhere, and it makes more sense. By the terminology you have linked to, the constant function $f(x)=c$ is an "increasing" function, yet it never increases. It is also a "decreasing" function, yet it never decreases. Curious terminology, no? $\endgroup$ – Reinstate Monica Oct 31 '19 at 21:12
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A simple counterexample for any $X_i$ with $P(X_i > 0) = 1$ is the function $f(x) = x^{-1}$. It reverses inequalities, hence $P(f(X_1) > f(X_2)) = P(X_1 < X_2)$, which is in general different from $P(X_1 > X_2)$.

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