# Effect of nonlinear transformations on the mean

Suppose I have a continuous random variable $$X$$ and a random variable $$Z = f(X)$$, where $$f$$ is a nonlinear monotonic transformation. How can I prove the following relation between the mean and the median if $$Z$$ is from a Gaussian distribution: $$X^{median} = f^{-1}(Z^{mean})$$ ?

I found it in this paper: Warped Gaussian Processes, but I don't see why this is obvious.

Thank you!

• You can't prove this without additional assumptions. The likeliest one is that for $Z$ "the median and mean lie at the same point" (see the text of the paper preceding equation 9).
– whuber
Jun 11 '20 at 14:45
• Thanks! I've edited my question since I'm mainly interested in the Gaussian case and I still don't know how to prove this Jun 11 '20 at 14:55
• it is true for the median, so, it is also true for the mean only if they coincide Jun 11 '20 at 14:55
• I think I understand now. thank you! Jun 11 '20 at 14:58

• Any monotonic transformation doesn't changes the ranks of the data (this directly comes from the definition of monotonicity: if $$x_1 < x_2$$ then $$f(x_1) \le f(x_2)$$)
• Hence, for any monotonic transformation $$f$$, the median of $$f(X)$$ is $$f(median_X)$$.
• For any invertible monotonic transformation $$f$$, its inverse $$f^{-1}$$ is also monotonic (as above, this also is directly implied by the definition of monotonicity)
Here you are: $$f(X)$$ is not gaussian, so you can't know anything about its mean, but its median is $$f(mean_X)$$, because $$X$$ is gaussian and its mean and median coincide.
• I believe your definition in the first bullet point is in error (or there are different definitions of monotonicity). Shouldn't it read "If $x_1 < x_2$ then $f(x_1) \leq f(x_2)$?" In other words "Non-decreasing" does not necessarily mean "increasing," and vice versa. As an example, in a monotonic function with horizontal regions, for certain values of $x_1 < x_2, f(x_1) = f(x_2)$. Also, this definition seems to apply only to monotonically increasing functions. For a monotonically decreasing function, I would expect if $x_1 < x_2$, then $f(x1)\ge f(x_2)$. Jun 12 '20 at 0:37
• $f()$ is a natural notation for anyone who has spent more time on calculus than on statistics. For any one the other way round, $f()$ is likely to seem bespoke as notation for probability density function. I suggest $T()$ as generic notation for a transformation, although no notation suits all (some economists would want to underline that time runs $t = 1, \dots, T$). Jun 12 '20 at 8:45
This certainly doesn't hold in general. For instance, assume that $$X$$ is lognormal with log-mean 0 and log-sd 1. Then its median is $$e^0=1$$, but the mean of $$X^2$$ (look in the "Properties" section of the Wikipedia page) is $$e^2$$.