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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!

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    $\begingroup$ 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). $\endgroup$
    – whuber
    Jun 11 '20 at 14:45
  • $\begingroup$ 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 $\endgroup$ Jun 11 '20 at 14:55
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    $\begingroup$ it is true for the median, so, it is also true for the mean only if they coincide $\endgroup$
    – carlo
    Jun 11 '20 at 14:55
  • $\begingroup$ I think I understand now. thank you! $\endgroup$ Jun 11 '20 at 14:58
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  • 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)
  • For gaussian distributions, the mean and the median coincide.

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.

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  • $\begingroup$ 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 $f(x1)\ge f(x_2)$. $\endgroup$
    – Alexis
    Jun 12 '20 at 0:37
  • $\begingroup$ added the equality condition, but I never wanted to provid a full proof, just to make my point clearer, extending that reasoning to decreasing functions is easy enough so that I can gloss over it. $\endgroup$
    – carlo
    Jun 12 '20 at 7:50
  • $\begingroup$ $f()$ 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 $t = 1, \dots, T$). $\endgroup$
    – Nick Cox
    Jun 12 '20 at 8:45
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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 $e^2$.

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