# If you know the central moments of the data $X$, find a function $f$ for which $f(X)$ has arbitrary central moments

Say you are given one-dimensional data $X$, with mean $\mu$ and central moments $a_n$ which you know. Can you construct a function $f(x)$ which transforms the data such that $f(X)$ has the central moments $b_n$, also given?

For instance, say we want our second moment (variance) to be $b_2$, then $f(x)=(x-\mu) \sqrt{\frac{b_2}{a_2}}$. Can we extend this function to higher orders as well, such that we can also determine skew, kurtosis or even higher central moments?

If possible, can we have this $f(x)$ monotonically increasing?

I am looking for this in the context of ANN, where such a function might be interesting to work as some kind of copula to make the data more gaussian-like (or maybe uniform?).

## 1 Answer

The function $$f(x)$$ should be monotonically increasing, such that its inverse $$g(y)=f^{-1}(y)$$ is defined. Then you can use the theorem about random variable transformations, which says that, when $$\varphi(x)$$ is the probability density of $$X$$, then $$Y=f(X)$$ has the probability density $$\frac{d}{dy}\Phi(g(y))=\varphi(g(y))\cdot\frac{dg(y)}{dx}$$ When you choose a parametrized ansatz for the function $$g(y)$$, your moment condition transforms into a system of equations for the parameters. Depending on the parametric ansatz and the density function $$\varphi$$, this system of equations will require a numeric solution, but this is merely a technical point. I would thus conclude that the answer to your question is yes.

• There seems to be a serious technical point to address: how do you know that both random variables have probability densities at all?? – whuber Apr 23 '19 at 14:54
• @whuber Yes, you are right: I made this assumption. Formally, this could be circumvented by setting $$\varphi(x) = \sum_{a\inX(\Omega)} P(X=a)\cdot(\delta(x-a)$$ – cdalitz Apr 23 '19 at 15:02
• I don't see how this works. Suppose, for example, that one set of moments gives a discrete distribution and the other set of moments gives a continuous distribution. There cannot possibly exist an invertible function $g$ that accomplishes what you claim in this case. – whuber Apr 23 '19 at 15:05
• @whuber Sorry, I pressed the "enter" key for adding a newline, but this closed my comment. Hiere is the complete comment: Yes, you are right: I made this assumption, and it might be that my suggestion only works for continuous random variables. Formally, this could be generalized by setting $$\varphi(x) = \sum_{a\in X(\Omega)} P(X=a)\cdot\delta(x-a)$$ for a discrede random variable, where $\delta$ is Dirac's delta distribution. This would transform the integral for computing the central moments into a sum and still yield a set of equations for the parameters. – cdalitz Apr 23 '19 at 15:10