Underlying distribution // Backing out from convolution

Question

Let' define $G_i^L$ as the cdf of the average of L independent draws from another cdf $G_i$. Let's further assume that I know $G_i^L$ (and the corresponding pdf/pmf $g_i^L$). Is it now possible from here on to back out the cdf $G_i$ for any $G_i^L$ and for any $integer L\in [2,\infty)$.

Answer 1

I assume that all the draws have the identical distribution, otherwise it is lack of information.

If the distribution is infinitely-divisible, then it will be easy as in such case the sample average will have the same family of distribution as each individual (just scaled), and you just need to manipulate the parameters accordingly.

In general it may not be easy. The following is a theoretical way to compute, but may only be computed numerically.

Denote $X_i$ be the individual draw, having the common CDF $G_X$ and $\displaystyle \bar{X} = \frac {1} {L} \sum_{i=1}^L X_i$ be the sample average which have the CDF $G_{\bar{X}}$

First we compute the characteristic function $\varphi_{\bar{X}}$ from the CDF $G_{\bar{X}}$ :

$$ \varphi_{\bar{X}}(t) = E[e^{it\bar{X}}] = \int_{-\infty}^{+\infty} e^{itx}dG_{\bar{X}} (x)$$

On the other hand,

$$ \begin{align} \varphi_{\bar{X}}(t) &= E[e^{it\bar{X}}] \\ &= E\left[\exp\left\{it\frac {1} {L} \sum_{i=1}^L X_i\right\}\right] \\ &= \prod_{i=1}^L E\left[\exp\left\{i\frac {t} {L} X_i\right\}\right] \\ &= \prod_{i=1}^L \varphi_{X_i}\left(\frac {t} {L}\right) \\ &= \varphi_{X_1}\left(\frac {t} {L}\right) \\ \end{align} $$

So we have

$$ \varphi_{X_1}\left(\frac {t} {L}\right) = \varphi_{\bar{X}}(t)^{1/L}$$

Finally we need to invert the characteristic function to obtain the CDF: https://en.wikipedia.org/wiki/Characteristic_function_(probability_theory)#Inversion_formulae

and they are closely related to the (inverse) Fourier Transform.

The above integrals may not have a closed form so numerically computation maybe needed.