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Let $X_i \sim \mathcal{N}(\mu_i, \sigma_i)$ be independent, normally-distributed random variables. Let

$$Y = a + b \max_i X_i$$

where $a \in \mathbb{R}$ and $b \in (0, 1)$. Which Gaussian distribution is closest to the distribution of $Y$? That is,

$$\operatorname*{argmin}_{\mu, \sigma} \text{KL}(F_Y, \mathcal{N}(\mu, \sigma))$$

where $F_Y$ and $f_Y$ denote the CDF and PDF of the distribution of $Y$, respectively. According to this answer we have

$$f_{\frac{Y-a}{b}}(y) = \left( \frac{1}{\sigma} \sum_i \frac{\phi \left( \frac{y-\mu_i}{\sigma} \right) }{\Phi \left( \frac{y-\mu_i}{\sigma} \right)} \right) \left( \prod_i \Phi \left( \frac{y-\mu_i}{\sigma} \right) \right)$$

where $\Phi$ and $\phi$ denote the CDF and PDF of the standard normal distribution, respectively. Unfortunately, the integral needed to compute the KL divergence (before taking its gradient to find the argmin) seems difficult. Does this problem have an analytic solution? If not, what's the best way to compute it numerically?

user76284
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  • Rather than trying to find the *Gaussian distribution* that best approximates the maximum of a set of Gaussian random variables, may I recommend that you instead read some material on extreme value theory, and look at the kinds of asymptotic distributions that constitute good approximations to a sample maximum. This is likely to give you a superior approximation with less difficulty than the way you are presently proceeding. – Ben Mar 29 '20 at 06:06
  • @Ben-ReinstateMonica Thanks. Do you know of a particular such distribution that might work well? This is related to my other question [here](https://stats.stackexchange.com/questions/457255/parametric-model-closed-under-translation-contraction-and-maximum), so the assumption of Gaussian random variables is flexible. – user76284 Mar 29 '20 at 06:09
  • The Gumbel distribution is the obvious choice. Have a look at some material on extreme value theory and you will see some relevant theorems. – Ben Mar 29 '20 at 10:43

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