Given a random vector $\bar{a}$ with uncorrelated random components of unit variance and zero mean, how do I calculate the expected value over the distribution $$\langle g(\bar{a}\cdot\bar{b})\rangle_{\bar{a}}\text{,}$$ if $\bar{b}$ is some fixed vector with unknown mean and variance and $g(x)$ a continuous function?

For example, how do you solve $$\bigg\langle \text{erf}\bigg(\frac{\bar{a}\cdot\bar{b}} {\sqrt{2}}\bigg)^2\bigg\rangle_{\bar{a}}\text{,}$$ which is $$\frac{1}{\pi}\arcsin\bigg(\frac{\bar{b}\cdot\bar{b}}{(1+\bar{b}\cdot\bar{b})}\bigg)$$ according to this paper.

As far as I understood, I can not use the law of the unconscious statistician, since I do not know the exact distribution of $\bar{a}$ but only its mean and variance. I also do not see how I could exploit the linearity of the expected value, since my random variable is part of the argument of a function.

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    $\begingroup$ If you only know the first two moments, you have insufficient information to answer your question. You might be able to derive some approximation or upper bound on a case by case basis. $\endgroup$ – Simon Boge Brant Dec 17 '19 at 10:49
  • $\begingroup$ @SimonBogeBrant Even if $\bar{b}$ is a fixed vector? I added a full example. Maybe I got something wrong? $\endgroup$ – Genius Dec 17 '19 at 12:45
  • $\begingroup$ Formula (8) in that paper has two more terms $\endgroup$ – Matt F. Dec 17 '19 at 13:11
  • $\begingroup$ @MattF. The other two terms are related to the second and third term of the quadratic loss and thus the problem can be reduced to the first term only (for the sake of readability) $\endgroup$ – Genius Dec 17 '19 at 13:31
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    $\begingroup$ @Genius, yes. I can't find what you are referring to in the paper, apart from the identity that $\int_{-\infty}^\infty\phi(x)erf(c_1 x)erf(c_2x) dx= \frac{1}{\pi} \text{arcsin}\left(\frac{2c_1c_2}{\sqrt{(1 + 2c_1^2)(1 + 2c_2^2)}}\right),$ where $\phi(x) = \frac{1}{\sqrt{2\pi}}e^{-x^2/2}.$ I guess you could use this to solve the expectation, but only if you assume that $\mathbf{a}$ is normally distributed. $\endgroup$ – Simon Boge Brant Dec 17 '19 at 14:36

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