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I am trying to calculate the following expression:

$$ Z = \mathbb{E}\left[\left| \langle \mathbf{a}, u \rangle \right| \right] = \left| \sum_{i=1}^d a_i u_i \right|, \quad \left\| u \right\| = 1 $$

where $a_i$ are i.i.d random variables sampled from the truncated normal distribution with endpoints $[- c\sigma, c\sigma]$, where $c$ is some positive constant.

Initially, I was attempting to find a closed form expression for the pdf of a linear combination of truncated normals, but it seems to be a tedious task (and I haven't been able to find anything relevant in the literature). For the purposes of my work, I would also be interested in a lower bound on the value of $Z$ without explicitly calculating it.

VHarisop
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  • Hint: the distribution of $\langle a_i, u\rangle$ [is identical to the distribution for any $u$ with $||u||=1$.](https://stats.stackexchange.com/questions/85916) Choose $u=(1,0,\ldots,0)$. The integration to obtain the expectation of this *Half-normal distribution* is demonstrated at https://stats.stackexchange.com/questions/11707. – whuber Jan 29 '18 at 23:48
  • @whuber: $u$ is not a random variable, but some fixed vector from the unit sphere. Does your hint still hold? – VHarisop Jan 29 '18 at 23:52
  • Yes, exactly as explained in my answer in the first link. After all, a fixed vector can always be considered a (degenerate) random variable. Everything comes down to the fact that $\mathbf a$ and $Q\mathbf a$ both have a multivariate Normal distribution for any $d\times d$ matrix $Q$. In the special case you seem to be in, where all the components have equal variance $\sigma$ and are uncorrelated, when $Q$ is orthogonal it doesn't change the distribution at all. – whuber Jan 29 '18 at 23:53
  • @whuber: Thanks! Consider making your comment an answer as well. – VHarisop Jan 29 '18 at 23:54

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