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How to calculate $E[\{\Phi(\alpha X)\}^{K}]$ for K being positive integer, $\Phi()$ is the CDF of standard normal distribution and $X\sim N(0,1)$. I tried using integration by part stuck.

Satya Prakash
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  • What is the distribution of $x$ here? – kimchi lover Aug 06 '17 at 14:30
  • You have not bothered to identify what the random variable is or its distribution and how this random variable relates to the argument of $E[\cdot]$. There is nothing random about $[\Phi(ax)]^K$ and so its expectation is just $[\Phi(ax)]^K$ regardless of what the random variable is. – Dilip Sarwate Aug 06 '17 at 14:32
  • See this question on stats.SE for an approach that might suggest some possibilities. – Dilip Sarwate Aug 06 '17 at 14:36
  • Indeed, due to the invariance of the standard normal distribution by the rotations, this can be converted into the task of computing the surface of a given sector of the unit sphere in $\mathbb R^{K+1}$ (as I might have explained elsewhere on the site). Little hope of getting explicit formulas in general. – Did Aug 06 '17 at 17:58
  • Any thought about the minimization of this expectation wrt to $0<\alpha<\infty$? What is the optimal value of $\alpha$? – Satya Prakash Aug 06 '17 at 18:04

1 Answers1

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In the case when $\alpha = 1$, the solution has an easy form:

$\Phi(X)$ is just the cdf of $X \sim N(0,1)$, which by the probability integral transform, has a standard Uniform distribution, so that the $\alpha = 1$ problem reduces to finding:

$$E[Z^k] \quad \text{ where } Z \sim \text{Uniform}(0,1)$$

... to which the answer is: $\large \frac{1}{1+k}$ .

wolfies
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