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For $X \sim N(x|\mu,\sigma)$ with probability density function $p(x)$, what is $E[p(x)]$?

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  • $\begingroup$ Please also read the self-study tag wiki before modifying your question $\endgroup$
    – Glen_b
    Apr 1 '16 at 4:02
  • $\begingroup$ @Glen_b ok so I wrongly understood the self-study tag, I didn't get this from a textbook or something, I'm actually writing a program that would use this result, so would it be fine if i just remove the self-study tag? $\endgroup$
    – dontloo
    Apr 1 '16 at 4:08
  • $\begingroup$ @Glen_b I added this tag because I was told before that this tag might be helpful for some entry-level questions (and it really is) here (stats.stackexchange.com/questions/199298/… $\endgroup$
    – dontloo
    Apr 1 '16 at 4:11
  • $\begingroup$ You'd likely be asked to add the tag for any routine textbook-style question, which (say) might be set as an exercise (which this certainly qualifies for). If you had not done so I'd have asked you to add it anyway. Asking about concepts is fine, as is seeking some level of guidance to a solution -- for example, the guidance "use the law of the unconscious statistician to write the expectation as an integral, combine the exponents giving a single squared term, recognize the Gaussian density, pull extra constants out the front of the integral; the density integrates to 1" should be okay $\endgroup$
    – Glen_b
    Apr 1 '16 at 4:50
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    $\begingroup$ However, if you could explain the application in your question a bit further there might be some argument to remove the tag and reopen (in spite of it then spoiling the value of a lot of future student's assignments) $\endgroup$
    – Glen_b
    Apr 1 '16 at 5:13
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There might be an easier way to do this using properties of the Normal distribution, but I present a direct answer from the definition of expectation. The pdf of $X \sim N(\mu, \sigma^2)$ is

$$p(x) = \dfrac{1}{\sqrt{2\pi \sigma^2}} \exp \left(-\dfrac{(x - \mu)^2}{2\sigma^2} \right) $$

The expectation of the pdf is then \begin{align*} E[p(x)] & = \int_{-\infty}^{\infty} \dfrac{1}{\sqrt{2\pi \sigma^2}} \exp \left(-\dfrac{(x - \mu)^2}{2\sigma^2} \right) \dfrac{1}{\sqrt{2\pi \sigma^2}} \exp \left(-\dfrac{(x - \mu)^2}{2\sigma^2} \right) dx\\ & = \dfrac{1}{\sqrt{2\pi \sigma^2}} \int_{-\infty}^{\infty} \dfrac{1}{\sqrt{2\pi \sigma^2}} \exp \left(-2\dfrac{(x - \mu)^2}{2\sigma^2} \right) dx\\ & = \dfrac{1}{\sqrt{2}\sqrt{2\pi \sigma^2}} \int_{-\infty}^{\infty}\dfrac{1}{\sqrt{2\pi \sigma^2/2}} \exp \left(-\dfrac{(x - \mu)^2}{2\sigma^2/2} \right) dx\\ & = \dfrac{1}{\sqrt{2}\sqrt{2\pi \sigma^2}}, \end{align*}

where the last step is because the integrand is the pdf for a $N(\mu, \sigma^2/2)$ random variable.

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  • $\begingroup$ and for multivariate Gaussians this would be $(4\pi)^{D/2}|\Sigma|^{1/2}$ right? $\endgroup$
    – dontloo
    Apr 1 '16 at 3:59
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    $\begingroup$ Please note that for self-study questions, the requested behavior is to provide hints and guidance to lead the asker in the correct direction rather than provide a complete solution. $\endgroup$
    – Glen_b
    Apr 1 '16 at 3:59
  • $\begingroup$ I mean the denominator. $\endgroup$
    – dontloo
    Apr 1 '16 at 8:19

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