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I am trying to compute the expected value, E$[x]$, of a random variable $X\sim\mathcal{N}(\mu,\sigma^2)$. The density function of the normal distribution is $$f_X(x)=\frac{1}{\sigma \sqrt{2\pi}}\exp\left(-\left(\frac{x-\mu}{\sigma\sqrt{2}}\right)^2\right), \ \ \ -\infty<x<\infty.$$

I am attempting to use the following substitution to help find the expected value: $$y=\frac{x-\mu}{\sigma\sqrt{2}}\implies dx=\sigma\sqrt{2} dy \tag{1}.$$

The expected value is computed as \begin{align} \text{E}[x]&=\int_{-\infty}^{\infty} x f_X(x) \ dx \\ &=\frac{1}{\sigma \sqrt{2\pi}}\int_{-\infty}^{\infty} x\exp\left(-\left(\frac{x-\mu}{\sigma\sqrt{2}}\right)^2\right) \ dx \\ &=\frac{1}{\sqrt{\pi}}\int_{-\infty}^{\infty}(\mu+\sigma\sqrt{2} y)\exp(-y^2)\ dy \ \ \ \ \text{(using substitution $(1)$)} \\ &=\frac{\mu}{\sqrt{\pi}}\int_{-\infty}^{\infty} \exp(-y^2) \ dy \ + \ \sigma\sqrt{\frac{2}{\pi}}\int_{-\infty}^{\infty} y\exp(-y^2) \ dy \\ &=\sigma\sqrt{\frac{2}{\pi}}\left(\left[-\frac{1}{2}\exp(-y^2)\right]_{-\infty}^{\infty}+\frac{1}{2}\int_{-\infty}^{\infty}\frac{\exp(-y^2)}{y} \ dy\right). \end{align} I am unsure what this simplifies to (specifically how to deal with the final integral). I've noticed that the integrand is odd (does the integral simply cancel?).

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  • $\begingroup$ Try $\endgroup$ – BruceET 2 days ago
  • $\begingroup$ @BruceET Thanks. I have read the solution and agree with the result. However, I'm wondering if my method will work. It seems the most intuitive to me. $\endgroup$ – M B 2 days ago
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Your substitution is terrible but it must work as well! the natural substitution is

$$Y=\frac{X-\mu}{\sigma}$$

Using this (in this case you are standardizing your Gaussian) the result is very easy.

Considering valid your procedure, reading your last but one passage, the sum is the following

$$\frac{\mu}{\sqrt{\pi}}\cdot \sqrt{\pi}+0=\mu$$

this because the first integral is the Gaussian integral (in the link you can find the easy proof too) and the second is the integral of an odd function over a symmetric domain around zero.

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  • $\begingroup$ Would you suggest that I instead use the substitution $$Y=\frac{X-\mu}{\sigma}?$$ If it's easier, it's probably a good idea. It never occurred to me. $\endgroup$ – M B 2 days ago
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    $\begingroup$ @MB the substitution I suggested you is the more natural but the passages are more or less the same as yours $\endgroup$ – tommik 2 days ago

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