Let $x\in\Bbb{R}$ be a normal random variable with mean $\bar{x}$ and variance $\sigma_x^2$. I'm curious about a new variable that is defined as the sign of $s$, say $s=\operatorname{sgn}(x)\in\{\pm1\}$. What is the expected value of this variable? Based on the answers below, $$ \Bbb{E}[s] = (-1)\cdot P(x<0) + 1\cdot P(x>0) = -P\left(\frac{x-\bar{x}}{\sigma_x}<-\frac{\bar{x}}{\sigma_x}\right) + P\left(\frac{x-\bar{x}}{\sigma_x}>-\frac{\bar{x}}{\sigma_x}\right) \implies \Bbb{E}[s] = -P\left(z < -\frac{\bar{x}}{\sigma_x}\right) + P\left(z > -\frac{\bar{x}}{\sigma_x}\right), $$ where $z\sim\mathcal{N}(0,1)$ is the standard normal variable. Thus, $$ \Bbb{E}[s] = -\Phi\left(-\frac{\bar{x}}{\sigma_x}\right) + 1 - \Phi\left(-\frac{\bar{x}}{\sigma_x}\right) = 1 - 2\Phi\left(-\frac{\bar{x}}{\sigma_x}\right), $$ where $\Phi$ is the cumulative distribution function of the standard normal variable. Thus, $$ \Bbb{E}[s] = \operatorname{erf} \left( \frac{\bar{x}}{\sqrt{2}\sigma_x} \right). $$

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    $\begingroup$ Observe that your expectation can only take nonnegative values. The third equality is wrong. $\endgroup$
    – drhab
    Nov 12 '17 at 12:23
  • $\begingroup$ I see, thank you. I'll get it done and return to update the question. $\endgroup$ Nov 12 '17 at 12:31
  • $\begingroup$ $\sqrt2$ must dissappear everywhere except in the last equality where function $\text{erf}$ comes in. Also $\Phi$ is a Cumulative Distribution Function (CDF). Not a cumulative "density" as you call it. $\endgroup$
    – drhab
    Nov 13 '17 at 9:00
  • $\begingroup$ @drhab thanks! $\sqrt{2}$ was due to copy-paste.. I'll correct it. Thanks for your time! $\endgroup$ Nov 13 '17 at 9:08

$$\mathsf {E(sgn}(X))=1\cdot\mathsf P(X>0)+(-1)\cdot\mathsf P(X<0)$$

  • $\begingroup$ Thank you for your answer. I have update the question accordingly. May you take a quick look? Thanks. $\endgroup$ Nov 13 '17 at 8:41

Check your last equation… it's wrong… ofc the symmetry of the density depends on $\overline{s}$ and additionally if we assume $\overline{s} = 0$ we would get $$\int_{-\infty}^{0}(-1)f(s)\mathrm{d}s = - \int_{0} ^\infty f(s)\mathrm{d}s$$


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