Conditional Mean of a Standard Normal Given It Is Positive

Worked Solution

How to Think About It: A conditional expectation given an event is the partial expectation over that event divided by the event's probability. The partial expectation is a Gaussian integral that the substitution $u = x^2/2$ kills instantly; the only trap is forgetting to divide by $P(X > 0) = 1/2$.

Quick Estimate: Given $X > 0$, the density is $2\phi(x)$ on $(0, \infty)$: decreasing and right-skewed, so its mean sits above its median $0.674$ (the standard normal's 75th percentile) but well below $1$. Something around $0.8$ is a good guess.

Formal Solution:

*Step 1 -- Write the definition.* $$E[X \mid X > 0] = \frac{E[X \mathbf{1}_{\{X>0\}}]}{P(X > 0)} = \frac{\int_0^{\infty} x\,\phi(x)\,dx}{1/2}, \qquad \phi(x) = \frac{1}{\sqrt{2\pi}} e^{-x^2/2}.$$

*Step 2 -- Evaluate the numerator.* Substitute $u = x^2/2$, $du = x\,dx$: $$\int_0^{\infty} \frac{x}{\sqrt{2\pi}} e^{-x^2/2}\,dx = \frac{1}{\sqrt{2\pi}} \int_0^{\infty} e^{-u}\,du = \frac{1}{\sqrt{2\pi}}.$$ Equivalently, note $\phi'(x) = -x\phi(x)$, so $\int_0^\infty x\phi(x)\,dx = \phi(0) - \phi(\infty) = \phi(0) = 1/\sqrt{2\pi}$.

*Step 3 -- Divide by the probability.* $$E[X \mid X > 0] = \frac{1/\sqrt{2\pi}}{1/2} = \frac{2}{\sqrt{2\pi}} = \sqrt{\frac{2}{\pi}} \approx 0.7979.$$

*Step 4 -- Cross-checks.* By symmetry $E[X \mid X < 0] = -\sqrt{2/\pi}$, and the law of total expectation gives $\tfrac12\sqrt{2/\pi} - \tfrac12\sqrt{2/\pi} = 0 = E[X]$. Also $E|X| = \sqrt{2/\pi}$, a standard half-normal fact. (A common slip is to report $1/\sqrt{2\pi} \approx 0.399$, which is $E[X^{+}] = E[\max(X,0)]$, not the conditional mean.)

Answer: $E[X \mid X > 0] = \sqrt{2/\pi} \approx 0.798$.