# Solved – Conditional expectation of a univariate Gaussian

Suppose I have a univariate Gaussian distribution with mean \$mu_X\$ and standard deviation \$sigma_X\$, and I know the random variable \$X\$ is least some positive value \$y\$: \$X geq y\$. What is the conditional expectation \$mathbb{E}[X | X geq y]\$ of \$X\$ given \$X geq y\$? Is there a closed-form expression for this?

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Let's assume you mean \$Xsimmathcal{N}(mu_X, sigma_X)\$ and that \$y\$ is a constant chosen independently of observing \$X\$. Reduce the problem to finding the conditional expectation of \$Z = X-y\$ conditional on \$Z ge 0\$: adding \$y\$ to that value gives the desired answer. (Whether \$y\$ is positive is immaterial.)

The governing property of conditional probability is the multiplicative relationship

\$\$Pr(Zinmathcal{A},|,Z ge 0)Pr(Z ge 0) = Pr(Zinmathcal{Acap[0,infty)})\$\$

for all measurable sets \$mathcal{A}\$. In particular, letting \$mathcal{A}=(z,infty)\$ for some \$zge 0\$, solve for the conditional probability:

\$\$Pr(Z gt z,|,Z ge 0) = frac{Pr(Z gt z)}{Pr(Z ge 0)}.\$\$

The left hand side is the conditional survival function while the numerator and denominator on the right are both in terms of the survival function of \$Z\$ itself. Write \$Phi(z; mu, sigma)\$ for the Normal distribution function with mean \$mu\$ and standard deviation \$sigma\$. Its complement \$1-Phi\$ is the survival function. Because \$Z\$ obviously is Normal with mean \$mu_X-y\$ and standard deviation \$sigma_X\$, the survival function of the positive part of \$Z\$, \$Z^{+}\$, is

\$\$S_{Z^{+}}(z) = frac{1-Phi(z; mu_X-y, sigma_X)}{1 – Phi(0; mu_X-y, sigma_X)}\$\$

for \$z ge 0\$. Its integral gives the conditional expectation. Add back \$y\$ to give the answer

\$\$y + frac{1}{1 – Phi(0; mu_X-y, sigma_X)}int_0^infty left(1 – Phi(z; mu_X-y, sigma_X)right)dz.\$\$

As the integral of a complementary error function, it has no simpler expression in general.

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