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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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  • $\begingroup$ Yes, you have to look at the joint distribution of $X$ and $\mathbb{I}(X\ge y)$ and then apply Bayes' rule. $\endgroup$ – Xi'an Jun 6 '16 at 9:28
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Let's assume you mean $X\sim\mathcal{N}(\mu_X, \sigma_X)$ 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

$$\Pr(Z\in\mathcal{A}\,|\,Z \ge 0)\Pr(Z \ge 0) = \Pr(Z\in\mathcal{A\cap[0,\infty)})$$

for all measurable sets $\mathcal{A}$. In particular, letting $\mathcal{A}=(z,\infty)$ for some $z\ge 0$, 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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