# Bounding the variance of a truncated Gaussian random variable

Suppose $$X_1, X_2, X_3 \sim N(0, 1)$$ are three independent standard normal random variables. I am interested in showing that:

$$\text{Var}[X_2\mid X_2 \geq X_1 - a, X_1 \leq X_3 + b] < 1,$$

where the inequality is strict. I've run some simulations and this seems to be true for various choices of $$a$$ and $$b$$, and it's also intuitively reasonable, since for any fixed value of $$X_1$$, this is just the variance of a normal random variable truncated to $$(-\infty, X_1 - a)$$. I've tried both using the law of total variance (to no avail) and also reparameterizing in terms of $$[X_2, X_1 - X_2, X_1 - X_3]$$ and using the closed form for the variance of a truncated multivariate normal (Eq. 16), also to no avail.

I'm happy with a loose bound as long as I can show it's strictly less than one!

• (i) $X_2$ is independent of $(X_1,X_3)$. So, the condition $X_1 \le X_3 + b$ can be omitted. (Or is there a typo in your post?) (ii) Are any conditions on $a$ and $b$, or at least on their signs, imposed? Jun 21, 2023 at 20:24
• @IosifPinelis $X_2$ is independent of $X_1$ and $X_3$, but it feels like there is a weird dependence between them introduced by the inequalities? For example, cases where $X_1 >> b$ and $X_2 >> X_1$ are very unlikely under both inequalities, but become more likely under just the one. It's possible I am making a logical error though. Jun 21, 2023 at 21:25
• @IosifPinelis For (ii) there is currently no bound on $a$ and $b$ or their sign, but I'd be happy with, e.g., something that says that they must both lay in some interval $[-C, C]$. Jun 21, 2023 at 21:27
• @BMerlot : You don't need to write $X_2>>X_1;$ you can write $X_2 \gg X_1,$ coded as X_2\gg X_1. MathJax is as good at some things as LaTeX is, and LaTeX is very very good. Just enter "LaTeX symbols" into a search engine and you can find things. Jun 22, 2023 at 18:24
• Not only does Iosif Pinelis have a point but omission of the condition $X_1\le X_3+b$ will make Monte Carlo simulations run faster. Jun 22, 2023 at 18:32

The inequality is a special case of the following claim.

Claim: If $$X = (X_1, \dotsc, X_d) \sim N(\mu, \Sigma)$$ is an $$\mathbb{R}^d$$-valued normal random variable with invertible covariance matrix $$\Sigma$$, $$L : \mathbb{R}^d \to \mathbb{R}$$ is linear, and $$K \subseteq \mathbb{R}^d$$ is convex and has a nonempty interior, then $$\operatorname{Var}(L(X) \mid X \in K) \le \operatorname{Var}(L(X))$$, with equality if and only if $$L(X)$$ and $$I(X \in K)$$ are independent. (Here $$I(X \in K)$$ is the indicator function of the event $$X \in K$$.)

Proof: We may assume that $$L \neq 0$$. After a linear change of coordinates, we may assume that $$X$$ is standard normal and $$L(X) = X_1$$. Then the PDF of $$X_1$$ is $$\frac{1}{\sqrt{2 \pi}} e^{-\frac{x^2}{2}}$$, and the PDF of $$Z = (X_1 \mid X \in K)$$ is $$\varphi(x) = C e^{-\frac{x^2}{2}} P(X' \in K_x)$$, where $$X' = (X_2, \dotsc, X_d)$$, $$K_x = \{(x_2, \dotsc, x_d) \in \mathbb{R}^{d-1}; \, (x, x_2, \dotsc, x_d) \in K\}$$ and $$C > 0$$. The function $$\psi \colon \mathbb{R} \to [0,1]$$, $$x \mapsto P(X' \in K_x)$$ is log-concave: if $$x, y \in \mathbb{R}$$ and $$\lambda \in (0,1)$$, then $$\psi((1-\lambda) x + \lambda y) \ge P(X' \in (1-\lambda) K_x + \lambda K_y) \ge \psi(x)^{1-\lambda} \psi(y)^{\lambda},$$ because normal distributions are log-concave (see https://en.wikipedia.org/wiki/Logarithmically_concave_measure). Note that $$X_1$$ and $$I(X \in K)$$ are independent if and only if $$\psi$$ is a constant function. The Claim follows from the following lemma.

Lemma: If $$Z$$ is a real random variable with PDF $$\varphi(x)$$ such that $$e^{\frac{x^2}{2}} \varphi(x)$$ is log-concave, then $$\operatorname{Var}(Z) \le 1$$, with equality if and only if $$Z \sim N(E[Z],1)$$.

Proof: We may add a constant to $$Z$$, so we may assume that $$\varphi$$ is maximal at $$0$$. Then $$\varphi$$ is decreasing in $$\mathbb{R}_{\ge 0}$$, increasing in $$\mathbb{R}_{\le 0}$$, and $$\varphi(x) \le \varphi(0) e^{-\frac{x^2}{2}}$$ for every $$x \in \mathbb{R}$$. We have \begin{align} & 1 - \operatorname{Var}(Z) \ge \int_{\mathbb{R}} (1-x^2) \varphi(x) \, dx \\[6pt] = {} & \int_0^\infty (1-x^2) \varphi(x) \, dx + \int_0^\infty (1-x^2) \varphi(-x) \, dx. \end{align} We prove that $$\int_0^\infty (1-x^2) \varphi(x) \, dx \ge 0$$. This is clear if $$\varphi(1) = 0$$, so let $$\varphi(1) > 0$$. There is a $$c \ge 0$$ such that $$\tilde{\varphi}(1) = \varphi(1)$$, where $$\tilde{\varphi}(x) = \varphi(0) e^{-c x - \frac{x^2}{2}}$$. Then $$\varphi(x) \ge \tilde{\varphi}(x)$$ for $$x \in [0,1]$$, and $$\varphi(x) \le \tilde{\varphi}(x)$$ for $$x \in [1,\infty)$$, so $$\int_0^\infty (1-x^2) \varphi(x) \, dx \ge \int_0^\infty (1-x^2) \tilde{\varphi}(x) \, dx = \varphi(0) \int_0^\infty (1-x^2) e^{-c x - \frac{x^2}{2}} \, dx$$. We can do the same for $$\int_0^\infty (1-x^2) \varphi(-x) \, dx$$. So all we need now is that $$\int_0^\infty (1-x^2) e^{-c x - \frac{x^2}{2}} \, dx > 0$$ for $$c > 0$$. We have $$\int_0^\infty e^{-c x - \frac{x^2}{2}} \, dx = \frac{\sqrt{\pi}}{\sqrt{2}} e^{\frac{c^2}{2}} \operatorname{erfc}(\frac{c}{\sqrt{2}})$$, and differentiating twice in $$c$$ we get $$\int_0^\infty x^2 e^{-c x - \frac{x^2}{2}} \, dx$$. In the end, the inequality boils down to $$c e^{\frac{c^2}{2}} \operatorname{erfc}(\frac{c}{\sqrt{2}}) < \frac{\sqrt{2}}{\sqrt{\pi}}$$. Equivalently, $$\frac{\sqrt{\pi}}{2} \operatorname{erfc}(x) < x^{-1} e^{-x^2}$$ for $$x > 0$$. This follows from $$(\frac{\sqrt{\pi}}{2} \operatorname{erfc}(x) - x^{-1} e^{-x^2})' = (1+x^{-2}) e^{-x^2} > 0$$, since the limits at $$x \to \infty$$ are $$0$$.

• How did you get $\phi(x) \ge \tilde{\phi}(x)$ for $x \in [0,1]$ and $\phi(x) \le \tilde{\phi}(x)$ for $x \in [1,\infty)$? Using the log-concavity of $\phi$, I can get $\phi(x)\ge\phi(0)e^{-cx-x/2}$ for $x\in[0,1]$ and $\phi(x)\le\phi(0)e^{-cx-x/2}$ for $x\in[1,\infty)$ -- but this is not quite what is needed. Jun 28, 2023 at 15:28
• @IosifPinelis $\phi(x) / \widetilde{\phi}(x)$ is log-concave, and it is $1$ at both $x = 0$ and at $x = 1$, so it is $\ge 1$ for $x \in [0,1]$, and $\le 1$ for $x \in [1,\infty)$. Jun 28, 2023 at 15:35
• I see. I forgot about the condition that $e^{x^2/2} \phi(x)$ is log concave, not (just) $\phi(x)$ is log concave. Very nice answer! Jun 28, 2023 at 15:41
• Concerning the latter displayed integral in your answer being $>0$, you can just note that it equals $c(1-cr(c))$ and use the well-known inequality $cr(c)<1$, where $r$ is the Mills ratio for the standard normal distribution. Jun 28, 2023 at 16:06
• @BMerlot We can transform X to standard normal first. Then we can use an orthogonal transformation, and get $L = a X_1$ for some nonzero $a$ (note that if $X$ is standard normal, and $A$ is an orthogonal matrix, then $A X$ is also standard normal). Finally, multiplying $L$ by a nonzero constant we can get $L = X_1$. Jun 29, 2023 at 17:37