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!

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    $\begingroup$ (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? $\endgroup$ Jun 21, 2023 at 20:24
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    $\begingroup$ @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. $\endgroup$
    – B Merlot
    Jun 21, 2023 at 21:25
  • $\begingroup$ @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 $[-C, C]$. $\endgroup$
    – B Merlot
    Jun 21, 2023 at 21:27
  • $\begingroup$ @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. $\endgroup$ Jun 22, 2023 at 18:24
  • $\begingroup$ 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. $\endgroup$ Jun 22, 2023 at 18:32

1 Answer 1


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 $0$.

  • $\begingroup$ 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. $\endgroup$ Jun 28, 2023 at 15:28
  • $\begingroup$ @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 $x \in [1,\infty)$. $\endgroup$
    – user42355
    Jun 28, 2023 at 15:35
  • $\begingroup$ 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! $\endgroup$ Jun 28, 2023 at 15:41
  • $\begingroup$ 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. $\endgroup$ Jun 28, 2023 at 16:06
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    $\begingroup$ @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 $L = X_1$. $\endgroup$
    – user42355
    Jun 29, 2023 at 17:37

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