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](https://arxiv.org/pdf/1206.5387.pdf) (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!



Bounding the variance of a truncated Gaussian random variable
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 $0$.


