Consider the double truncated standard normal distribution:

$\phi_T(x) = \frac{\mathbb{1}_{[\alpha, \beta]}(x) \phi(x)}{\Phi(\beta) - \Phi(\alpha)}, \alpha < \beta$,

where $\phi$ is the standard normal PDF and $\Phi$ is the standard normal CDF.

I'm trying to show that the variance of this distribution, say $\gamma_T^2$, is always less than 1.

I've been able to prove this for some special cases, like $\alpha = -\beta$, $\alpha < 0, \beta > 0$, but not in its full generality, i.e., for arbitrary $\alpha, \beta$.

Things that I've tried so far:

  • Stare directly at the formula for $\gamma_T^2$
  • Directly use the definition of variance. This leads to $ \gamma_T^2 < 1/\mu$, where $\mu = \Phi(\beta) - \Phi(\alpha) < 1$.
  • Use Popoviciu's inequality on variances: $\gamma_T^2 < (\beta - \alpha)^2 / 4$. This leads to $\gamma_T^2 < 1$ for $(\beta - \alpha)^2 < 4$, which is again a special case
  • Try using results on second order stochastic dominance, but that requires proving $\int_{-\infty}^y\Phi(x)dx > \int_{-\infty}^y \Phi_T(x)dx$ for all y, where $\Phi_T$ is the CDF of the truncated normal distribution, which also, I was unable to prove.

Is this a known result? Any help will be appreciated!

  • $\begingroup$ For $X\sim N(0,1)$, we have $$\operatorname{Var}(X\mid\alpha<X<\beta)=1+\frac{\alpha\phi(\alpha)-\beta\phi(\beta)}{\Phi(\beta)-\Phi(\alpha)}-\left[\frac{\phi(\alpha)-\phi(\beta)}{\Phi(\alpha)-\Phi(\beta)}\right]^2$$ Is this the formula you ended up with and want to show this is less than $1$? $\endgroup$ Commented Aug 22, 2018 at 19:06
  • $\begingroup$ @StubbornAtom Yes! If it helps, $\alpha\phi(\alpha) = -\phi'(\alpha)$ $\endgroup$ Commented Aug 22, 2018 at 19:08

1 Answer 1


As shown in this article (pp. 5-6), instead of focusing immediately on $\text{N}(0,1)$ the desired result can be obtained as a corollary to the following theorem:

Theorem: If $Y\sim\text{N}(\eta,1)$ with any real $\eta$, then for any real $c>0,\ \ \mathbb{V}(Y\mid -c<Y<c) < 1.$


Corollary: If $X\sim\text{N}(\mu,1)$ with any real $\mu$, then for any real $a<b,\ \ \mathbb{V}(X\mid a<X<b) < 1.$

Proof of corollary:
$$\begin{align}\mathbb{V}(X\mid a<X<b)&=\mathbb{V}\left(X-{a+b\over 2}\ \ {\LARGE\mid}\ \ a<X<b\right)\\ &=\mathbb{V}\left(X-{a+b\over 2}\ \ {\LARGE\mid}\ \ a-{a+b\over 2}<X-{a+b\over 2}<b-{a+b\over 2}\right)\\ &=\mathbb{V}\left(X-{a+b\over 2}\ \ {\LARGE\mid}\ \ -{b-a\over 2}<X-{a+b\over 2}<{b-a\over 2}\right)\\ &< 1 \end{align}$$ where the last line follows from the theorem, with $Y=X-{a+b\over 2},\ \ \eta=\mu-{a+b\over 2}$ and $c={b-a\over 2}.$

Proof of theorem: (See p. 6 of the linked article.) Sketch: Because the normal distribution is a member of the exponential family of distributions, it is straighforward to show that $$\begin{align}\mathbb{V}(Y\mid -c<Y<c)&={d\over d\eta}\mathbb{E}(Y\mid -c<Y<c)\\ &=1-{d\over d\eta}h(\eta)\\ \end{align}$$ where $$h(\eta)={\phi(\eta-c)-\phi(\eta+c)\over \Phi(\eta+c)-\Phi(\eta-c)}. $$ The result follows upon observing that $h$ is a continuous odd function, increasing in $\eta$.

Aside: The above-linked article conjectures that $\mathbb{V}(X\mid\alpha<X<\beta)<\mathbb{V}(X)$ holds for any distribution if $0<\mathbb{P}(\alpha<X<\beta)<1$. I show, however, that the conjecture can fail spectacularly for Lognormal distributions.

  • $\begingroup$ This is exactly what I needed. Thank you! $\endgroup$ Commented Aug 23, 2018 at 14:44

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