# Variance of the truncated normal distribution (truncated from below) increases in $\sigma$?

I'm wondering whether the variance of the truncated normal distribution increases in $$\sigma$$ (which seems to hold numerically), where the untruncated normal distribution is $$N(\mu,\sigma^2)$$ and the truncated normal distribution is truncated from zero.

I found that there were some discussions on the relationship between the mean of the truncated normal distribution and $$\mu$$ (Is the mean of the truncated normal distribution monotone in $$\mu$$?), and the relationship between the mean of the truncated normal distribution and $$\sigma$$ (Effect of variance on truncated normal) but couldn't find any discussion on the relationship between the variance (or standard deviation) of the truncated normal distribution and $$\sigma$$.

The variance of the truncated normal distribution (truncated from below) is:

$$Var(X|X>0)=\sigma^2 \left[1+\frac{\left(-\frac{\mu}{\sigma}\right) \phi\left(-\frac{\mu}{\sigma}\right)}{1-\Phi\left(-\frac{\mu}{\sigma}\right)} -\left( \frac{\phi\left(-\frac{\mu}{\sigma}\right)}{1-\Phi\left(-\frac{\mu}{\sigma}\right)} \right)^2\right]$$

$$\Phi,\phi$$ are cdf and pdf of the standard normal distribution.

Is there any proved claim that $$Var(X|X>0)$$ increases in $$\sigma$$? Or can we prove it? Any information or insight would greatly help.

A straightforward (computer) symbolic integration shows that the variance of the truncated distribution for arbitrary $$\mu$$ gives:

$$\sigma \left(-\frac{2 \sigma e^{-\frac{\mu ^2}{\sigma ^2}}}{\pi \left(\text{erf}\left(\frac{\mu }{\sqrt{2} \sigma }\right)+1\right)^2}+\frac{\sqrt{\frac{2}{\pi }} \mu e^{-\frac{\mu ^2}{2 \sigma ^2}}}{\text{erfc}\left(\frac{\mu }{\sqrt{2} \sigma }\right)-2}+\sigma \right).$$

Here's a graph (for fixed $$\mu = 1$$):

Here's the derivative of the variance with respect to $$\sigma$$:

$$-\frac{4 \sqrt{2} \mu e^{-\frac{3 \mu ^2}{2 \sigma ^2}}}{\pi ^{3/2} \left(\text{erf}\left(\frac{\mu }{\sqrt{2} \sigma }\right)+1\right)^3}-\frac{2 e^{-\frac{\mu ^2}{\sigma ^2}} \left(3 \mu ^2+2 \sigma ^2\right)}{\pi \sigma \left(\text{erf}\left(\frac{\mu }{\sqrt{2} \sigma }\right)+1\right)^2}+\frac{\sqrt{\frac{2}{\pi }} \mu e^{-\frac{\mu ^2}{2 \sigma ^2}} \left(\mu ^2+\sigma ^2\right)}{\sigma ^2 \left(\text{erfc}\left(\frac{\mu }{\sqrt{2} \sigma }\right)-2\right)}+2 \sigma$$

Here's a graph of the variance with respect to $$\sigma$$ and $$\mu$$: always monotonic:

• You do not give any proofs whether the variance of the truncated normal distribution (truncated from below at zero) increases in $\sigma$... The variance formula has been already given in this question. Commented Nov 23, 2018 at 7:57
• @sndwec It is and old post. But I wonder whether you could prove it or not. Commented Aug 29, 2023 at 15:51