I am wondering whether the variance of the left truncated normal distribution is always decreasing in $\alpha$ (lower bound)?
The untruncated distribution of x is $\mathcal{N}(\mu,\sigma^2)$. The variance of the left truncated distribution is given by
$Var(x|x>a)=\sigma^2(1-\delta (\alpha))$ where $\alpha=\frac{a-\mu}{\sigma}$
$\delta (\alpha)=\lambda (\alpha) [\lambda (\alpha)-\alpha], \,\,\, where \,\,\, 0<\delta (\alpha)<1$ and $\lambda (\alpha)=\frac{\phi(\alpha)}{1-\Phi(\alpha)}$
$\frac{d Var(x|x>\alpha)}{d\alpha}=\lambda[(\lambda−\alpha)^2+\lambda(\lambda−\alpha)−1]$
Intuitively, this should be negative: Increase the lower bound, variance of the left truncated distribution decreases. But I can not prove it.