Calculating the variance of a truncated normally distributed RV given the prob. of a range Let's say that I have a Truncated Normal Distribution (from a to b) with mean $m$  (where $m$ is the mean of the truncated Normal) and not known std. I know the probability for the range $m-d$ and $m+d$ (this range is within the truncation points, of course). How do I proceed in calculating the variance of this RV? 
Thank you. 
 A: Let $\phi(y) = \frac{1}{\sqrt{2\pi}}e^{\frac{-y^2}{2}}$
And $\Phi(y) = P(Y \leq y)$
If $X \sim N(\mu, \sigma^2)$, it is possible to write it as $X = \mu + \sigma Y$ where $Y \sim N(0,1)$
Then, $f_X(x) = \frac{1}{\sigma} f(\frac{x-\mu}{\sigma})$
Let $Z$ represent the truncated normal bounded between $[a,b]$. Then $f_Z(z)$ is given by:
$$
f_Z(z | a,b) = \frac{\frac{1}{\sigma} \phi(\frac{z-\mu}{\sigma})}{\Phi(\frac{b-\mu}{\sigma})-\Phi(\frac{a-\mu}{\sigma})}
$$
It is easy to since the denominator represents the probability of a normal r.v. being in the range $[a,b]$
Now consider the moment generating function(MGF): $M(t)=E[e^{tZ}]$
\begin{align*}
E[e^{tZ}] &=  \frac{1}{\sigma} \frac{1}{\Phi(\frac{b-\mu}{\sigma})-\Phi(\frac{a-\mu}{\sigma})} \int_a^b  e^{tz}\phi(\frac{z-\mu}{\sigma})dz\\
\int_a^b  e^{tz}\phi(\frac{z-\mu}{\sigma})dz &= \int_{a}^b \frac{1}{\sqrt{2\pi}}e^{tz}e^{\frac{-(z-\mu)^2}{2\sigma^2}} \\
&= \frac{1}{\sqrt{2\pi}} \int_a^b e^{\frac{\sigma^2tz-(z-\mu)^2}{2\sigma^2}}\\
&=\frac{1}{\sqrt{2\pi}} \int_a^b e^{\frac{-\mu^2+(\sigma^2t +\mu)^2}{2\sigma^2}} e^{\frac{-(z-(\sigma^2t +\mu))^2}{2\sigma^2}}\\
&= \frac{1}{\sqrt{2\pi}} e^{\mu t + \frac{\sigma^2t^2}{2}} \big({\Phi(\frac{z-\mu'}{\sigma})-\Phi(\frac{a-\mu'}{\sigma})}  \big)
\end{align*}
where $\mu' = \mu+\sigma^2 t$
Use this relation to find $M(t)$. Then:
$var(Z) = M''(t)|_{t=0} - (M'(t)|_{t=0})^2$
A: Given: $X \sim \text{DoublyTruncatedNormal}(a,b)$, constructed from a $N(\mu, \sigma^2)$ parent, where:


*

*the bounds $a$ and $b$ are known

*the parent distribution parameters $\mu$ and $\sigma$ are NOT known

*the mean $m$ of $X$ is known (apparently from data)

*the data is NOT available. 

*Finally, $P(m-d < X < m+d)$ is known (again, not stated how??), for a given constant $d$. 


The question is to find the $\text{Var}(X)$, without the data, knowing only $m= E[X]$ and $P(m-d < X < m+d)$.
A suggested path may be:


*

*We know that (see for instance Wiki Truncated Normal):


$$m = \mu + \sigma \frac{\phi(\frac{a-\mu}{\sigma}) - \phi(\frac{b-\mu}{\sigma})}{\Phi(\frac{b-\mu}{\sigma}) - \Phi(\frac{a-\mu}{\sigma})}$$
This provides one equation with two unknowns ($\mu$ and $\sigma$).


*It is straightforward to calculate a symbolic expression for  $P(m-d < X < m+d)$, where $d$ is a number. Let $q = P(m-d < X < m+d)$; you assert that $q$ is a known number. This provides a second equation in two unknowns ($\mu$ and $\sigma$). 


We thus have two equations in two unknowns, which should be solvable, at least numerically for $\mu$ and $\sigma$. You can then substitute the estimated values of $\mu$ and $\sigma$ into the expression for the symbolic expression for $\text{Var}(X)$ which can also be found on the wiki page referenced above.
