Variance of marginals of truncated bivariate normal distribution I have a truncated bivariate normal distribution
$$f(x,y)=\begin{cases}\frac{1}{2\pi \sigma_{1}\sigma_{2}\sqrt{1-\rho^{2}}}\exp\left(-\frac{z}{2(1-\rho^{2})}\right) &, |x| \leq a\\0 &, |x| > a\end{cases}$$
, where $$z=\frac{x^2}{\sigma_{1}^2}+\frac{y^2}{\sigma_{2}^2}-\frac{2\rho xy}{\sigma_{1}\sigma_{2}}$$
moreover, in case $f(x,y)$ would be not truncated, it would be normalized.
Is there a simple way to calculate variance of marginal distributions ?
I know that if the distribution is not truncated, variance of marginal distributions are $\sigma_{1}^2$ and $\sigma_{2}^2$, but what if the distribution is truncated ?
 A: It depends what one means by simple ...
Given  $\mu = (0,0)$,  $\quad \Sigma =\left(
\begin{array}{cc}
 \sigma _1^2 & \rho  \sigma _1 \sigma _2 \\
 \rho  \sigma _1 \sigma _2 & \sigma _2^2 \\
\end{array}
\right) \quad $ and $\quad a>0$.
Unconditional model
Let $(X,Y) \sim N( \mu, \Sigma)$ with pdf $f(x,y)$ and cdf $F(x,y)$:

Normalising constant
Let $c = P(-a<X<a) = F(a,\infty)-F(-a,\infty) =2 \big[ F(a,\infty) -\frac12 \big] = \text{Erf}\left(\frac{a}{\sqrt{2} \sigma _1}\right)$
where Erf denotes the error function. Or, automating it:

Doubly Truncated model 
Let $\big(X \big| -a<X<a,Y\big)$ have pdf $g(x,y)$. Then:

Then, the variance-covariance matrix for $(X,Y)$ when $X$ is truncated above at $a$ and below at $-a$ is given by:

where I am using the Varcov function from the mathStatica package for Mathematica to automate the nitty gritties. 


*

*The top left element denotes $\text{Var}(X)$ for the conditional model $g$ and yields a closed-form solution. An interesting feature is that the conditional variance of $X$ does not depend on $\rho$ - this appears to be because the truncation is symmetrical around the mean 0.

*The bottom right entry (with the integral sign) denotes $\text{Var}(Y)$ for conditional model $g$: since the software is unable to find a closed-form solution, the integration does not appear to be 'easy'. 

*The off-diagonal elements denote $\text{Cov}(X,Y)$ for conditional model $g$.
Hope this helps.
