Defining priors fom past years' data I'm trying to define a prior from published research.  Lets say several studies have collected data and then estimated some statistic $\hat{\theta}$, as well as the statistic's uncertainty $\hat{\mathrm{Var}}\left(\hat{\theta}\right)$.  For the sake of the argument, lets say there were 3 published studies, so 
$$
\hat{\theta}_1 \mbox{ and } \hat{\mathrm{Var}}\left(\hat{\theta}_1\right)\\
\hat{\theta}_2 \mbox{ and } \hat{\mathrm{Var}}\left(\hat{\theta}_2\right)\\
\hat{\theta}_3 \mbox{ and } \hat{\mathrm{Var}}\left(\hat{\theta}_3\right)
$$
I want to define a prior distribution that takes into account this information, and lets say I know that this distribution is normally distributed a priori; $\theta \sim N(\mu,\sigma^2)$.
 A: No, this is not how you should be finding the mean and variance. The prior precision should be the sum of the individual study precisions and the prior mean should be the precision weighted average of the individual study means. 
Suppose that your published studies truly had normal posteriors and that your stated estimates and variances are actually posterior means and variances, i.e. 
$$
p_i(\theta|y_i) = N(\hat{\theta}_i, \hat{\tau}_i^2)
$$
where $\hat{\tau}_i^2 = \hat{Var}(\hat{\theta}_i)$ for $i=1,\ldots,n$ where $n=3$ in your example. Then your prior should be the combination of each of these individual posteriors. If each of your individual studies are independent of each other and each had a uniform prior for $\theta$, i.e. $p_i(\theta)\propto 1$, and you had a normal prior for $\theta$, say $\theta\sim N(m,C)$, then 
$$ \begin{array}{rl}
p(\theta|y_1,\ldots,y_n) &\propto \left[\prod_{i=1}^n p(y_i|\theta) \phantom{p_i(\theta)}\,\,\right] p(\theta) \\
&\propto \left[\prod_{i=1}^n p_i(\theta|y_i)p_i(\theta)\right] p(\theta)  \\
&\propto \left[\prod_{i=1}^n p_i(\theta|y_i)\phantom{p_i(\theta)}\,\,\right] p(\theta)  \\
&\propto N(\theta|\mu,\sigma^2)
\end{array} $$
where 
$$
\sigma^2 = \left[ \frac{1}{C} + \sum_{i=1}^n \frac{1}{\hat{\tau}_i^2} \right]^{-1}
$$
and 
$$
\mu = \sigma^2 \sum_{i=1}^n \left[ \frac{m}{C} + \frac{\hat{\theta}_i}{\hat{\tau}_i^2}\right].
$$
Letting $C\to \infty$ recovers the precision weighted average of means and posterior precision being the sum of the individual study precisions. 
If the posteriors for the previous studies are not well approximated by normal distributions, then the answer will be more complicated. But this should be better than the equal-weighted average of means and simple average of variances. 
Note that 


*

*as you have more data or have a study that provides an accurate estimate of the parameter, then the sum of the precisions will accurately reflect this increased certainty whereas a simple average will not and

*the precision weighted average of the mean will not be poor when a study has a bizarre estimate but huge uncertainty while a simple average of the means will. 

