Help me understand why 


$\int\_{-\infty}^{\infty}\frac{1}{2}[1+\operatorname{erf}(\frac{\theta-x}{\sqrt{2q^2}})]\frac{1}{\sqrt{2\pi\sigma^2}}{\exp(-\frac{(x-\mu)^2}{2\sigma^2})}dx \approx \frac{1}{2}[1+\operatorname{erf}(\frac{\theta-\mu}{\sqrt{2(q^2+\sigma^2)}})]$


This transformation used by Mark E. Glickma in "Parameter estimation in large dynamic paired
comparison experiments" [1](http://www.glicko.net/research/glicko.pdf), but i can't find why it is so.



Integral over error function and normal distribution
This is too long to be a comment.


Let $X$ and $Y$ be independent ${\rm N}(\mu,\sigma\_2)$ and ${\rm N}(0,q^2)$ rv's, respectively.
Since $X+Y \sim {\rm N}(\mu,q^2+\sigma^2)$, it is equal in distribution to $Z + \mu$, where $Z \sim {\rm N}(0,q^2+\sigma^2)$. Hence,  

$$
{\rm P}(X + Y \le \theta ) = {\rm P}(Z \le \theta - \mu ) = \frac{1}{{\sqrt {2\pi (\sigma ^2 + q^2 )} }}\int\_{ - \infty }^{\theta - \mu } {e^{ - z^2 /[2(\sigma ^2 + q^2 )]} \,{\rm d}z} .
$$
On the other hand, by the law of total probability (conditioning on $X$), we have
$$
{\rm P}(X + Y \le \theta ) = \int\_{ - \infty }^\infty {{\rm P}(Y \le \theta - x)\frac{1}{{\sqrt {2\pi \sigma ^2 } }}e^{ - (x - \mu )^2 /(2\sigma ^2 )} \,{\rm d}x}.
$$
Therefore,
$$ 
{\rm P}(X + Y \le \theta ) = \int\_{ - \infty }^\infty {\bigg[\int\_{ - \infty }^{\theta - x} {\frac{1}{{\sqrt {2\pi q^2 } }}e^{ - y^2 /(2q^2 )} \,{\rm d}y\bigg]\frac{1}{{\sqrt {2\pi \sigma ^2 } }}e^{ - (x - \mu )^2 /(2\sigma ^2 )} \,{\rm d}x} }. 
$$
So,
$$
\int\_{ - \infty }^\infty {\bigg[\int\_{ - \infty }^{\theta - x} {\frac{1}{{\sqrt {2\pi q^2 } }}e^{ - y^2 /(2q^2 )} \,{\rm d}y\bigg]\frac{1}{{\sqrt {2\pi \sigma ^2 } }}e^{ - (x - \mu )^2 /(2\sigma ^2 )} \,{\rm d}x} },
$$
which may correspond to the left-hand side expression in the question, is equal to
$$
\frac{1}{{\sqrt {2\pi (\sigma ^2 + q^2 )} }}\int\_{ - \infty }^{\theta - \mu } {e^{ - z^2 /[2(\sigma ^2 + q^2 )]} \,{\rm d}z},
$$
which may correspond to the right-hand side expression in the question (where $s^2$ should be $\sigma^2$).


