# Integral of an integral with variable limits

I'd like to prove the following but not sure where to start:

$$\int_{-\infty}^\infty\int_{-\infty}^a\frac{1}{\sqrt{2\pi}}\exp\left(-\frac{x^2}{2}\right)dx\frac{1}{\sqrt{2\pi\sigma^2}}\exp\left(-\frac{(a-\mu)^2}{2\sigma^2}\right)da\\ =\int_{-\infty}^{\frac{\mu}{\sqrt{1+\sigma^2}}}\frac{1}{\sqrt{2\pi}}\exp\left(-\frac{x^2}{2}\right)dx$$

Note that the hard part is due to the fact that the $a$ is a variable limit,

If it helps, the first term inside the integral is $\mathcal{N}(x;0,1)$ the second term is, $\mathcal{N}(a;\mu,\sigma^2)$

• You can integrate independently over $\large x$: The result is $\large 1$. The "$\large a$" as an integration variable must be something else, like for example $\large y$. – Felix Marin Sep 24 '13 at 2:06
• I don't think so, anyway I found the proof in the answer below I posted. – sachinruk Sep 24 '13 at 6:38

The LHS is $P[A\geqslant X]$ where $A$ is gaussian with mean $\mu$ and variance $\sigma^2$, $X$ is standard gaussian, and $(X,A)$ is independent.
Thus, $A=\mu+\sigma Y$ where $(X,Y)$ is standard gaussian, and the LHS is $P[X-\sigma Y\leqslant\mu]$.
Now, $X-\sigma Y$ is centered gaussian with variance $1+\sigma^2$ hence $X-\sigma Y=\sqrt{1+\sigma^2}Z$ where $Z$ is standard gaussian, and the LHS is $P[Z\leqslant z]$ with $z=\mu/\sqrt{1+\sigma^2}$.
The RHS is $\Phi(z)$ hence the LHS and the RHS coincide.