Exercise question (from Timo Koski's book Lecture Notes: Probability and Random Processes at KTH):
Let $(X_{1},X_{2})'\in N(\pmb{\mu},\pmb{C})$, where
$$\pmb{\mu}=\begin{pmatrix} 0\\ 0 \end{pmatrix}$$ and $$\pmb{C}=\begin{pmatrix} 1&\rho \\ \rho&1 \end{pmatrix}$$
a) We want to find the distribution of the random variable $X_{1}|X_{2}\leq a$. Show that
$$\pmb{P}(X_{1}\leq x|X_{2}\leq a)=\frac{1}{\Phi(a)}\int_{-\infty}^{x}\phi(u)\Phi\left(\frac{a-\rho u}{\sqrt{1-\rho^2}}\right)du,$$
where $\Phi(x)$ is the distribution function of $N(0,1)$ and $\phi(x)$ the p.d.f. (probability density function) od $N(0,1)$, i.e., $\frac{d}{dx}\Phi(x)=\phi(x)$. We sketch two different solutions.
Aid 1. We need to find
$$\pmb{P}(X_{1}\leq a|X_{2}\leq a)=\frac{\pmb{P}(\{X_{1}\leq x\}\cap \{X_{2}\leq a\})}{\pmb{P}(X_{2}\leq a)}.$$
Then
$$\pmb{P}(\{X_{1}\leq x\}\cap \{X_{2}\leq a\})=\int_{-\infty}^{x}\int_{-\infty}^{a}f_{X_{1},X_{2}}(u,v)dudv=\\ =\int_{-\infty}^{x}f_{X_{2}}(v)\int_{-\infty}^{a}f_{X_{1}|X_{2}=v}(u)dudv.$$
Now find $f_{X_{2}}(v)$ and $f_{X_{1}|X_{2}=v}(u)$ and make a change of variable in $\int_{-\infty}^{a}f_{X_{1}|X_{2}=v}(u)du$.
Aid 2. Write $(X_{1},X_{2})'$ as a linear transformation of the standard Gaussian vector $\pmb{Z}=(Z_{1},Z_{2})'\in N(\pmb{0},\pmb{I})$ as
$$\begin{pmatrix} X_{1}\\ X_{2} \end{pmatrix}=\pmb{B} \begin{pmatrix} Z_{1}\\ Z_{2} \end{pmatrix}.$$
Here we have
$$\pmb{B}=\begin{pmatrix} \sigma_{1}&0 \\ \sigma_{2}\rho& \sigma_{2}\sqrt{1-\rho^2} \end{pmatrix}.$$
Then you can, since $\pmb{B}$ is invertible, write the event
$$\{X_{1}\leq x\}\cap \{X_{2}\leq a\}$$
as an event using (the innovations) $Z_1$ and $Z_2$ and then compute the desired probability using the joint distribution of $Z_1$ and $Z_2$.
Now, these aids don't help me that much and I'm in no better position to solve this problem. How am I supposed to go about it? I did try writing out the $X_1$ and $X_2$ and inserting the innovations into the event, but from there I do not know what to do. My intuition tells me nothing.
Proceed with your first method.
Simply use the law of total probability for the joint CDF of $(X_1,X_2)$:
\begin{align} P(X_1\le x,X_2\le a)&=\int_{-\infty}^x P(X_2\le a\mid X_1=u)f_{X_1}(u)\,du \\&=\int_{-\infty}^x \Phi\left(\frac{a-\rho u}{\sqrt{1-\rho^2}}\right)\phi(u)\,du\quad, \end{align}
where we have used that $X_2$ given $X_1=u$ has a $N(\rho u,1-\rho^2)$ distribution, and of course $X_1\sim N(0,1)$.