# Conditional probability of multivariate gaussian

I'm unsure regarding my (partial) solution/approach to the below problem. Any help/guidance regarding approach would be much appreciated.

Let $$\mathbf{X} = (X_1, X_2)' \in N(\mu, \Lambda )$$ , where

\begin{align} \mu &= \begin{pmatrix} 1 \\ 1 \end{pmatrix} \end{align} \begin{align} \Lambda &= \begin{pmatrix} 3 \quad 1\\ 1 \quad 2 \end{pmatrix} \end{align} We are tasked with computing: $$P(X_1 \geq 2 \mid X_2 +3X_1=3)$$

I here begin by doing a transformation, $$\mathbf{Y} = (Y_1, Y_2)', \qquad Y_1 = X_1, \qquad Y_2 = X_2 + 3X_1$$ We now are interested in the probability, $$P(Y_1 \geq 2 \mid Y_2 = 3)$$ Since we can write that $$\mathbf{Y = BX}$$, it follows that, $$\mathbf{Y} \in \mathcal{N}(\mathbf{B\mu, B\Lambda B')})$$ where $$\mathbf{B}= \begin{pmatrix} 1 \quad 0\\ 3 \quad 1 \end{pmatrix} \rightarrow \quad \mathbf{B \mu} = \begin{pmatrix} 1 \\ 4 \end{pmatrix}, \quad \mathbf{B\Lambda B'}= \begin{pmatrix} 1 \quad 0\\ 3 \quad 1 \end{pmatrix} \begin{pmatrix} 3 \quad 1\\ 1 \quad 2 \end{pmatrix} \begin{pmatrix} 1 \quad 3\\ 0 \quad 1 \end{pmatrix} = \begin{pmatrix} 3 \quad 10\\ 10 \; \; 35 \end{pmatrix}$$

We thereafter know that we can obtain the conditional density function by, $$f_{Y_1\mid Y_2 = 3} (y_1) = \frac{f_{Y_1,Y_2}(y_1, 3)}{f_{Y_2}(3)} \tag 1$$

The p.d.f. of the bivariate normal distribution,

$$f_{Y_1, Y_2}(y_1, y_2) = \frac{1}{2\pi \sigma_1 \sigma_2 \sqrt{1-\rho^2}} e^{\frac{1}{2(1-\rho^2)}(\frac{(y_1 - \mu_1)^2}{\sigma_1^2} - \frac{2 \rho (y_1 - \mu_1)(y_2 - \mu_2)}{\sigma_1 \sigma_2} + \frac{(y_1 - \mu_1)^2}{\sigma_2^2})}$$

The marginal probability density of $$Y_2$$, $$f_{Y_2}(y_2) = \frac{1}{\sqrt{2\pi} \sigma_2} e^{-(y_2 - \mu_2)^2 / (2\sigma_2^2)}$$ Given that, $$\sigma_1 = \sqrt{3}, \quad \sigma_2 = \sqrt{35}, \quad \rho = \frac{10}{\sigma_1 \sigma_2 } = \frac{10}{\sqrt{105}}$$ we are ready to determine (1). However, the resulting expression, which I then need to integrate as follows,

$$Pr(Y_1 \geq 2 \mid Y_2 = 3) = \int_2^\infty f_{Y_1\mid Y_2 = 3} (y_1) \, dy_1$$ becomes quite ugly, making me unsure whether I've approached the problem in the wrong way?

https://en.wikipedia.org/wiki/Multivariate_normal_distribution#Conditional_distributions Just apply the result after you obtain the distribution of $\mathbf{Y}$
The covariance between $$X_1 + \lambda (3 X_1 + X_2)$$ and $$3 X_1 + X_2$$ is $$10 + 35 \lambda$$, therefore if we take $$\lambda = -2/7$$, we get $$\operatorname{P}(X_1 \geq 2 \mid 3 X_1 + X_2 = 3) = \operatorname{P} \left( X_1 -\frac 2 7 (3 X_1 + X_2 - 3) \geq 2 \mid 3 X_1 + X_2 = 3 \right) = \\ \operatorname{P} \left( X_1 -\frac 2 7 (3 X_1 + X_2 - 3) \geq 2 \right),$$ and $$X_1/7 -2 X_2/7 \sim \mathcal N(-1/7, 1/7)$$.
From where you are, simply use the relation $$(a=3)$$ $$Y_1 | Y_2=a \sim \mathcal{N} \left( \mu_1+ \rho\frac{\sigma_1}{\sigma_2} (a-\mu_2), (1-\rho^2) \sigma_1^2 \right)$$ and then compute the probability as $$1-\Phi_{Y1|Y_2}(2)$$ using the cdf of $$Y_1|Y_2$$