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?
Thanks in advance!
Best Answer
https://en.wikipedia.org/wiki/Multivariate_normal_distribution#Conditional_distributions Just apply the result after you obtain the distribution of $\mathbf{Y}$