In the given problem, we are dealing with a multivariate normal (MVN) distribution of a random vector $\mathbf{Y} = (Y_1, Y_2, Y_3)'$ where:

$\mathbf{Y} \sim \text{MVN} \left( \begin{pmatrix} 4 \\ 6 \\ 10 \end{pmatrix}, \begin{pmatrix} 8 & 5 & 0 \\ 5 & 12 & 4 \\ 0 & 4 & 9 \end{pmatrix} \right)$

We are tasked with finding the distribution of the linear combination $X_1 = Y_1 - Y_2 + Y_3$.

Step 1: Distribution of a Linear Combination of MVN Variables

If $\mathbf{Y} \sim \text{MVN}(\mu_Y, \Sigma_Y)$, and $X_1 = \mathbf{a}^T \mathbf{Y}$ is a linear combination of the elements of $\mathbf{Y}$, where $\mathbf{a} = [1, -1, 1]$, then:

$X_1 \sim \mathcal{N}( \mu_{X_1}, \sigma^2_{X_1})$

Mean of $X_1$

The mean of $X_1$ is given by:

$\mu_{X_1} = \mathbf{a}^T \mu_Y = [1, -1, 1] \begin{pmatrix} 4 \\ 6 \\ 10 \end{pmatrix}$

$\mu_{X_1} = 1 \cdot 4 - 1 \cdot 6 + 1 \cdot 10 = 4 - 6 + 10 = 8$

Variance of $X_1$

The variance of $X_1$ is given by:

$\sigma^2_{X_1} = \mathbf{a}^T \Sigma_Y \mathbf{a}$

Let's compute this in detail. We have:

$\Sigma_Y = \begin{pmatrix} 8 & 5 & 0 \\ 5 & 12 & 4 \\ 0 & 4 & 9 \end{pmatrix}$

And $\mathbf{a} = [1, -1, 1]$.

First, let's compute $\Sigma_Y \mathbf{a}$: