Finding max of $var(a^TY)$.

Question

Let $X=(X_1,X_2,X_3)^T$ be a multivariate random variable with the standard Gaussian distribution on $\mathbb{R}^3$. Define the multivariate random variable $Y=(Y_1,Y_2,Y_3)^T$ by

$$\begin{pmatrix} Y_1 \\\ Y_2 \\\ Y_3 \end{pmatrix} = \begin{pmatrix} -\frac{2}{3} & \frac{2}{3} & \frac{1}{3} \\\ \frac{1}{3} &-\frac{2}{3} & \frac{2}{3} \\\ -\frac{2}{3} & \frac{1}{3} &\frac{1}{3}\end{pmatrix} \begin{pmatrix}X_1 \\\ 2X_2 \\\ X_3 \end{pmatrix}$$

How do I find the maximum value of for $var(a^TY)$ given that $a^Ta = 1$?

I am really struggling with this question, if anyone could help me , I would greatly appreciate it.

Answer 1

For the sake of simplicity, I solve the more general problem. Let $Y = MX$ with

• $X \in \mathbb{R}^{n}$ a standard multivariate Gaussian random variable
• $M\in \mathbb{R}^{n\times n}$ a matrix

We find $a \in \mathbb{R}^{n}$ such that the expression $V(a^TY)$ reaches its maximum with the constraint $a^Ta = 1$. Remark: for your question, $n = 3$, $a = (a_1,a_2,a_3)$ and $$M =\begin{pmatrix} -\frac{2}{3} & \frac{2}{3} & \frac{1}{3} \\\ \frac{2}{3} &-\frac{4}{3} & \frac{4}{3} \\\ -\frac{2}{3} & \frac{1}{3} &\frac{1}{3}\end{pmatrix} $$

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We have $$ \begin{align} V(a^TY) &=V((a^TM)X) \\ &=V((M^Ta)^TX) \\ &=(M^Ta)^T(M^Ta) \\ &=(M^Ta)^T(M^Ta) \\ &=a^T (MM^T)a \tag{1} \\ \end{align} $$

The expression $V(a^TY)$ can reach its maximum or not depend on whether the matrix $MM^T$ is positive definite or not. For the matrix in your example, $MM^T$ is well positive definite. So, we can diagonalize this matrix as follows

$$MM^T = U^T D U$$ with $U$ the eigenvectors matrix and $D$ a diagonal matrix of eigeinvalues of $MM^T$.

Then from $(1)$ $$ \begin{align} V(a^TY) &= a^T ( U^T D U)a \\ &= (Ua)^T D(Ua) \end{align} $$

The epression $V(a^TY)$ reaches its maximum of the maximum eigenvalue of $M^TM$.

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