Find combination of $Y=a_1X_1 + a_2X_2$ to get $Y \sim N_1(0,1)$ distribution

Question

Could you please assist me on this example ?

We suppose that $X = (X_1, X_2) \sim N_2(\mu, \Sigma) $ where: $$ \mu = (1, -\frac{1}{2}) $$

$$ \Sigma = \begin{bmatrix} 1 & -3 \\ -3 & 2 \end{bmatrix} $$

I need to find combination $$ Y = a_1 X_1 + a_2 X_2 $$ and Y must be

$$ Y \sim N_1(0,1) $$

My observations:

I was resolving similar issue where I knew the constants $a_1, a_2$ and didn't know the resulting distribution and it's values. According to mentioned example, I would now substitute for $X_1 = 1$ and $X_2 = -\frac{1}{2}$ based on $\mu$ values: $$ Y = a_1 (1) + a_2 (-\frac{1}{2}) $$

In the previous issue I used this formula $ Y \sim N_p(\mu, A \Sigma A^T) $ where $A$ were constants $\begin{bmatrix} a_1 & a_2 \end{bmatrix}$ and for $\mu$ I just computed $Y = a_1 (1) + a_2 (-\frac{1}{2})$ with actual constant values.

I assume I should also achieve to get this matrix: $$ \Sigma = \begin{bmatrix} 1 & \rho \\ \rho & 1 \end{bmatrix} $$

But I don't know how.

Could you please explain me how to resolve this ? I just don't see the logic inside.

Answer 1

If $Y$ is the sum of two scalar random variables, it can only have a univariate distribution. $$Y\sim\mathcal N_1(0,1)$$

So you are looking at $\mathsf E(a_1X_1+a_2X_2)= 0$ and $\mathsf {Var}(a_1X_1+a_2X_2)=1$.

Now you know the mean vector, $\mu=(1, -\tfrac 12)$ gives us an equation of two unknowns: $$a_1-\tfrac{a_2}2 =0 \tag{1}$$

So you must use the covariance matrix $\Sigma =\left[\begin{smallmatrix}~~1 & -3 \\[1ex] -3 & ~~2\end{smallmatrix}\right]$ to obtain the second of two simultaneous equations.

$$\color{orange}{\boxed{\qquad\qquad?}} = 1 \tag{2}$$

You can do this.

$\Diamond$

Big hint, by definition, $\Sigma := \begin{bmatrix}\mathsf{Var}(X_1) & \mathsf{Cov}(X_1, X_2)\\[1ex]\mathsf{Cov}(X_1, X_2) & \mathsf {Var}(X_2)\end{bmatrix}$, and $\mathsf{Var}(a_1X_1+a_2X_2)= \ldots$

Answer 2

If you use the properties for the mean and variance for random variables, you can do the next thing:

$E[Y]=E[a_1X_1]+E[a_2X_2]=a_1E[X_1]+a_2E[X_2] =a_1-\frac{1}{2}a_2=0\rightarrow a_1=\frac{1}{2}a_2$

and

$VAR(Y)=VAR(a_1X_1+a_2X_2)=a_1^2VAR(X_1)+a_2^2VAR(X_2)+2a_1a_2COV(X_1,X_2)=a_1^2+2a_2^2-6a_1a_2=1\rightarrow a_1^2+8a_1^2-12a_1^2=1\rightarrow a_1=\pm i\frac{1}{\sqrt{3}}$

I do not know if your coefficients $a_1,a_2$ have to be real, but for the case you gave, only for complex values $a_1,a_2$ can be found a solution.