Let the pair $(X, Y)$ have the bivariate normal distribution. Show that $aX + bY$ has a univariate normal distribution Let the pair $(X, Y)$ have the bivariate normal distribution of (6.76), and let $a, b \in \mathbb{R}$.
Show that $aX + bY$ has a univariate normal distribution, possibly with zero variance.
$$g(x,y)=\frac{1}{2\pi\sigma_1\sigma_2\sqrt{1-\rho^2}}e^{-\frac{1}{2}Q(x,y)} \text{ for } x,y \in \mathbb{R}, \tag{6.76} $$
where
$$Q(x,y)=\frac{1}{1-\rho^2}\left[\left(\frac{x-\mu_1}{\sigma_1} \right)^2-2\rho \left(\frac{x-\mu_1}{\sigma_1}\right) \left(\frac{y-\mu_2}{\sigma_2}\right) + \left(\frac{y-\mu_2}{\sigma_2}\right)^2 \right]$$ for $\mu_1,\mu_2 \in \mathbb{R}, \sigma_1,\sigma_2 >0, -1<\rho<1$.
I have no idea how to do this problem. I was thinking maybe trying to split $g(x,y)$ into separate marginal density functions but that math seems way too complicated considering what $g$ is.
 A: We do the following reversible transformation on the random vector
$$
\left[\begin{array}{l}
U \\
Y
\end{array}\right]=\left[\begin{array}{cc}
a & b \\
-b & a
\end{array}\right]\left[\begin{array}{l}
X \\
Y
\end{array}\right] 
$$
where require
$$
\left|\begin{array}{cc}
a & b \\
-b & a
\end{array}\right|=a^{2}+b^{2} \neq 0
$$
This means $a$ and $b$ are not zero at the same time. Now what we know is the PDF of  random vector $\left[\begin{array}{l}
X \\
Y
\end{array}\right] $
,that is $g(x,y)$ as mentioned above.
Then we can get the PDF of  random vector $\left[\begin{array}{l}
U \\
V
\end{array}\right] $ , denote it by $q(u,v)$.
Note the Jacobi Determinant $|J| = \frac{1}{a^2+b^2}$. To simplify the result, without the generality, we can assume $a = \cos\alpha, b=\sin\alpha$ where $0\le\alpha\le2\pi$ , after compulation, we can get the following fomular
$$
\begin{aligned}
q(u, v) &=g(u \cos \alpha-v \sin \alpha, u \sin \alpha+v \cos \alpha) \\
&=\frac{1}{2 \pi \sigma_{1} \sigma_{2} \sqrt{1-\rho^{2}}} \exp \left\{-\frac{1}{2\left(1-\rho^{2}\right)}\left(A u^{2}-2 B u v+C v^{2}\right)\right\}
\end{aligned}
$$
where
$$
\begin{array}{l}
A=\frac{\cos ^{2} \alpha}{\sigma_{1}^{2}}-2 \rho \frac{\cos \alpha \sin \alpha}{\sigma_{1} \sigma_{2}}+\frac{\sin ^{2} \alpha}{\sigma_{2}^{2}} \\
B=\frac{\cos \alpha \sin \alpha}{\sigma_{1}^{2}}-\rho \frac{\sin ^{2} \alpha-\cos ^{2} \alpha}{\sigma_{1} \sigma_{2}}-\frac{\cos \alpha \sin \alpha}{\sigma_{2}^{2}} \\
C=\frac{\sin ^{2} \alpha}{\sigma_{1}^{2}}+2 \rho \frac{\cos \alpha \sin \alpha}{\sigma_{1} \sigma_{2}}+\frac{\cos ^{2} \alpha}{\sigma_{2}^{2}}
\end{array}
$$
Now we get the following conclusion:
1. The random vector obtained by the coordinate rotation transformation of the two-dimensional normal vector still obeys the normal distribution.
Now what we want to know is the PDF of random variable $U = aX+bY$ which is just Marginal probability density function of $g(x,y)$ , the Join Probability Density Fuction
There is another conclusion about Normal Distribution as following.
2. The marginal distribution of the bivariate normal distribution is still the normal distribution
That what we just want to get!!!
Hope my answer can help you!!!
