Vrftjkry

Let $X_1, X_2,ldots, X_n$ be i.i.d with the distribution $N(mu, sigma^2)$.

Prove $Y =sum_{i=1}^{n} a_iX_i$ and $Z =sum_{i=1}^{n} b_iX_i$ are independent iff $sum a_ib_i=0$.

I have proved it by Basu's theorem.

I am searching for another methods.
For example, if I could prove that $(Y,Z)sim N(mu_1,mu_2,sigma_1^2,sigma_2^2,rho)$, it seems easier to show their independence.

In general if $X_1,dots,X_n$ are independent and all have normal distribution then random vector $mathbf X:=(X_1,dots,X_n)^T$ has normal distribution.

To get light on this realize that there is a common PDF that is the product of the PDF's of the $X_i$.

Then also in general random vector $Amathbf X$ has normal distribution where $A$ is an $mtimes n$ matrix.

Now note that $(Y,Z)^T$ can be written as $Amathbf X$ where $A$ is an $2times n$ matrix.

First row of $A$ is $(a_1,dots, a_n)$ and second row of $A$ is $(b_1,dots, b_n)$.

Here $Y$ and $Z$ have a joint normal distribution and in that case they are independent if they are uncorrelated.

So it suffices to prove that their covariance equals $0$.

Using bilinearity of $mathsf{Cov}$ we find:$$mathsf{Cov}(Y,Z)=sum_{i=1}^nsum_{j=1}^na_ib_jmathsf{Cov}(X_i,X_j)=sigma^2sum_{i=1}^na_ib_i$$

So that $$mathsf{Cov}(Y,Z)=0iffsum_{i=1}^na_ib_i=0$$

Since your variables are jointly Gaussian, then the independence of $Y$ and $Z$ is given by their decoration. Observe that
begin{align*}
text{Cov}(Y,Z) &= text{Cov}(sum_{i=1}^n a_i X_i, sum_{j=1}^n b_j X_j)\
&=sum_{i=1}^nsum_{j=1}^n a_i b_j^asttimes text{Cov}( X_i, X_j)\
&=sum_{i=1}^nsum_{j=1}^n a_i b_j^asttimes sigma^2mathbf{1} (i=j)\
&=sigma^2 sum_{i=1}^n a_i b_j^ast
end{align*}

So $X$ and $Y$ are independents if and only if $text{Cov}(Y,Z)=0$ which is if and only if $sum_{i=1}^n a_i b_j^ast=0$.