Showing $\sum a_iX_i$ and $\sum b_iX_i$ are independent iff $\sum a_ib_i=0$ where $X_i's$ are i.i.d $N(\mu,\sigma^2)$

Question

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.

Answer 1

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 $A\mathbf X$ has normal distribution where $A$ is an $m\times n$ matrix.

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

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

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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:$$\mathsf{Cov}(Y,Z)=\sum_{i=1}^n\sum_{j=1}^na_ib_j\mathsf{Cov}(X_i,X_j)=\sigma^2\sum_{i=1}^na_ib_i$$

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

Answer 2

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}^n\sum_{j=1}^n a_i b_j^\ast\times \text{Cov}( X_i, X_j)\\ &=\sum_{i=1}^n\sum_{j=1}^n a_i b_j^\ast\times \sigma^2\mathbf{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 $\sum_{i=1}^n a_i b_j^\ast=0$.