Computing the variance of a series of random variables I have the following random variables: $U$ is distributed uniformly with variance $1/3$ and mean $0$; $G$ is distributed normally with variance $1$ and mean $0$. Both random variables are independent of one another. I have the following series: $$\frac{2}{T}\sum_{j=1}^{T/2}G_{2j}+\frac{2}{T}\sum_{i=1}^{T/2}U_{2i-1}$$
(where e.g. $G_{2j}$ means that the $2j^ { \space th}$ term is the random variable $G$—and similarily for $U_{2i-1}$).
I want to compute the variance of this series. Is the following correct?
$$\begin{array}{l}
\left(\frac{2}{T}\right)^{2} \sum_{j=1}^{T / 2} \operatorname{var}\left(G_{2 j}\right)+\left(\frac{2}{T}\right)^{2} \sum_{i=1}^{T / 2} \operatorname{var}\left(U_{2 i-1}\right)\\=
\left(\frac{2}{T}\right)^{2}\left(\frac{T}{2}\right)+\left(\frac{2}{T}\right)^{2}\left(\frac{1}{3} \frac{T}{2}\right)=\frac{2}{T}+\frac{2}{T} \frac{1}{2}=\frac{2}{T}+\frac{2}{3 T}=\\
\frac{6+2}{3 T}=\frac{8}{3 T}
\end{array}$$
Thank you.

Good job, your answer is correct, we do not need the properties that they are uniform or normal.
We just need their variance and the fact that they are independent to perform the computation.
$$\begin{array}{l}
\left(\frac{2}{T}\right)^{2} \sum_{j=1}^{T / 2} \operatorname{var}\left(G_{2 j}\right)+\left(\frac{2}{T}\right)^{2} \sum_{i=1}^{T / 2} \operatorname{var}\left(U_{2 i-1}\right)\\=
\left(\frac{2}{T}\right)^{2}\left(\frac{T}{2}\right)+\left(\frac{2}{T}\right)^{2}\left(\frac{1}{3} \frac{T}{2}\right)=\frac{2}{T}+\frac{2}{T} \frac{1}{\color{blue}3}=\frac{2}{T}+\frac{2}{3 T}=\\
\frac{6+2}{3 T}=\frac{8}{3 T}
\end{array}$$
As $T$ increases, the variance decreases.
Assumption: not only that $G$ and $U$ are independent, we also neethat each pair of the random variable being independent
