I want to evaluate and compare the following two expressions:
$$ T_1 = \mathbb{E}\left\{ \sum_{p=1}^P \sum_{m=1}^M \left( x_{m}^p \right)^2 \right\} \\ T_2 = \mathbb{E}\left\{ \sum_{p=1}^P \sum_{m=1}^M \left( \sum_{n=1}^N d_n x_{N(m-1)+n}^p \right)^2 \right\}, \quad \text{with} \quad \sum_{n=1}^N d_n = 1 $$
where $x_i$ are iid. Gaussian $\sim \mathcal{N}(0,\sigma^2)$ ($i=m$ for the first case and $i=N(m-1)+n$ for the second).
I can show experimentally (in MATLAB) that $T_2 > T_1$ but I fail to show this with algebra.
I think $T_1$ is pretty straight forward:
$$ T_1 = \sum_{p=1}^P \sum_{m=1}^M \mathbb{E}\left\{ x_{m}^{2p} \right\} = M\sum_{p=1}^P \sigma^{2p} (2p-1)!! $$
For the second case my my attempt is:
$$ T_2 = M \sum_{p=1}^P \mathbb{E}\left\{ \left( \sum_{n=1}^N d_n x_{N(m-1)+1}^p \right)^2 \right\} =\\ M \sum_{p=1}^P \mathbb{E}\left\{ \sum_{n=1}^N \left(d_n x_{N(m-1)+n}^p \right)^2 + 2\sum_{n<m} d_n x_{N(m-1)+n}^p d_m x_{N(m-1)+m}^p \right\} $$
Since the variables are iid I assume that the second term is zero and I have:
$$ T_2 = M \sum_{p=1}^P \sum_{n=1}^N d_n^2 \mathbb{E}\left\{ x_{N(m-1)+n}^{2p} \right\} = T_1 \sum_{n=1}^N d_n^2 $$
But since $\sum_{n=1}^N d_n=1$, $\sum_{n=1}^N d_n^2 < 1$ and hence $T_2 < T_1$.
Is there a mistake in this derivation?