Vtyjkyuk

Problem: Assume we have an i.i.d data set $barx subset mathbbR^n$, sampled from an $n-$dimensional normal distribution where each dimension is independent, with zero mean and $sigma^2$ variance (i.e. $P(barx) = Pi_k mathcalN(x_k| 0, sigma^2)$). Prove that $var(fracVert barx^i - barx^j Vert_2^2mathbbE[Vert barx^i - barx^j Vert_2^2]) = fracn + 2n - 1$.

Attempt at solution: knowing that $barx^i$ and $barx^j$ are i.i.d. and normal, we know $bary = barx^i - barx^j$ is also normal, and in fact, $P(y_k) = mathcalN(y_k|0, 2 sigma^2)$. So I can simplify the target expression to $$var(fracVert bary Vert_2^2mathbbE[Vert bary Vert_2^2]) = var(fracsum_k y_k^2mathbbE[sum_k y_k^2]) = (frac1sum_k mathbbE[y_k^2])^2 sum_k var(y_k^2) \
= (frac1sum_k mathbbE[y_k^2])^2 sum_k mathbbE[y_k^4] - mathbbE[y_k^2]^2 \
= (frac1sum_k mathbbE[y_k^2])^2 sum_k 12 sigma^4 - mathbbE[y_k^2]^2 $$

where in the second line I used a hint for the problem that $mathbbE[y_i^2 y_j^2] = mathbbE[y_i^2] mathbbE[y_j^2]$, and in third line, I used $mathbbE[y_k^4] = 12sigma^4$ from a hint for the problem. But now I arrive at a wrong result. Any help is appreciated!

Substituting $E[y_k^4] = 12 sigma^4$ and $E[y_k^2] = 2 sigma^2$ into your last line yields
$$fracn(12 - 4) sigma^4(n cdot 2 sigma^2)^2 = frac2n$$
which is the answer you are looking for.