Expected squared dot product between IID Gaussian vectors? Suppose $x,y$ are IID samples from a Gaussian distribution in $\mathbb{R}^d$. The following seems true:
$$2\ \mathbb{E}\left[\langle x, y\rangle^2\right] = \mathbb{E}\left[\|x\|^4\right]-\mathbb{E}\left[\|x\|^2\right]^2+2\ \|\mathbb{E}x\|^4$$
Is this a well known identity? How do I prove this?
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Write $x = (x_i)$ and $y = (y_i)$ so that $\langle x,y\rangle = \sum x_iy_i.$  Further, for the moments write $E[x_ix_j]=\sigma_{ij}.$  The Normality assumption implies
$$E[x_i^4] = 3\sigma_{ii}^2\ \text{ and }\ E[x_i^2x_j^2] = \sigma_{ii}\sigma_{jj} + 2\sigma_{ij}^2.$$
(The right hand formula reduces to the left hand one when $i=j$ anyway.)
Assume temporarily that the mean is zero.  Because $x,y$ are independent and expectation is linear, we have
$$\begin{aligned}
E\left[\right(\sum x_iy_i\left)^2\right] &= E\left[\sum x_iy_i\sum x_jy_j\right] = \sum_{i,j} E\left[x_iy_ix_jy_j\right]\\&= \sum_{i,j} E\left[x_ix_j\right]E\left [y_iy_j\right] \\
&= \sum_{i,j}\sigma_{ij}^2.
\end{aligned}\tag{*}$$
On the other hand,
$$\begin{aligned}
E\left[||x||^4\right] &= E\left[\left(\sum x_i^2\right)^2\right] = E\left[\sum x_i^2\sum x_j^2\right] = \sum_{i,j} E\left[x_i^2x_j^2\right] = \sum_{i,j} \sigma_{ii}\sigma_{jj} + 2\sigma_{ij}^2
\end{aligned}$$
and
$$\begin{aligned}
E\left[||x||^2\right]^2 &= E\left[\sum x_i^2\right]^2 = \sum_{i,j} \sigma_{ii}\sigma_{jj}.
\end{aligned}$$
Subtracting this from the previous result leaves twice $(*),$ demonstrating the equality for zero-mean variables.
The reason why we needn't consider the case where the mean $\mu=(\mu_i)$ is nonzero is because
$$E\left[\sum x_iy_i\right] = \sum E[x_i] E[y_i] = \sum \mu_i^2 = ||\mu||^2$$
shows
$$E\left[\right(\sum x_iy_i\left)^2\right] - ||E[x]||^4 =E\left[\right(\sum x_iy_i\left)^2\right] - E\left[\sum x_iy_i\right]^2 = \operatorname{Var}\left(\sum x_iy_i\right)$$
and, clearly, $E\left[||x||^4\right] - E\left[||x||^2\right]^2 = \operatorname{Var}(||x||^2).$  Thus, your identity equates two variances and, since variances are invariant under change of location, the mean does not affect them, QED.
