I'm wondering about the following quantity $$E\left[\left(\frac{w_1^\intercal w_2}{w_1^\intercal w_1}\right)^2\right] $$ where both $w_1,w_2 \in \mathbb{R}^d \sim N(0,\sigma^2)$. Empirically, the expectation seems to approach zero as the dimensionality $d$ grows (irrespective of the choice of $\sigma^2$) as $O(1/d)$ asymptotically but I wonder if there is a closed form expression (or easy upper bound) for this quantity. Or more generally, what distribution this ratio follows.
I'd be grateful for any hint. Thanks in advance :)
$$E\left[\left(\frac{w_1^\intercal w_2}{w_1^\intercal w_1}\right)^2 \mid w_1\right ]=\frac{1}{{(w_1^\intercal w_1)^2}} w_1^\intercal E(w_2 w_2^\intercal)w_1= \frac{1}{{(w_1^\intercal w_1)^2}} w_1^\intercal \sigma^2 I w_1=\frac{\sigma^2}{w_1^\intercal w_1} $$
Now, let $Z = w_1^\intercal w_1$. Then $E(Z)=d \sigma^2$ and $Var(Z)=2d \sigma^4$.
For large $d$ we can approximate (eg) :
$$ \begin{align} E[1/Z] &\approx \frac{1}{E[Z]} + \frac{1}{2} Var(Z) \frac{2}{(E[Z])^3} +\cdots\\ &=\frac{1}{d \sigma^2} + \frac{2 }{d^2 \sigma^2} + \cdots \end{align}$$
Hence
$$E\left[\left(\frac{w_1^\intercal w_2}{w_1^\intercal w_1}\right)^2 \right] = E \left[E\left[\left(\frac{w_1^\intercal w_2}{w_1^\intercal w_1}\right)^2 \mid w_1\right ] \right ] \approx \frac{1}{d} + \frac{2 }{d^2 } +\cdots \approx \frac{1}{d}$$
The exact value seems to be $ \frac{1}{d-2}$