A random vector $V \in \mathbb{R}^n$ is said to be spherically symmetric. If for every orthogonal matrix $A$ the distribution of $AV$ is the same as the distribution of $V$.
My question: Let $W$ be i.i.d. zero mean Gaussian random variable. Can we show that for a spherically symmetric $V$ the conditional expectation \begin{align} E[V|V+W=t]= g(\|t\|). \end{align} That is the conditional expectation is only a function of the magnitude.
My attempt: \begin{align} E[V|V+W=t]= \int V \frac{f_W(t-v)}{f_{V+W}(t)} dP_V(v)= \frac{\int v f_W(t-v) dP_V(v)}{f_{V+W}(t)}= \frac{E[ V f_W(t-V) ]}{f_{V+W}(t)} \end{align}
Therefore, we can show that the two functions in the above ratio only depend on $\|t\|$.
Observe that we can write $t= \|t\| u_t$ where $u_t$ is a unit vector. Then, for any orthogonal matrix $A$ \begin{align} f_{V+W}(t)=f_{V+W}(\|t\|, u_t)&= \frac{1}{(2\pi)^{n/2}} E[ e^{-\frac{\|t-V \|^2}{2}} ]\\ &= \frac{1}{(2\pi)^{n/2}} E[ e^{-\frac{\|t-AV \|^2}{2}} ]\\ &= \frac{1}{(2\pi)^{n/2}} E[ e^{-\frac{\|tA^{-1}-V \|^2}{2}} ]\\ &= \frac{1}{(2\pi)^{n/2}} E[ e^{-\frac{\|t A^T-V \|^2}{2}} ]\\ &= f_{V+W}(tA^T)\\ &= f_{V+W}(\|t\|, u_t A^T) \end{align} Since, $A$ was arbitrary we have that $f_{V+W}(t)=h(\|t\|)$(i.e., only depends on the magnitute).
Now I am not sure how to show that $E[ V f_W(t-V) ]$ only depends on $\|t\|$?
Any function $f:\mathbb R^n\to\mathbb R^n$ satisfying $Af(t)=f(At)$ for all orthogonal $A$ is of the form $f(t)=g(\|t\|)u_t.$ In other words, $f(ru)=g(r)u$ for all $r\geq 0$ and unit vectors $u.$ To see this, first note that $f(ru)$ must be fixed by all orthogonal transformations fixing $u,$ and must therefore be parallel to $u,$ and so of the form $g(r,u)u.$ But then $g(r,Au)Au=f(rAu)=Af(ru)=Ag(r,u)u=g(r,u)Au,$ so $g(r,u)$ is independent of $u.$
For any orthogonal $A,$ $$AE[V\mid V+W=t]=E[AV\mid V+W=t]=E[V\mid A^{-1}V+W=t]=E[V\mid V+W=At]$$ where I have used that $V$ and $W$ are independent, that $V$ has the same distribution as $A^{-1}V,$ and that $W$ has the same distribution as $AW.$ This shows that the function $f$ defined by $f(t)=E[V\mid V+W+t]$ satisfies $Af(t)=f(At).$