Consider a multivariate, centered gaussian $X$. Define $Y:= \|X\|$.
I am wondering how to describe the conditional distribution of $X\mid Y$. Intuitively, by $X$ being centered, one would expect that conditional on $Y=c$, $X$ is uniformly distributed on the circle of radius $c$. That is, I suspect that we have $$\mathbb{E}[\mathbb{I}_A[X]\mid Y] = \int \mathbb{I}_A[Z(u) Y(\omega)] \, d\mathbb{P}(u)$$
where $Z$ is a random variable $Z: \Omega \to \mathbb{R}^n$, independent of $X$, having the property that $Z \sim \operatorname{Unif}(S^n)$, where I don't know exactly how to define this uniformity on the circle correctly. (I guess we would want that $\phi(X) \sim \operatorname{Unif}([0,2\pi))$ for any $\phi: \mathbb{R}^n \to \mathbb{R}$ s.t. $\phi(S^n) = [0,2\pi)$. )
Am I correct in my intuition? If so, how to correctly define $Z$ above? And is there some easy proof?
Later revision:
I will assume you meant $X\sim\operatorname N_{n+1}(\mathbf 0, \sigma^2 I_{n+1}),$ where $\mathbf 0\in\mathbb R^{(n+1)\times1}$ and $I_{n+1}\in\mathbb R^{(n+1)\times(n+1)}$ is the identity matrix. It is well known that $Y^2/\sigma^2\sim\chi^2_{n+1}.$
If $G$ is an $(n+1)\times(n+1)$ orthogonal matrix then $GX\sim\operatorname N_{n+1}(G\mathbf 0, \sigma^2 G(I_{n+1})G^T)$ $= \operatorname N_{n+1}(\mathbf 0, \sigma^2 I_{n+1}).$ Thus the distribution of $GX$ is the same as that of $X.$
Now observe that $\|GX\|^2 = (GX)^T(GX)$ $= (X^TG^T)(GX)$ $= X^T(G^TG)X$ $=X^TX$ $= \|X\|^2.$ Thus the conditional distribution of $X$ given $\|X\|$ is the same as the conditional distribution of $X$ given $\|GX\|,$ and that is the same as the conditional distribution of $GX$ given $\|GX\|,$ since $GX$ has the same distribution as $X.$
The only distribution on $S^n$ that is invariant under all mappings of the form $x\mapsto Gx$ is the uniform distribution.
You need an additional hypothesis that the variance of $X$ is a scalar multiple of the identity matrix rather than some other nonnegative-definite matrix. Thus the scalar components of $X$ should be assumed to be uncorrelated and all to have the same variance. You didn't say that.My first thought is this. To say that $Y\in\mathbb R^{n+1}$ is uniformly distributed on the sphere $S^n \subseteq\mathbb R^{n+1}$ is equivalent to the conjunction of $(1)$ $\Pr(Y\in S^n) = 1$ and $(2)$ the distribution of $Y$ being the same as that of $GY$ for every $(n+1)\times(n+1)$ orthogonal matrix $G$.
Now suppose $U$ is uniformly distributed on $S^n$ and $V\sim \sigma^2\chi^2_{n+1}$ and $U,V$ are independent. Show that then $U\sqrt V$ has just the Gaussian distribution that you have in mind.