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Question

Let $X_1,\cdots, X_n,$ be independently and identically distributed from $N(0,1)$, and try to find, given $X_1^2+X_2^2+\cdots+X_n^2=1$ condition, the joint distribution of $(X_1,\ldots,X_n)$.

$$ f(x_1,\ldots,x_n \mid X_1^2+\cdots+X_n^2=1) $$

Is it possible to use the polar coordinate transformation?

\begin{cases} x_{1}=r\cos\varphi _{1} \\[1ex] x_{2}=r\sin\varphi_{1}\cos\varphi_{2} \\[1ex] x_{3}=r\sin\varphi_{1}\sin\varphi_{2} \\[1ex] \cdot\cdot\cdot\cdot\cdot\cdot \\[1ex] x_{n-1}=r\sin\varphi_{1}\cdot\cdot\cdot \sin\varphi_{n-2}\cos\varphi_{n-1} \\[1ex] x_{n}=r\sin\varphi_{1}\cdot\cdot\cdot \sin\varphi_{n-2}\sin\varphi_{n-1} \end{cases}

$$0\leq r\leq +\infty \\ 0\leq\varphi_{1}\leq \pi \\ 0\leq\varphi_{2}\leq \pi \\ \cdot\cdot\cdot 0\leq\varphi_{n-1}\leq\ 2\pi$$ Jacobi determinant of the transformation :$\mathcal{J}$ $$\mathcal{J}=r^{n-1}\mathcal{J_{1}}= r^{n-1}\sin^{n-2}\varphi_{1}\sin^{n-3}\varphi_{2}...\sin\varphi_{n-2}$$

StubbornAtom
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okko
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1 Answers1

1

More of a comment than an answer...

Let $X=(X_1,X_2,\ldots,X_n)$ be the standard multivariate normal vector.

Then $\frac{X}{\lVert X \rVert}$ is independent of $\lVert X \rVert$. Check this and its linked posts for details.

Therefore, the conditional distribution of $X$ given $\lVert X \rVert=1$ is same as the distribution of $\frac{X}{\lVert X \rVert}$:

Let $S^{n-1}=\{x\in \mathbb R^n: \lVert x \rVert=1\}$ be the unit $(n-1)$-sphere.

Then, for $u \in S^{n-1}$,

$$ P(X\le u\mid \lVert X \rVert=1)=P\left(\frac{X}{\lVert X \rVert}\le u \mid \lVert X \rVert=1 \right) =P\left(\frac{X}{\lVert X \rVert}\le u \right) $$

And the distribution of the normalized vector $\frac{X}{\lVert X \rVert}$ is uniform on $S^{n-1}$ (see this and this).

StubbornAtom
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