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Suppose $Z$ is a $p\times n$ matrix and $v$ is a $n\times 1$ vector. Both have entries that are independent, standard normals (in particular $Z$ and $v$ themselves are independent). What is the distribution of the vector $w=Zv$?

I had read that it has the distribution of $$w \sim \chi_n \cdot \chi_p \cdot U_p$$ where $\chi^2_k$ is a chi-squared variable with $k$ degrees of freedom and $U$ is a vector uniform on the surface of a $p-1$ dimensional sphere. I wasn't sure how this was obtained.

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  • $\begingroup$ have you tried writing out the matrix multiplication terms explicitly $\endgroup$ Mar 2, 2021 at 17:37

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First, the distribution of $v \in N(0, I_n)$ is spherically symmetric. For suppose $A^T A = I_n$. Then $(Av)_i = \sum_j A_{ij} v_j$ is a sum of independent normal random variables, so it's normal with variance $\sum_j A_{ij}^2 = 1$. It follows that $v/|v| \sim U_n$. Moreover, $|v| \sim \chi_n$ by definition, so $v \sim \chi_n \cdot U_n$. For essentially the same reason, $w/|w| \sim U_p$, so all that remains is to show $|w| \sim \chi_n \cdot \chi_p$.

Let $Z^i$ be the $i$th row of $Z$. For any particular value of $v$, we may pick an orthogonal matrix $A$ depending on $v$ such that $Av = |v|e_1$. Let $Y^i := AZ^i \sim N(0, I_n)$, which remains spherically symmetric for the same reasons as before. Thus

\begin{align*} w_i &= \langle Z^i, v\rangle = \langle AZ^i, Av\rangle \\ &= \langle Y^i, |v|e_1\rangle \\ &= |v|Y_i^1 \sim \chi_n \cdot N(0, 1). \end{align*}

Hence \begin{align*} |w| &= \left(\sum_i w_i^2\right)^{1/2} \\ &= \left(\sum_i |v|^2 (Y_i^1)^2\right)^{1/2} \\ &= |v| \left(\sum_i (Y_i^1)^2\right)^{1/2} \sim \chi_n \cdot \chi_p. \end{align*}

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