# Distribution of product of normal random variables

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.

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

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*}