Let $Z\stackrel{d}{=}\mu + \xi A U$ where $\xi$ is a non specified (one dimensional) random variable with $\mathbb{E}(\xi^2)=n$ independent of $U$, $U$ is uniformly distributed on the $n$-dimensional sphere $\mathbb{S^{n-1}}$ and A is a $n\times n$ matrix. $Z$ and $\mu$ are vectors of length $n$.

Denote with $A_i$ the i-th row of $A$ and with $Z_i, \mu_i, U_i$ the i-th entry.

Also I know that $\mathbb{E}(U_i)=0$ for all $i$ and the covariance matrix of U equals $\frac{1}{n}\mathbb{I}$, where $\mathbb{I}$ denotes the identity matrix.

Now I want to prove that the convariance of $Z$ equals $A A^T$.

It is obvious that $\mathbb{E}(Z_i)=\mu_i$. Now I calculate $\mathbb{E}(Z_iZ_j)$ and $\mathbb{E}(Z_i^2)$. For this reason we have

$$\mathbb{E}(Z_iZ_j)=\mathbb{E}((\mu_i+\xi A_i U)(\mu_j+\xi A_j U))$$ or

$$\mathbb{E}(Z_i^2)=\mathbb{E}((\mu_i+\xi A_i U)(\mu_i+\xi A_i U))$$

but now I'm stuck what to do next.

Of course I can solve the two brackets but then I get something like

$$\mathbb{E}(\xi A_i U\xi A_i U)$$ I can't just switch the vectors or can I??


1 Answer 1


Splitting in scalar components may make it unnecessarily complicated. I would calculate the covariance matrix for the random vector ${\bf z}$ as a whole, using the dyadic product: $$E({\bf z} ~ {\bf z}^T) = E[({\bf \mu}+ \xi {\bf A} \cdot {\bf u}) ({\bf \mu}+ \xi {\bf A} \cdot {\bf u})^T ].$$ Then $E[({\bf z}-{\bf \mu}) ({\bf z}-{\bf \mu})^T) = E[(\xi {\bf A} \cdot {\bf u})(\xi {\bf u}^T \cdot {\bf A}^T)] = E({\xi}^2) {\bf A} \cdot E({\bf u} {\bf u}^T) \cdot {\bf A}^T$ = $n{\bf A} \cdot ((1/n) {\bf I}) \cdot {\bf A}^T$ = ${\bf A} \cdot {\bf A}^T$.


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