Let $\mathbf{Y}$ be a $p$-dimensional random vector with $\mathbb{E}|\mathbf{Y}|^2<\infty$. Show that $$\text{Cov}(\mathbf{Y})=\mathbb{E}[(\mathbf{Y}-\mathbb{E}\mathbf{Y})\mathbf{Y}^T]$$
I tried using the fact that $$\text{Cov}(\mathbf{Y})=\mathbb{E}[(\mathbf{Y}-\mathbb{E}\mathbf{Y})(\mathbf{Y}-\mathbb{E}\mathbf{Y})^T]$$
and then expanding it and simplifying it, but I couldn't derive the former.
How do I go about with this?
The difference between the two terms is $E[(Y-EY) (EY)^\top]$. Since $(EY)^\top$ is not random, you can pull it out to obtain $E[(Y-EY)] (EY)^\top$. Can you conclude from here?
The only thing you have to remind is $E[E[Y]] = E[Y]$.
By definition, $$ \begin{align} \text{Cov}(Y) &= E[(Y-E[Y])(Y-E[Y])^\top]\\ &= E[YY^\top - YE[Y]^\top - E[Y] Y^\top +E[Y]E[Y]^\top]\\ &= E[YY^\top]-E[Y]E[Y]^\top \end{align} $$
The form you wrote is $$ E[(Y - E[Y])Y^\top] = E[YY^\top - E[Y]Y^\top] = E[YY^\top] - E[Y]E[Y]^\top $$ which gives what you want.
