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I am trying to answer the following question:

\begin{aligned} &\text { Let } \mathbf{V} \text { be a vector random variable with mean vector } E(\mathbf{V})=\boldsymbol{\mu}_{\mathbf{v}} \text { and covariance matrix }\\ &E\left(\mathbf{V}-\boldsymbol{\mu}_{\mathbf{v}}\right)(\mathbf{V}-\boldsymbol{\mu} \mathbf{v})^{\prime}=\boldsymbol{\Sigma}_{\mathbf{v}} . \text { Show that } E\left(\mathbf{V} \mathbf{V}^{\prime}\right)=\mathbf{\Sigma}_{\mathbf{v}}+\boldsymbol{\mu}_{\mathbf{v}} \boldsymbol{\mu}_{\mathbf{v}}^{\prime} \end{aligned}

I have no idea where to start. Any guidance would be greatly appreciated!

username97
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  • Welcome to MSE. Please type your questions rather than posting images. Images can't be browsed, and are not accessible to those using screen readers. If you need help formatting math on this site, here's a tutorial To begin with, surround all math expressions (including numbers,) with $ signs. Use ^ for exponents and _ for subscripts. $x_1^{2/3}$ shows up as $x_1^{2/3}$. – saulspatz Apr 29 '21 at 15:16
  • Thank you, I have done so! – username97 Apr 29 '21 at 15:20
  • I think $E\left(\mathbf{V}-\boldsymbol{\mu}{\mathbf{v}}\right)(\mathbf{V}-\boldsymbol{\mu} \mathbf{v})^{\prime}$ should be $E\left(\mathbf{V}-\boldsymbol{\mu}{\mathbf{v}}\right)(\mathbf{V}-\boldsymbol{\mu}_\mathbf{v})^{\prime}$ That is, the last $\mathbf v$ is supposed to be a subscript, right? – saulspatz Apr 29 '21 at 15:27

1 Answers1

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$$\begin{align} \Sigma_V&=E(V-\mu_v)(V-\mu_v)'\\&=E(V-\mu_v)(V'-\mu_v')\\ &=E(VV'-\mu_vV'-V\mu_v'+\mu_v\mu_v')\\ &=E(VV')-E(\mu_vV')-E(V\mu_v')+E(\mu_v\mu_v') \end{align}$$ by linearity of expectation.

Obviously, $E(\mu_v\mu_v')=\mu_v\mu_v'$, and you're done once you justify $$E(\mu_vV')=\mu_vE(V')=\mu_v\mu_v'\\ E(V\mu_v')=E(V)\mu_v'=\mu_v\mu_v'$$

saulspatz
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