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The variance of two linear combinations can be expressed as $\rm{var}({{\bf{a}}^{\rm{T}}}{\bf{X}})={{\bf{a}}^{\rm{T}}}\rm{var}(\bf{X}) {\bf{a}}$ where ${\bf{X}}$ is a vector of random variables and $\bf{a}^\rm{T}$ is a row vector as coefficients. I want to know is there a similar expression for co-variance ${\mathop{\rm cov}} ({\bf{a}}_{\bf{1}}^{\rm{T}}{\bf{X,a}}_{\bf{2}}^{\rm{T}}{\bf{Y}})$. Thank you.

supplement: I have searched the internet and what I can find is this, and I find I am unable to convert it into a vector form.

\begin{align} \sigma(aX+bY, cW+dV) &= ac\,\sigma(X,W)+ad\,\sigma(X,V)+bc\,\sigma(Y,W)+bd\,\sigma(Y,V) \end{align}

Tony
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1 Answers1

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Since $\mathbf {a^TX}$ and $\mathbf{b^TY}$ are scalar random variables, you can apply the elementary definition of covariance (i.e., ${\rm cov}(U,V)=E[(U-EU)(V-EV)]$ for scalar $U$, $V$), and use matrix algebra to simplify: $$\begin{align} \mathrm {cov}(\mathbf{a^TX,b^TY})&=E\left[ (\mathbf{a^TX}-E(\mathbf{a^TX}))(\mathbf{b^TY}-E(\mathbf{b^TY}))\right]\\ &\stackrel{(*)}=E\left[(\mathbf{a^T X}-E(\mathbf{a^T X}))(\mathbf {Y^Tb}-E(\mathbf {Y^Tb}))\right]\\ &=E\left[\mathbf{a^T}(\mathbf X-E\mathbf X)\mathbf{( Y^T-E Y^T) b}\right]\\ &=\mathbf a^TE\left[(\mathbf X-E\mathbf X)\mathbf{( Y-EY)^T}\right]\mathbf b \end{align} $$ In step (*) we use the fact that $\mathbf{b^TY}$ is a scalar, so $\mathbf{b^TY} = \mathbf{[b^TY]^T} = \mathbf{Y^Tb}$.

grand_chat
  • 38,951