is this the right method of proving:

\begin{align} Cov(a_0+a_1R_1+a_2R_2, Q_1)=a_1Cov(R_1,Q_1)+a_2(R_2,Q_1) \end{align}


\begin{align} =Cov(a_0+a_1R_1+a_2R_2, Q_1) \end{align}

\begin{align} =Cov(a_1R_1+a_2R_2, Q_1) \end{align}

\begin{align} =E[((a_1R_1+a_2R_2)-E(a_1R_1+a_2R_2))(Q-EQ_1)] \end{align}

\begin{align} =a_1Cov(R_1,Q_1)+a_2(R_2,Q_1) \end{align}


\begin{align} E[(X-EX)(Y-EY)] = Cov(X,Y) \end{align}


1 Answer 1


I do not totally understand your first argument $$" =E(a_0+a_1R_1+a_2R_2,Q_1)".$$ I think this should be proven by stating beginning with the definition $$Q_2\equiv a_0+a_1R_1+a_2R_2$$ and assuming that $a_0=E[Q_1]=E[Q_2]=0$. This assumption is without loss of generality because of the demeaning in the definition of the covariance. Then we can writte $$Cov(Q_2,Q_1)=E[Q_1Q_2]=E[a_1R_1Q_1+a_2R_2Q_1]\\=a_1E[R_1Q_1]+a_2E[R_2Q_1]\\=a_1Cov[R_1,Q_1]+a_2Cov[R_2,Q_1]$$ where the first equality follows from the definition, the second equality is just inserting for $Q_2$ and the second line follows from the linearity of the expectation. In the third line the definition of the covariance is used again.

  • $\begingroup$ Thank you. I have corrected the first argument, it was meant to say $Cov$. Is it always to define the terms with another variable, especially when one of the variable is very long? $\endgroup$ Nov 25, 2016 at 14:28
  • $\begingroup$ Your are welcome. In general it is often helpful to do so. $\endgroup$
    – Michael L.
    Nov 25, 2016 at 14:42
  • $\begingroup$ Would following the process for calculating the covariance always hold, for most random variables defined, even when we use the summation sum as well? $\endgroup$ Nov 25, 2016 at 14:45

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