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The random variables $X_1$ and $X_2$ have means 1 and 2 respectively, with same variance 4 and covariance 0.8. Let $Y = X_1 − 3X_2$ and $Z = \alpha X_1 + X_2$. What value of the constant α makes Y and Z uncorrelated?

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It is given that covariance of $X_1$ and $X_2$ is $0.8$ Hence $EX_1X_2-(EX_1) (EX_2)=0.8$. So $EX_1X_2 =0.8+(1)(2)=2.8$. Now can you proceed?

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  • $\begingroup$ Now, how can I relate $E[X_1X_2]$ with $E[YZ]$? $\endgroup$ – Mark Jacon Jan 24 at 13:36
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    $\begingroup$ $EYZ=E(X_1-3X_2)(\alpha X_1+X_2)=\alpha EX_1^{2}-3\alpha EX_1X_2+EX_1X_2-3EX_2^{2}$ $\endgroup$ – Kavi Rama Murthy Jan 24 at 23:13
  • $\begingroup$ Ok thank you, now the last question, how can I calculate $E[X_1^2]$ and $E[X_2^2]$, then I can complete the question :) $\endgroup$ – Mark Jacon Jan 25 at 8:17
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    $\begingroup$ $EX_1^{2}=var(X_1)+(EX_1)^{2}=4+1=5$. similarly, $EX_2^{2}=4+4=8$. $\endgroup$ – Kavi Rama Murthy Jan 25 at 8:19
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Solve:

$E[X_1]=1$, $E[X_2]=2$, $\operatorname{Var}(X_1)=4$, $\operatorname{Var}(X_2)=4$ and $\operatorname{Cov}(X_1,X_2)=0.8$

$Y, Z$ are uncorrelated if $\operatorname{Cov}(Y,Z)=0$ where $\operatorname{Cov}(Y,Z) = E[YZ] − E[Y]E[Z]$.

$E[Y]=E[X_1]-3E[X_2]=1-3*2=-5$

$E[Z]=\alpha E[X_1]+E[X_2]=\alpha+2$

$Cov(X_1,X_2)=E[X_1X_2]-E[X_1]*E[X_2]=0.8$ so $E[X_1X_2]=0.8+2*1=2.8$

$E[X_1^2]=Var(X_1)+(E[X_1])^2=4+1^2=5$ and $E[X_2^2]=Var(X_2)+(E[X_2])^2=4+2^2=8$

$E[YZ]=E[(X_1 − 3X_2)(\alpha X_1 + X_2)]=\alpha E[X_1^2]+(1-3 \alpha)E[X_1X_2]-3E[X_2^2]=5\alpha+(1-3\alpha)*2.8-3*8$

$Cov(Y,Z)=E[YZ] − E[Y]E[Z]=0$, $5\alpha+(1-3\alpha)*2.8-3*8-(-5)*(2+\alpha)=0$, $\alpha=7$

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