# What value of the constant $\alpha$ makes $Y$ and $Z$ uncorrelated?

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?

## 2 Answers

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?

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

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 $$E[X_2^2]=Var(X_2)+(E[X_2])^2=4+2^2=8$$

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

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