Mixing
What value of the constant $alpha$ makes $Y$ and $Z$ uncorrelated?
$begingroup$
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?
probability
asked Jan 24 at 12:07
Mark JaconMark Jacon
1127
$endgroup$
add a comment |
$begingroup$
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?
probability
asked Jan 24 at 12:07
Mark JaconMark Jacon
1127
$endgroup$
add a comment |
$begingroup$
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?
probability
asked Jan 24 at 12:07
Mark JaconMark Jacon
1127
$endgroup$
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?
probability
probability
asked Jan 24 at 12:07
Mark JaconMark Jacon
1127
asked Jan 24 at 12:07
Mark JaconMark Jacon
1127
edited Jan 25 at 8:35
Mark Jacon
asked Jan 24 at 12:07
Mark JaconMark Jacon
1127
asked Jan 24 at 12:07
Mark JaconMark Jacon
1127
asked Jan 24 at 12:07
Mark JaconMark Jacon
1127
1127
add a comment |
add a comment |
2 Answers
2
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oldest
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$begingroup$
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?
answered Jan 24 at 12:18
Kavi Rama MurthyKavi Rama Murthy
67.6k53067
$endgroup$
$begingroup$
Now, how can I relate $E[X_1X_2]$ with $E[YZ]$?
$endgroup$
– Mark Jacon
Jan 24 at 13:36
1
$begingroup$
$EYZ=E(X_1-3X_2)(alpha X_1+X_2)=alpha EX_1^{2}-3alpha EX_1X_2+EX_1X_2-3EX_2^{2}$
$endgroup$
– 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
1
$begingroup$
$EX_1^{2}=var(X_1)+(EX_1)^{2}=4+1=5$. similarly, $EX_2^{2}=4+4=8$.
$endgroup$
– Kavi Rama Murthy
Jan 25 at 8:19
add a comment |
$begingroup$
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 $E[X_2^2]=Var(X_2)+(E[X_2])^2=4+2^2=8$
$E[YZ]=E[(X_1 − 3X_2)(alpha X_1 + X_2)]=alpha E[X_1^2]+(1-3 alpha)E[X_1X_2]-3E[X_2^2]=5alpha+(1-3alpha)*2.8-3*8$
$Cov(Y,Z)=E[YZ] − E[Y]E[Z]=0$, $5alpha+(1-3alpha)*2.8-3*8-(-5)*(2+alpha)=0$, $alpha=7$
answered Jan 25 at 8:40
Mark JaconMark Jacon
1127
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
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?
answered Jan 24 at 12:18
Kavi Rama MurthyKavi Rama Murthy
67.6k53067
$endgroup$
$begingroup$
Now, how can I relate $E[X_1X_2]$ with $E[YZ]$?
$endgroup$
– Mark Jacon
Jan 24 at 13:36
1
$begingroup$
$EYZ=E(X_1-3X_2)(alpha X_1+X_2)=alpha EX_1^{2}-3alpha EX_1X_2+EX_1X_2-3EX_2^{2}$
$endgroup$
– 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
1
$begingroup$
$EX_1^{2}=var(X_1)+(EX_1)^{2}=4+1=5$. similarly, $EX_2^{2}=4+4=8$.
$endgroup$
– Kavi Rama Murthy
Jan 25 at 8:19
add a comment |
$begingroup$
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?
answered Jan 24 at 12:18
Kavi Rama MurthyKavi Rama Murthy
67.6k53067
$endgroup$
$begingroup$
Now, how can I relate $E[X_1X_2]$ with $E[YZ]$?
$endgroup$
– Mark Jacon
Jan 24 at 13:36
1
$begingroup$
$EYZ=E(X_1-3X_2)(alpha X_1+X_2)=alpha EX_1^{2}-3alpha EX_1X_2+EX_1X_2-3EX_2^{2}$
$endgroup$
– 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
1
$begingroup$
$EX_1^{2}=var(X_1)+(EX_1)^{2}=4+1=5$. similarly, $EX_2^{2}=4+4=8$.
$endgroup$
– Kavi Rama Murthy
Jan 25 at 8:19
add a comment |
$begingroup$
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?
answered Jan 24 at 12:18
Kavi Rama MurthyKavi Rama Murthy
67.6k53067
$endgroup$
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?
answered Jan 24 at 12:18
Kavi Rama MurthyKavi Rama Murthy
67.6k53067
answered Jan 24 at 12:18
Kavi Rama MurthyKavi Rama Murthy
67.6k53067
answered Jan 24 at 12:18
Kavi Rama MurthyKavi Rama Murthy
67.6k53067
answered Jan 24 at 12:18
Kavi Rama MurthyKavi Rama Murthy
67.6k53067
67.6k53067
$begingroup$
Now, how can I relate $E[X_1X_2]$ with $E[YZ]$?
$endgroup$
– Mark Jacon
Jan 24 at 13:36
1
$begingroup$
$EYZ=E(X_1-3X_2)(alpha X_1+X_2)=alpha EX_1^{2}-3alpha EX_1X_2+EX_1X_2-3EX_2^{2}$
$endgroup$
– 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
1
$begingroup$
$EX_1^{2}=var(X_1)+(EX_1)^{2}=4+1=5$. similarly, $EX_2^{2}=4+4=8$.
$endgroup$
– Kavi Rama Murthy
Jan 25 at 8:19
add a comment |
$begingroup$
Now, how can I relate $E[X_1X_2]$ with $E[YZ]$?
$endgroup$
– Mark Jacon
Jan 24 at 13:36
1
$begingroup$
$EYZ=E(X_1-3X_2)(alpha X_1+X_2)=alpha EX_1^{2}-3alpha EX_1X_2+EX_1X_2-3EX_2^{2}$
$endgroup$
– 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
1
$begingroup$
$EX_1^{2}=var(X_1)+(EX_1)^{2}=4+1=5$. similarly, $EX_2^{2}=4+4=8$.
$endgroup$
– Kavi Rama Murthy
Jan 25 at 8:19
$begingroup$
Now, how can I relate $E[X_1X_2]$ with $E[YZ]$?
$endgroup$
– Mark Jacon
Jan 24 at 13:36
$begingroup$
Now, how can I relate $E[X_1X_2]$ with $E[YZ]$?
$endgroup$
– Mark Jacon
Jan 24 at 13:36
1
1
$begingroup$
$EYZ=E(X_1-3X_2)(alpha X_1+X_2)=alpha EX_1^{2}-3alpha EX_1X_2+EX_1X_2-3EX_2^{2}$
$endgroup$
– Kavi Rama Murthy
Jan 24 at 23:13
$begingroup$
$EYZ=E(X_1-3X_2)(alpha X_1+X_2)=alpha EX_1^{2}-3alpha EX_1X_2+EX_1X_2-3EX_2^{2}$
$endgroup$
– 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
$begingroup$
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
1
1
$begingroup$
$EX_1^{2}=var(X_1)+(EX_1)^{2}=4+1=5$. similarly, $EX_2^{2}=4+4=8$.
$endgroup$
– Kavi Rama Murthy
Jan 25 at 8:19
$begingroup$
$EX_1^{2}=var(X_1)+(EX_1)^{2}=4+1=5$. similarly, $EX_2^{2}=4+4=8$.
$endgroup$
– Kavi Rama Murthy
Jan 25 at 8:19
add a comment |
$begingroup$
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 $E[X_2^2]=Var(X_2)+(E[X_2])^2=4+2^2=8$
$E[YZ]=E[(X_1 − 3X_2)(alpha X_1 + X_2)]=alpha E[X_1^2]+(1-3 alpha)E[X_1X_2]-3E[X_2^2]=5alpha+(1-3alpha)*2.8-3*8$
$Cov(Y,Z)=E[YZ] − E[Y]E[Z]=0$, $5alpha+(1-3alpha)*2.8-3*8-(-5)*(2+alpha)=0$, $alpha=7$
answered Jan 25 at 8:40
Mark JaconMark Jacon
1127
$endgroup$
add a comment |
$begingroup$
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 $E[X_2^2]=Var(X_2)+(E[X_2])^2=4+2^2=8$
$E[YZ]=E[(X_1 − 3X_2)(alpha X_1 + X_2)]=alpha E[X_1^2]+(1-3 alpha)E[X_1X_2]-3E[X_2^2]=5alpha+(1-3alpha)*2.8-3*8$
$Cov(Y,Z)=E[YZ] − E[Y]E[Z]=0$, $5alpha+(1-3alpha)*2.8-3*8-(-5)*(2+alpha)=0$, $alpha=7$
answered Jan 25 at 8:40
Mark JaconMark Jacon
1127
$endgroup$
add a comment |
$begingroup$
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 $E[X_2^2]=Var(X_2)+(E[X_2])^2=4+2^2=8$
$E[YZ]=E[(X_1 − 3X_2)(alpha X_1 + X_2)]=alpha E[X_1^2]+(1-3 alpha)E[X_1X_2]-3E[X_2^2]=5alpha+(1-3alpha)*2.8-3*8$
$Cov(Y,Z)=E[YZ] − E[Y]E[Z]=0$, $5alpha+(1-3alpha)*2.8-3*8-(-5)*(2+alpha)=0$, $alpha=7$
answered Jan 25 at 8:40
Mark JaconMark Jacon
1127
$endgroup$
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 $E[X_2^2]=Var(X_2)+(E[X_2])^2=4+2^2=8$
$E[YZ]=E[(X_1 − 3X_2)(alpha X_1 + X_2)]=alpha E[X_1^2]+(1-3 alpha)E[X_1X_2]-3E[X_2^2]=5alpha+(1-3alpha)*2.8-3*8$
$Cov(Y,Z)=E[YZ] − E[Y]E[Z]=0$, $5alpha+(1-3alpha)*2.8-3*8-(-5)*(2+alpha)=0$, $alpha=7$
answered Jan 25 at 8:40
Mark JaconMark Jacon
1127
answered Jan 25 at 8:40
Mark JaconMark Jacon
1127
answered Jan 25 at 8:40
Mark JaconMark Jacon
1127
answered Jan 25 at 8:40
Mark JaconMark Jacon
1127
1127
add a comment |
add a comment |
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