Bivariate normal: Expected value and variance
$begingroup$
I have this simple exercise to do and I'm new to this topic.
Seeing the slides of my professor, I would solve the problem in this way:
$E(Y)=c_1 μ_1+c_2 μ_2$
$E(Y)=-54.2424$
$Var(Y)=σ_{22}$
$Var(Y)=0.65$
I don't have the solutions so I take this opportunity to know if my reasoning has errors or not. If there are, how should I proceed to solve it?
calculus variance expected-value bivariate-distributions
asked Dec 10 '18 at 18:04
Martina MartyMartina Marty
225
$endgroup$
add a comment |
$begingroup$
I have this simple exercise to do and I'm new to this topic.
Seeing the slides of my professor, I would solve the problem in this way:
$E(Y)=c_1 μ_1+c_2 μ_2$
$E(Y)=-54.2424$
$Var(Y)=σ_{22}$
$Var(Y)=0.65$
I don't have the solutions so I take this opportunity to know if my reasoning has errors or not. If there are, how should I proceed to solve it?
calculus variance expected-value bivariate-distributions
asked Dec 10 '18 at 18:04
Martina MartyMartina Marty
225
$endgroup$
add a comment |
$begingroup$
I have this simple exercise to do and I'm new to this topic.
Seeing the slides of my professor, I would solve the problem in this way:
$E(Y)=c_1 μ_1+c_2 μ_2$
$E(Y)=-54.2424$
$Var(Y)=σ_{22}$
$Var(Y)=0.65$
I don't have the solutions so I take this opportunity to know if my reasoning has errors or not. If there are, how should I proceed to solve it?
calculus variance expected-value bivariate-distributions
asked Dec 10 '18 at 18:04
Martina MartyMartina Marty
225
$endgroup$
I have this simple exercise to do and I'm new to this topic.
Seeing the slides of my professor, I would solve the problem in this way:
$E(Y)=c_1 μ_1+c_2 μ_2$
$E(Y)=-54.2424$
$Var(Y)=σ_{22}$
$Var(Y)=0.65$
I don't have the solutions so I take this opportunity to know if my reasoning has errors or not. If there are, how should I proceed to solve it?
calculus variance expected-value bivariate-distributions
calculus variance expected-value bivariate-distributions
asked Dec 10 '18 at 18:04
Martina MartyMartina Marty
225
asked Dec 10 '18 at 18:04
Martina MartyMartina Marty
225
asked Dec 10 '18 at 18:04
Martina MartyMartina Marty
225
asked Dec 10 '18 at 18:04
Martina MartyMartina Marty
225
asked Dec 10 '18 at 18:04
Martina MartyMartina Marty
225
225
add a comment |
add a comment |
1 Answer
1
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$begingroup$
Your expression for $E(Y)$ is correct. The variance of $Y$ won't be $sigma_{22}$ - that is the variance of $X_2$. For the variance of $Y$ use the equation
$$Var(Y) = Var(c_1 X_1 + c_2 X_2) = c_1^2 Var(X_1) + c_2^2 Var(X_2) + 2c_1 c_2 Cov(X_1,X_2)$$
which is the general equation for the variance of a linear combination of two random variables.
answered Dec 10 '18 at 18:12
ODFODF
1,486510
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add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Your expression for $E(Y)$ is correct. The variance of $Y$ won't be $sigma_{22}$ - that is the variance of $X_2$. For the variance of $Y$ use the equation
$$Var(Y) = Var(c_1 X_1 + c_2 X_2) = c_1^2 Var(X_1) + c_2^2 Var(X_2) + 2c_1 c_2 Cov(X_1,X_2)$$
which is the general equation for the variance of a linear combination of two random variables.
answered Dec 10 '18 at 18:12
ODFODF
1,486510
$endgroup$
add a comment |
$begingroup$
Your expression for $E(Y)$ is correct. The variance of $Y$ won't be $sigma_{22}$ - that is the variance of $X_2$. For the variance of $Y$ use the equation
$$Var(Y) = Var(c_1 X_1 + c_2 X_2) = c_1^2 Var(X_1) + c_2^2 Var(X_2) + 2c_1 c_2 Cov(X_1,X_2)$$
which is the general equation for the variance of a linear combination of two random variables.
answered Dec 10 '18 at 18:12
ODFODF
1,486510
$endgroup$
add a comment |
$begingroup$
Your expression for $E(Y)$ is correct. The variance of $Y$ won't be $sigma_{22}$ - that is the variance of $X_2$. For the variance of $Y$ use the equation
$$Var(Y) = Var(c_1 X_1 + c_2 X_2) = c_1^2 Var(X_1) + c_2^2 Var(X_2) + 2c_1 c_2 Cov(X_1,X_2)$$
which is the general equation for the variance of a linear combination of two random variables.
answered Dec 10 '18 at 18:12
ODFODF
1,486510
$endgroup$
Your expression for $E(Y)$ is correct. The variance of $Y$ won't be $sigma_{22}$ - that is the variance of $X_2$. For the variance of $Y$ use the equation
$$Var(Y) = Var(c_1 X_1 + c_2 X_2) = c_1^2 Var(X_1) + c_2^2 Var(X_2) + 2c_1 c_2 Cov(X_1,X_2)$$
which is the general equation for the variance of a linear combination of two random variables.
answered Dec 10 '18 at 18:12
ODFODF
1,486510
answered Dec 10 '18 at 18:12
ODFODF
1,486510
answered Dec 10 '18 at 18:12
ODFODF
1,486510
answered Dec 10 '18 at 18:12
ODFODF
1,486510
1,486510
add a comment |
add a comment |
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