Expectation of the product of three normal variables
$begingroup$
Let $(X_1, X_2, X_3)sim N(mu,Sigma)$ be a three-dimensional random variable where each coordinates are dependent (i.e. $Sigma$ has non-zero values outside of its diagonal)
I want to know how to compute
$E(X_1 X_2^2)$
$E(X_1 X_2 X_3)$
Thanks.
EDIT : I think I know how to do it.
Using Isserlis Theorem, we know that $E((X_1-mu_1) (X_2-mu_2) (X_3-mu_3))=0$.
Also, using the identity $E(X_i X_j) = E(X_i)E(X_j) + Cov(X_i,X_j)$
I expand the expression $E((X_1-mu_1) (X_2-mu_2) (X_3-mu_3))=0$ and use the covariance identity to get:
$$E(X_1 X_2 X_3)=E(X_1) E(X_2) E(X_3)
+E(X_1)Cov(X_2,X_3)
+E(X_2)Cov(X_1,X_3)
+E(X_3)Cov(X_1,X_2)$$
or put differently
$$E(X_1 X_2 X_3)=mu_1mu_2mu_3
+mu_1sigma_{23}
+mu_2sigma_{13}
+mu_3sigma_{12}$$
To get $E(X_1 X_2^2)$, Can I use the derived result and say that $X_3=X_2$ all the time ?
I simply replace the "3"s by 2 in the formula to get
$$E(X_1 X_2^2)=E(X_1 X_2 X_2)=mu_1mu_2mu_2
+mu_1sigma_{22}
+mu_2sigma_{12}
+mu_2sigma_{12}
=mu_1(mu_2^2
+sigma_2^2)
+2mu_2sigma_{12}
$$
Can somebody confirm that I am not doing something wrong ?
Thank you.
normal-distribution covariance expected-value
asked Dec 13 '18 at 21:20
quantguyquantguy
835
$endgroup$
add a comment |
$begingroup$
Let $(X_1, X_2, X_3)sim N(mu,Sigma)$ be a three-dimensional random variable where each coordinates are dependent (i.e. $Sigma$ has non-zero values outside of its diagonal)
I want to know how to compute
$E(X_1 X_2^2)$
$E(X_1 X_2 X_3)$
Thanks.
EDIT : I think I know how to do it.
Using Isserlis Theorem, we know that $E((X_1-mu_1) (X_2-mu_2) (X_3-mu_3))=0$.
Also, using the identity $E(X_i X_j) = E(X_i)E(X_j) + Cov(X_i,X_j)$
I expand the expression $E((X_1-mu_1) (X_2-mu_2) (X_3-mu_3))=0$ and use the covariance identity to get:
$$E(X_1 X_2 X_3)=E(X_1) E(X_2) E(X_3)
+E(X_1)Cov(X_2,X_3)
+E(X_2)Cov(X_1,X_3)
+E(X_3)Cov(X_1,X_2)$$
or put differently
$$E(X_1 X_2 X_3)=mu_1mu_2mu_3
+mu_1sigma_{23}
+mu_2sigma_{13}
+mu_3sigma_{12}$$
To get $E(X_1 X_2^2)$, Can I use the derived result and say that $X_3=X_2$ all the time ?
I simply replace the "3"s by 2 in the formula to get
$$E(X_1 X_2^2)=E(X_1 X_2 X_2)=mu_1mu_2mu_2
+mu_1sigma_{22}
+mu_2sigma_{12}
+mu_2sigma_{12}
=mu_1(mu_2^2
+sigma_2^2)
+2mu_2sigma_{12}
$$
Can somebody confirm that I am not doing something wrong ?
Thank you.
normal-distribution covariance expected-value
asked Dec 13 '18 at 21:20
quantguyquantguy
835
$endgroup$
add a comment |
$begingroup$
Let $(X_1, X_2, X_3)sim N(mu,Sigma)$ be a three-dimensional random variable where each coordinates are dependent (i.e. $Sigma$ has non-zero values outside of its diagonal)
I want to know how to compute
$E(X_1 X_2^2)$
$E(X_1 X_2 X_3)$
Thanks.
EDIT : I think I know how to do it.
Using Isserlis Theorem, we know that $E((X_1-mu_1) (X_2-mu_2) (X_3-mu_3))=0$.
Also, using the identity $E(X_i X_j) = E(X_i)E(X_j) + Cov(X_i,X_j)$
I expand the expression $E((X_1-mu_1) (X_2-mu_2) (X_3-mu_3))=0$ and use the covariance identity to get:
$$E(X_1 X_2 X_3)=E(X_1) E(X_2) E(X_3)
+E(X_1)Cov(X_2,X_3)
+E(X_2)Cov(X_1,X_3)
+E(X_3)Cov(X_1,X_2)$$
or put differently
$$E(X_1 X_2 X_3)=mu_1mu_2mu_3
+mu_1sigma_{23}
+mu_2sigma_{13}
+mu_3sigma_{12}$$
To get $E(X_1 X_2^2)$, Can I use the derived result and say that $X_3=X_2$ all the time ?
I simply replace the "3"s by 2 in the formula to get
$$E(X_1 X_2^2)=E(X_1 X_2 X_2)=mu_1mu_2mu_2
+mu_1sigma_{22}
+mu_2sigma_{12}
+mu_2sigma_{12}
=mu_1(mu_2^2
+sigma_2^2)
+2mu_2sigma_{12}
$$
Can somebody confirm that I am not doing something wrong ?
Thank you.
normal-distribution covariance expected-value
asked Dec 13 '18 at 21:20
quantguyquantguy
835
$endgroup$
Let $(X_1, X_2, X_3)sim N(mu,Sigma)$ be a three-dimensional random variable where each coordinates are dependent (i.e. $Sigma$ has non-zero values outside of its diagonal)
I want to know how to compute
$E(X_1 X_2^2)$
$E(X_1 X_2 X_3)$
Thanks.
EDIT : I think I know how to do it.
Using Isserlis Theorem, we know that $E((X_1-mu_1) (X_2-mu_2) (X_3-mu_3))=0$.
Also, using the identity $E(X_i X_j) = E(X_i)E(X_j) + Cov(X_i,X_j)$
I expand the expression $E((X_1-mu_1) (X_2-mu_2) (X_3-mu_3))=0$ and use the covariance identity to get:
$$E(X_1 X_2 X_3)=E(X_1) E(X_2) E(X_3)
+E(X_1)Cov(X_2,X_3)
+E(X_2)Cov(X_1,X_3)
+E(X_3)Cov(X_1,X_2)$$
or put differently
$$E(X_1 X_2 X_3)=mu_1mu_2mu_3
+mu_1sigma_{23}
+mu_2sigma_{13}
+mu_3sigma_{12}$$
To get $E(X_1 X_2^2)$, Can I use the derived result and say that $X_3=X_2$ all the time ?
I simply replace the "3"s by 2 in the formula to get
$$E(X_1 X_2^2)=E(X_1 X_2 X_2)=mu_1mu_2mu_2
+mu_1sigma_{22}
+mu_2sigma_{12}
+mu_2sigma_{12}
=mu_1(mu_2^2
+sigma_2^2)
+2mu_2sigma_{12}
$$
Can somebody confirm that I am not doing something wrong ?
Thank you.
normal-distribution covariance expected-value
normal-distribution covariance expected-value
asked Dec 13 '18 at 21:20
quantguyquantguy
835
asked Dec 13 '18 at 21:20
quantguyquantguy
835
edited Dec 14 '18 at 16:55
quantguy
asked Dec 13 '18 at 21:20
quantguyquantguy
835
asked Dec 13 '18 at 21:20
quantguyquantguy
835
asked Dec 13 '18 at 21:20
quantguyquantguy
835
835
add a comment |
add a comment |
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