$E(X^2|X-Y) E(X^3|X-2Y)$ for Gaussians?
$begingroup$
For independent gaussians with following the normal distribution with expectation zero and variance one, how do I compute:
$E(X^2|X-2Y), E(X^3|X-2Y)$
I know that $X-2Y$,$X+2Y$ are independent. However, this does not seem to be enough to deduce the result, without a restriction such as:
$E(X^2|X-2Y)=-2E(Y^2|X-2Y)$
(I am not sure if this holds).
conditional-expectation
asked Nov 27 '18 at 9:32
DoleDole
901514
$endgroup$
add a comment |
$begingroup$
For independent gaussians with following the normal distribution with expectation zero and variance one, how do I compute:
$E(X^2|X-2Y), E(X^3|X-2Y)$
I know that $X-2Y$,$X+2Y$ are independent. However, this does not seem to be enough to deduce the result, without a restriction such as:
$E(X^2|X-2Y)=-2E(Y^2|X-2Y)$
(I am not sure if this holds).
conditional-expectation
asked Nov 27 '18 at 9:32
DoleDole
901514
$endgroup$
add a comment |
$begingroup$
For independent gaussians with following the normal distribution with expectation zero and variance one, how do I compute:
$E(X^2|X-2Y), E(X^3|X-2Y)$
I know that $X-2Y$,$X+2Y$ are independent. However, this does not seem to be enough to deduce the result, without a restriction such as:
$E(X^2|X-2Y)=-2E(Y^2|X-2Y)$
(I am not sure if this holds).
conditional-expectation
asked Nov 27 '18 at 9:32
DoleDole
901514
$endgroup$
For independent gaussians with following the normal distribution with expectation zero and variance one, how do I compute:
$E(X^2|X-2Y), E(X^3|X-2Y)$
I know that $X-2Y$,$X+2Y$ are independent. However, this does not seem to be enough to deduce the result, without a restriction such as:
$E(X^2|X-2Y)=-2E(Y^2|X-2Y)$
(I am not sure if this holds).
conditional-expectation
conditional-expectation
asked Nov 27 '18 at 9:32
DoleDole
901514
asked Nov 27 '18 at 9:32
DoleDole
901514
edited Nov 27 '18 at 9:38
Dole
asked Nov 27 '18 at 9:32
DoleDole
901514
asked Nov 27 '18 at 9:32
DoleDole
901514
asked Nov 27 '18 at 9:32
DoleDole
901514
901514
add a comment |
add a comment |
1 Answer
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$begingroup$
$X=frac {(X+2Y)+(X-2Y)} 2$. Compute $X^{2}$ and $X^{3}$ in terms of $X+2Y$ and $X-2Y$ from this. [ $$E(X^{2}|X-2Y)=frac {E((X+2Y)^{2}|X-2Y)+E((X-2Y)^{2}|X-2Y)+2E((X+2Y)E((X-2Y)|X-2Y)} 4$$ $$=[E((X+2Y)^{2} +(X-2Y)^{2}+2(X-2Y)E(X+2Y)] /4=[E((X+2Y)^{2} +(X-2Y)^{2}] /4$$ $$=frac 5 4+frac {(X-2Y)^{2}} 4.$$
This answer was written assuming that the OP was right is saying that $X+2Y$ and $X-2Y$ are independent. They are not, so a slight modification is required. See my comment below for the modification.
answered Nov 27 '18 at 9:38
Kavi Rama MurthyKavi Rama Murthy
56k42158
$endgroup$
$begingroup$
Wait now I am thinking I made a mistake... $X-2Y$ may not actually be independent from $X+2Y$. It should probably be $2X+Y$ that is independent from $X-2Y$.
$endgroup$
– Dole
Nov 27 '18 at 11:02
1
$begingroup$
Use $X=frac {(X-2Y)+2(2X+Y)} 5$.
$endgroup$
– Kavi Rama Murthy
Nov 27 '18 at 11:49
add a comment |
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$begingroup$
$X=frac {(X+2Y)+(X-2Y)} 2$. Compute $X^{2}$ and $X^{3}$ in terms of $X+2Y$ and $X-2Y$ from this. [ $$E(X^{2}|X-2Y)=frac {E((X+2Y)^{2}|X-2Y)+E((X-2Y)^{2}|X-2Y)+2E((X+2Y)E((X-2Y)|X-2Y)} 4$$ $$=[E((X+2Y)^{2} +(X-2Y)^{2}+2(X-2Y)E(X+2Y)] /4=[E((X+2Y)^{2} +(X-2Y)^{2}] /4$$ $$=frac 5 4+frac {(X-2Y)^{2}} 4.$$
This answer was written assuming that the OP was right is saying that $X+2Y$ and $X-2Y$ are independent. They are not, so a slight modification is required. See my comment below for the modification.
answered Nov 27 '18 at 9:38
Kavi Rama MurthyKavi Rama Murthy
56k42158
$endgroup$
$begingroup$
Wait now I am thinking I made a mistake... $X-2Y$ may not actually be independent from $X+2Y$. It should probably be $2X+Y$ that is independent from $X-2Y$.
$endgroup$
– Dole
Nov 27 '18 at 11:02
1
$begingroup$
Use $X=frac {(X-2Y)+2(2X+Y)} 5$.
$endgroup$
– Kavi Rama Murthy
Nov 27 '18 at 11:49
add a comment |
$begingroup$
$X=frac {(X+2Y)+(X-2Y)} 2$. Compute $X^{2}$ and $X^{3}$ in terms of $X+2Y$ and $X-2Y$ from this. [ $$E(X^{2}|X-2Y)=frac {E((X+2Y)^{2}|X-2Y)+E((X-2Y)^{2}|X-2Y)+2E((X+2Y)E((X-2Y)|X-2Y)} 4$$ $$=[E((X+2Y)^{2} +(X-2Y)^{2}+2(X-2Y)E(X+2Y)] /4=[E((X+2Y)^{2} +(X-2Y)^{2}] /4$$ $$=frac 5 4+frac {(X-2Y)^{2}} 4.$$
This answer was written assuming that the OP was right is saying that $X+2Y$ and $X-2Y$ are independent. They are not, so a slight modification is required. See my comment below for the modification.
answered Nov 27 '18 at 9:38
Kavi Rama MurthyKavi Rama Murthy
56k42158
$endgroup$
$begingroup$
Wait now I am thinking I made a mistake... $X-2Y$ may not actually be independent from $X+2Y$. It should probably be $2X+Y$ that is independent from $X-2Y$.
$endgroup$
– Dole
Nov 27 '18 at 11:02
1
$begingroup$
Use $X=frac {(X-2Y)+2(2X+Y)} 5$.
$endgroup$
– Kavi Rama Murthy
Nov 27 '18 at 11:49
add a comment |
$begingroup$
$X=frac {(X+2Y)+(X-2Y)} 2$. Compute $X^{2}$ and $X^{3}$ in terms of $X+2Y$ and $X-2Y$ from this. [ $$E(X^{2}|X-2Y)=frac {E((X+2Y)^{2}|X-2Y)+E((X-2Y)^{2}|X-2Y)+2E((X+2Y)E((X-2Y)|X-2Y)} 4$$ $$=[E((X+2Y)^{2} +(X-2Y)^{2}+2(X-2Y)E(X+2Y)] /4=[E((X+2Y)^{2} +(X-2Y)^{2}] /4$$ $$=frac 5 4+frac {(X-2Y)^{2}} 4.$$
This answer was written assuming that the OP was right is saying that $X+2Y$ and $X-2Y$ are independent. They are not, so a slight modification is required. See my comment below for the modification.
answered Nov 27 '18 at 9:38
Kavi Rama MurthyKavi Rama Murthy
56k42158
$endgroup$
$X=frac {(X+2Y)+(X-2Y)} 2$. Compute $X^{2}$ and $X^{3}$ in terms of $X+2Y$ and $X-2Y$ from this. [ $$E(X^{2}|X-2Y)=frac {E((X+2Y)^{2}|X-2Y)+E((X-2Y)^{2}|X-2Y)+2E((X+2Y)E((X-2Y)|X-2Y)} 4$$ $$=[E((X+2Y)^{2} +(X-2Y)^{2}+2(X-2Y)E(X+2Y)] /4=[E((X+2Y)^{2} +(X-2Y)^{2}] /4$$ $$=frac 5 4+frac {(X-2Y)^{2}} 4.$$
This answer was written assuming that the OP was right is saying that $X+2Y$ and $X-2Y$ are independent. They are not, so a slight modification is required. See my comment below for the modification.
answered Nov 27 '18 at 9:38
Kavi Rama MurthyKavi Rama Murthy
56k42158
edited Nov 27 '18 at 11:55
answered Nov 27 '18 at 9:38
Kavi Rama MurthyKavi Rama Murthy
56k42158
answered Nov 27 '18 at 9:38
Kavi Rama MurthyKavi Rama Murthy
56k42158
answered Nov 27 '18 at 9:38
Kavi Rama MurthyKavi Rama Murthy
56k42158
56k42158
$begingroup$
Wait now I am thinking I made a mistake... $X-2Y$ may not actually be independent from $X+2Y$. It should probably be $2X+Y$ that is independent from $X-2Y$.
$endgroup$
– Dole
Nov 27 '18 at 11:02
1
$begingroup$
Use $X=frac {(X-2Y)+2(2X+Y)} 5$.
$endgroup$
– Kavi Rama Murthy
Nov 27 '18 at 11:49
add a comment |
$begingroup$
Wait now I am thinking I made a mistake... $X-2Y$ may not actually be independent from $X+2Y$. It should probably be $2X+Y$ that is independent from $X-2Y$.
$endgroup$
– Dole
Nov 27 '18 at 11:02
1
$begingroup$
Use $X=frac {(X-2Y)+2(2X+Y)} 5$.
$endgroup$
– Kavi Rama Murthy
Nov 27 '18 at 11:49
$begingroup$
Wait now I am thinking I made a mistake... $X-2Y$ may not actually be independent from $X+2Y$. It should probably be $2X+Y$ that is independent from $X-2Y$.
$endgroup$
– Dole
Nov 27 '18 at 11:02
$begingroup$
Wait now I am thinking I made a mistake... $X-2Y$ may not actually be independent from $X+2Y$. It should probably be $2X+Y$ that is independent from $X-2Y$.
$endgroup$
– Dole
Nov 27 '18 at 11:02
1
1
$begingroup$
Use $X=frac {(X-2Y)+2(2X+Y)} 5$.
$endgroup$
– Kavi Rama Murthy
Nov 27 '18 at 11:49
$begingroup$
Use $X=frac {(X-2Y)+2(2X+Y)} 5$.
$endgroup$
– Kavi Rama Murthy
Nov 27 '18 at 11:49
add a comment |
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