How to find $E(|X+Y|^3)$?
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0
down vote
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Supposed that $X,Y$ are
independent random variables, and $Xsim N(mu,sigma^2),Ysim N(-mu,sigma^2)$.
I would like to culculate the value of $E(|X+Y|^3)$.
I have thought of some methods but I was caught into troubles when culculate the integration of $intint vert x+y vert ^3, dxdy$.
I also want to find it by some way like $E((X+Y)^3)=E(X^3)+E(Y^3)+3E(X^2Y)+3E(XY^2)$ but I am not sure that it may really work after some trasformation.
probability expected-value
asked yesterday
Maxius Xu
62
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up vote
0
down vote
favorite
Supposed that $X,Y$ are
independent random variables, and $Xsim N(mu,sigma^2),Ysim N(-mu,sigma^2)$.
I would like to culculate the value of $E(|X+Y|^3)$.
I have thought of some methods but I was caught into troubles when culculate the integration of $intint vert x+y vert ^3, dxdy$.
I also want to find it by some way like $E((X+Y)^3)=E(X^3)+E(Y^3)+3E(X^2Y)+3E(XY^2)$ but I am not sure that it may really work after some trasformation.
probability expected-value
asked yesterday
Maxius Xu
62
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Supposed that $X,Y$ are
independent random variables, and $Xsim N(mu,sigma^2),Ysim N(-mu,sigma^2)$.
I would like to culculate the value of $E(|X+Y|^3)$.
I have thought of some methods but I was caught into troubles when culculate the integration of $intint vert x+y vert ^3, dxdy$.
I also want to find it by some way like $E((X+Y)^3)=E(X^3)+E(Y^3)+3E(X^2Y)+3E(XY^2)$ but I am not sure that it may really work after some trasformation.
probability expected-value
asked yesterday
Maxius Xu
62
Supposed that $X,Y$ are
independent random variables, and $Xsim N(mu,sigma^2),Ysim N(-mu,sigma^2)$.
I would like to culculate the value of $E(|X+Y|^3)$.
I have thought of some methods but I was caught into troubles when culculate the integration of $intint vert x+y vert ^3, dxdy$.
I also want to find it by some way like $E((X+Y)^3)=E(X^3)+E(Y^3)+3E(X^2Y)+3E(XY^2)$ but I am not sure that it may really work after some trasformation.
probability expected-value
probability expected-value
asked yesterday
Maxius Xu
62
asked yesterday
Maxius Xu
62
asked yesterday
Maxius Xu
62
asked yesterday
Maxius Xu
62
asked yesterday
Maxius Xu
62
62
add a comment |
add a comment |
1 Answer
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If $Xsim N(mu,sigma^2)$ and $Ysim N(-mu,sigma^2)$ are independent, then $X+Ysim N(0,2sigma^2)$. So we can compute
begin{align}
mathbb{E}[lvert X+Yrvert^3]&=2mathbb{E}[((X+Y)^+)^3]\
&=2(2sigma^2)^{3/2}mathbb{E}[(Z^3)^+]&&Zsim N(0,1)\
&=frac{2(2sigma^2)^{3/2}}{sqrt{2pi}}int_0^infty z^3exp(-z^2/2),mathrm{d}z\
&=frac{8}{sqrt{pi}}sigma^3\
end{align}
answered yesterday
user10354138
6,189523
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
If $Xsim N(mu,sigma^2)$ and $Ysim N(-mu,sigma^2)$ are independent, then $X+Ysim N(0,2sigma^2)$. So we can compute
begin{align}
mathbb{E}[lvert X+Yrvert^3]&=2mathbb{E}[((X+Y)^+)^3]\
&=2(2sigma^2)^{3/2}mathbb{E}[(Z^3)^+]&&Zsim N(0,1)\
&=frac{2(2sigma^2)^{3/2}}{sqrt{2pi}}int_0^infty z^3exp(-z^2/2),mathrm{d}z\
&=frac{8}{sqrt{pi}}sigma^3\
end{align}
answered yesterday
user10354138
6,189523
add a comment |
up vote
0
down vote
If $Xsim N(mu,sigma^2)$ and $Ysim N(-mu,sigma^2)$ are independent, then $X+Ysim N(0,2sigma^2)$. So we can compute
begin{align}
mathbb{E}[lvert X+Yrvert^3]&=2mathbb{E}[((X+Y)^+)^3]\
&=2(2sigma^2)^{3/2}mathbb{E}[(Z^3)^+]&&Zsim N(0,1)\
&=frac{2(2sigma^2)^{3/2}}{sqrt{2pi}}int_0^infty z^3exp(-z^2/2),mathrm{d}z\
&=frac{8}{sqrt{pi}}sigma^3\
end{align}
answered yesterday
user10354138
6,189523
add a comment |
up vote
0
down vote
up vote
0
down vote
If $Xsim N(mu,sigma^2)$ and $Ysim N(-mu,sigma^2)$ are independent, then $X+Ysim N(0,2sigma^2)$. So we can compute
begin{align}
mathbb{E}[lvert X+Yrvert^3]&=2mathbb{E}[((X+Y)^+)^3]\
&=2(2sigma^2)^{3/2}mathbb{E}[(Z^3)^+]&&Zsim N(0,1)\
&=frac{2(2sigma^2)^{3/2}}{sqrt{2pi}}int_0^infty z^3exp(-z^2/2),mathrm{d}z\
&=frac{8}{sqrt{pi}}sigma^3\
end{align}
answered yesterday
user10354138
6,189523
If $Xsim N(mu,sigma^2)$ and $Ysim N(-mu,sigma^2)$ are independent, then $X+Ysim N(0,2sigma^2)$. So we can compute
begin{align}
mathbb{E}[lvert X+Yrvert^3]&=2mathbb{E}[((X+Y)^+)^3]\
&=2(2sigma^2)^{3/2}mathbb{E}[(Z^3)^+]&&Zsim N(0,1)\
&=frac{2(2sigma^2)^{3/2}}{sqrt{2pi}}int_0^infty z^3exp(-z^2/2),mathrm{d}z\
&=frac{8}{sqrt{pi}}sigma^3\
end{align}
answered yesterday
user10354138
6,189523
answered yesterday
user10354138
6,189523
answered yesterday
user10354138
6,189523
answered yesterday
user10354138
6,189523
6,189523
add a comment |
add a comment |
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