## Re: Characteristic Function Question

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• ... I have never done this calculation before, so my level of confidence is not too high, but here is what I would do in the case when X~N (0,1) [unfortunately
Message 1 of 21 , Aug 2, 2003
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> Can someone show me how to obtain the characteristic
> function for X^2, where X ~ N(mu,sigma^2)?

I have never done this calculation before, so my level of confidence
is not too high, but here is what I would do in the case when X~N
(0,1) [unfortunately I can easily extend it to N(0,s^2), but I don't
have an answer for the general case N(m,s^2)]

Given u in R, we want E[exp(iuX^2)] where X~N(0,1), i.e. we want the
integral:

I=(1/sqrt(2pi))int_R exp(iux^2-x^2/2)dx
=(1/sqrt(2pi))int_R exp(-cx^2/2)dx

where c is the complex number c=1-2iu. Using Fubini theorem, we have:

I^2=(1/2pi)int_R^2 exp(-c(x^2+y^2)/2)dxdy

and after switching to polar coordinates and integrating w.r. to the
angle:

I^2=int_0^+oo rexp(-cr^2/2)dr=1/c=(1+2iu)/(1+4u^2) (*)

The complex number 1+2iu has two roots:

+ or - [((q(u)+1)/2)^(1/2)+isgn(u)((q(u)-1)/2)^1/2] (**)

where q(u)=sqrt(1+4u^2) and sgn(u)=u/|u| (for u<>0) is the sign of u.

It is not too difficult to convince yourself that for u>0, the
imaginary part of I should be positive. From this fact and (*) and
(**) we obtain:

I=(1/q(u))[((q(u)+1)/2)^(1/2)+isgn(u)((q(u)-1)/2)^1/2]

I am sorry I can't do better that this for the time being.

Regards Noel.
• The generalized version of this question is in fact to find the characteristic function of Y=X1^2+X2^2+...+Xn^2 where X1, X2,....are independent normal random
Message 2 of 21 , Aug 2, 2003
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The generalized version of this question is in fact to
find the characteristic function of
Y=X1^2+X2^2+...+Xn^2
where X1, X2,....are independent normal random
variable N(Ui, 1) i=1,2,....n. This question arises
from my course work in graduate multivariate analysis.
Now here the question is simpler because n=1. So
direct derivation like the way Noel does makes the
best sense and is possible. But when n is arbitrary,
you need more work and that's something you need from
a book like "Introduction to Multivariate Analysis" by
T. Anderson. I am not gonna type the long derivation.
But I believe the solution should be:

(1-2*i*t*s^2)^(-1/2)*exp{i*t*s^2*mu^2*(1-2*i*t*s^2)^(-1)}

--- Noel Vaillant <vaillant@...>
的郵件內容：
---------------------------------
> Can someone show me how to obtain the characteristic
> function for X^2, where X ~ N(mu,sigma^2)?

I have never done this calculation before, so my level
of confidence
is not too high, but here is what I would do in the
case when X~N
(0,1) [unfortunately I can easily extend it to
N(0,s^2), but I don't
have an answer for the general case N(m,s^2)]

Given u in R, we want E[exp(iuX^2)] where X~N(0,1),
i.e. we want the
integral:

I=(1/sqrt(2pi))int_R exp(iux^2-x^2/2)dx
=(1/sqrt(2pi))int_R exp(-cx^2/2)dx

where c is the complex number c=1-2iu. Using Fubini
theorem, we have:

I^2=(1/2pi)int_R^2 exp(-c(x^2+y^2)/2)dxdy

and after switching to polar coordinates and
integrating w.r. to the
angle:

I^2=int_0^+oo rexp(-cr^2/2)dr=1/c=(1+2iu)/(1+4u^2) (*)

The complex number 1+2iu has two roots:

+ or - [((q(u)+1)/2)^(1/2)+isgn(u)((q(u)-1)/2)^1/2]
(**)

where q(u)=sqrt(1+4u^2) and sgn(u)=u/|u| (for u<>0) is
the sign of u.

It is not too difficult to convince yourself that for
u>0, the
imaginary part of I should be positive. From this fact
and (*) and
(**) we obtain:

I=(1/q(u))[((q(u)+1)/2)^(1/2)+isgn(u)((q(u)-1)/2)^1/2]

I am sorry I can't do better that this for the time
being.

Regards Noel.