Characteristic function (Probability)

[Zaare]

How can I show that if \phi(t) is a characteristic function for some distribution, then |\phi(t)|^2 is also a characteristic function?

 

[Hurkyl]

Maybe you could directly work out the distribution for it?

Do you have some sort of theorem on what functions are characteristic functions?

(My text on this stuff is at work. )

 

[Zaare]

It is given that \phi_{X} (t) = E(e^{itX}), and I know that E(e^{itX}) = \int_{-\infty}^{\infty}{e^{itx}f_X (x) dx}
f_X (x) is the probability density function.
I'm trying to find the distribution for it as you said, but I haven't succeeded yet.

 

[matt grime]

Right, here are the useful results google gave me in the first hit.

Let Y be another r.v. with distribtuion as -X, then its characteristic function out to be the complex conjugate of phi, I think. At least that seems plausible.

Given two distributions X and Y, the char function of their sum is the product of their char functions.

That is for sure, though I'm not sure about the first bit - you should try the integration

 

[Zaare]

Let Y be another r.v. with distribtuion as -X, then its characteristic function out to be the complex conjugate of phi, I think. At least that seems plausible.

I think I can show this, but how does that help me?

Given two distributions X and Y, the char function of their sum is the product of their char functions.

X and Y have to be independent for this to be ture, right? And I have to show that this is true for X and X even though X and X are not independent.

 

[matt grime]

Yes, so what;s the problem with having Y independent? I didn't say it was -X i said its distribution was -X.

 

[Zaare]

Ok, if Y has destribution as -X and X and Y are independent, I can show this:
<br /> \left| {\phi _X \left( t \right)} \right|^2 = \left| {\phi _X \left( t \right)} \right| \times \left| {\overline {\phi _X \left( t \right)} } \right| = \left| {\phi _X \left( t \right)} \right| \times \left| {\phi _Y \left( t \right)} \right| = \left| {\phi _{X + Y} \left( t \right)} \right|<br />
But is it enough? I don't know how to interpret this.

 

[Zaare]

By the way, this shows your first statement, right?

<br /> \left. \begin{array}{l}<br /> \phi _Y \left( t \right) = E\left[ {e^{itY} } \right] = E\left[ {e^{ - itX} } \right] = E\left[ {\cos x - i\sin x} \right] = E\left[ {\cos x} \right] - iE\left[ {\sin x} \right] \\ <br /> \phi _X \left( t \right) = E\left[ {e^{itX} } \right] = E\left[ {\cos x + i\sin x} \right] = E\left[ {\cos x} \right] + iE\left[ {\sin x} \right] \\ <br /> \end{array} \right\} \Rightarrow \underline{\underline {\phi _Y \left( t \right) = \overline {\phi _X \left( t \right)} }} <br />

Let Y be another r.v. with distribtuion as -X, then its characteristic function out to be the complex conjugate of phi, I think. At least that seems plausible.

 

[matt grime]

Yeah, that's it, though I was thinking of going for a change of variable in the integral, that was all.

If X is an r.v, so is Y, and so is X+Y... that's sufficient

 

[Zaare]

Thank you for all the help. :)