- #1
Zaare
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How can I show that if [tex]\phi(t)[/tex] is a characteristic function for some distribution, then [tex]|\phi(t)|^2[/tex] is also a characteristic function?
Characteristic function (Probability)
- Context: Graduate
- Thread starter Zaare
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Discussion Overview
The discussion revolves around the properties of characteristic functions in probability theory, specifically exploring whether the square of the modulus of a characteristic function is also a characteristic function. Participants delve into the definitions, relationships between random variables, and the implications of independence in distributions.
Discussion Character
- Technical explanation
- Mathematical reasoning
- Debate/contested
Main Points Raised
- One participant asks how to demonstrate that if \(\phi(t)\) is a characteristic function, then \(|\phi(t)|^2\) is also a characteristic function.
- Another participant suggests working out the distribution directly and inquires about theorems regarding characteristic functions.
- A participant references the definition of the characteristic function as \(E(e^{itX})\) and attempts to find the distribution related to \(|\phi(t)|^2\) but reports difficulty.
- One participant mentions that if \(Y\) is a random variable with distribution as \(-X\), then its characteristic function should be the complex conjugate of \(\phi\), which they find plausible.
- There is a discussion about the requirement of independence for the sum of two distributions' characteristic functions to equal the product of their characteristic functions.
- A participant expresses uncertainty about the implications of their findings regarding the relationship between \(|\phi_X(t)|^2\) and the characteristic function of the sum of independent random variables.
- Another participant confirms that the relationship between \(\phi_Y(t)\) and \(\phi_X(t)\) holds, providing a detailed breakdown of the expressions involved.
- One participant considers using a change of variable in the integral to further explore the properties of the characteristic functions.
Areas of Agreement / Disagreement
Participants express various viewpoints regarding the properties and implications of characteristic functions, with no clear consensus reached on the sufficiency of the arguments presented or the interpretation of the results.
Contextual Notes
Some discussions involve assumptions about independence and the nature of the random variables involved, which may not be fully resolved. The relationship between the characteristic functions and their distributions is explored but remains complex and nuanced.
- #2
Hurkyl
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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.
)
Do you have some sort of theorem on what functions are characteristic functions?
(My text on this stuff is at work.
)- #3
Zaare
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It is given that [tex]\phi_{X} (t) = E(e^{itX})[/tex], and I know that [tex]E(e^{itX}) = \int_{-\infty}^{\infty}{e^{itx}f_X (x) dx}[/tex]
[tex]f_X (x)[/tex] is the probability density function.
I'm trying to find the distribution for it as you said, but I haven't succeeded yet.
[tex]f_X (x)[/tex] is the probability density function.
I'm trying to find the distribution for it as you said, but I haven't succeeded yet.
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- #4
matt grime
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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
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
- #5
Zaare
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I think I can show this, but how does that help me?matt grime said:
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 said:
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- #6
matt grime
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Yes, so what;s the problem with having Y independent? I didn't say it was -X i said its distribution was -X.
- #7
Zaare
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Ok, if Y has destribution as -X and X and Y are independent, I can show this:
[tex] \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|[/tex]
But is it enough? I don't know how to interpret this.
[tex] \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|[/tex]
But is it enough? I don't know how to interpret this.
- #8
Zaare
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By the way, this shows your first statement, right?
[tex] \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)} }} [/tex]
[tex] \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)} }} [/tex]
matt grime said:
- #9
matt grime
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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
If X is an r.v, so is Y, and so is X+Y... that's sufficient
- #10
Zaare
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Thank you for all the help. :)
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