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Combinations of characteristic functions: $\alpha\phi_1+(1-\alpha)\phi_2$

If $\phi(t)$ is the characteristic function of a random variable $X$, then $\Re(\phi(t))$ is also a characteristic function of some random variable $Y$ (This can be easily seen by Polya's criterion). How can we find $Y$? ($\Re(\cdot)$ detones the real part of a complex number)

I have already read this related question but I cannot figure out the solution. How can we find the characteristic function of $AX+(1-A)(-X)$?

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marked as duplicate by Did, Alexander Gruber, cardinal, Sasha, Davide Giraudo Jan 13 '13 at 19:57

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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Think probabilistically about it. (Hint: Introduce an independent r.v. uniform on $\{-1,+1\}$.) –  cardinal Jan 13 '13 at 16:25
    
Isn't the characteristic function explicitly given in did's answer to question you link to? –  Stefan Hansen Jan 13 '13 at 16:33
    
Yes. My problem is how can we calculate $\mathbb{E}e^{it(AX+(1-A)(-X))}$, in order to show that the characteristic function is indeed $\Re(\phi(t))$. –  Nick Papadopoulos Jan 13 '13 at 16:42
    
Nick: Say, what prevented you to ask this on the other page? –  Did Jan 13 '13 at 17:26
    
I thought that since the other question was answered and the OP was satisfied with the answer, it would be better to ask a new question linking back to the other. Should I have asked my question in the comments of the old answer? –  Nick Papadopoulos Jan 14 '13 at 12:11

1 Answer 1

up vote 3 down vote accepted

Let $A$ be independent of $X$ with $P(A=1)=P(A=0)=\frac{1}{2}$. Then $$ E\left[e^{it\{AX+(1-A)(-X)\}}\right]=\frac{1}{2}E\left[e^{itX}\right]+\frac{1}{2}E\left[e^{it(-X)}\right]=\frac{\phi(t)+\phi(-t)}{2}, $$ but using that $\cos$ is even and $\sin$ is odd, we obtain $$ \phi(-t)=E\left[e^{i(-t)X}\right]=E[\cos(-tX)]+iE[\sin(-tX)]=E[\cos(tX)]-iE[\sin(tX)] $$ and so $\phi(t)+\phi(-t)=2E[\cos(tX)]=2\Re(\phi(t))$, which yields the result.

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So, we have that: $\mathbb{E}\left[e^{it\{AX+(1-A(-X)}\right]=\mathbb{E}\left[e^{it\{AX+(1-A(-X)\}‌​}\cdot I_B\right]+\mathbb{E}\left[e^{it\{AX+(1-A(-X)}\cdot I_{B^c}\right]=\mathbb{E}\left[e^{itX}\cdot I_B\right]+\mathbb{E}\left[e^{it(-X)}\cdot I_{B^c}\right]=\mathbb{P}(B)\mathbb{E}\left[e^{itX}\right]+ \mathbb{P} (B^c)\mathbb{E}\left[e^{it(-X)}\right]$ where $B=\{\omega: A(\omega)=1\}$. Right? –  Nick Papadopoulos Jan 13 '13 at 17:00
    
Yes, exactly. Here we use the independence of $A$ and $X$. –  Stefan Hansen Jan 13 '13 at 17:01
    
Thanks a lot for your answer! –  Nick Papadopoulos Jan 13 '13 at 17:05
    
(+1) Good exposition. –  cardinal Jan 13 '13 at 17:05

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