I am given that $X$ has c.f. $\phi(t)$, I need to find the random variable whose c.f. is equal to $\frac{1}{2-\phi(t)}$ in terms of $X$.
My idea is that express $\frac{1}{2-\phi(t)}$ as a series, since $|\phi(t)| \leq 1$ so we have
$$
\frac{1}{2-\phi(t)} = \sum_{n = 0}^{\infty}\frac{\phi(t)^n}{2^{n+1}}
$$
From the question asked here, I am guessing that this c.f. corresponds to the random variable (I may be wrong):
$$
Z = \sum_{n = 0}^{\infty}I(A = n)Z_n
$$
where $P(A = n) = \frac{1}{2^{n+1}}, \ n = 0,1,2,…$ and $Z_n = \sum_{i = 1}^{n}Y_i$, $Z_0 = 0$, $Y_i$ are iid r.v.'s s.t. $Y_i \sim X$.
But I don't know how to prove this, can anyone point out a general direction? Thanks so much!
Best Answer
You already got the answer, you are just not completely writing that out. In probability context we usually write
$$\displaystyle Z = \sum_{i=1}^A Y_i$$
as a random sum, with the convention that $Z = 0$ when $A = 0$. We assumed $A$ is independent of $Y_i$ also, and they are defined as what you written. Then by law of total expectation, the characteristic function of $Z$ is
$$ \begin{align} \phi_Z(t) &= E\left[\exp\left\{itZ\right\}\right] \\ &= E\left[\exp\left\{it\sum_{i=1}^A Y_i\right\}\right] \\ &= \sum_{n=0}^{\infty} E\left[\exp\left\{it\sum_{i=1}^A Y_i\right\} \Bigg| A=n \right] \Pr\{A = n\} \\ &= \Pr\{A = 0\} + \sum_{n=1}^{\infty} E\left[\exp\left\{it\sum_{i=1}^n Y_i\right\}\right] \Pr\{A = n\} \\ &= \frac {1} {2} + \sum_{n=1}^{\infty} E\left[\prod_{i=1}^n\exp\left\{it Y_i\right\}\right] \frac {1} {2^{n+1}} \\ &= \frac {1} {2} + \sum_{n=1}^{\infty} \prod_{i=1}^n E\left[\exp\left\{it Y_i\right\}\right] \frac {1} {2^{n+1}} \\ &= \frac {1} {2} + \sum_{n=1}^{\infty} \phi_X(t)^n \frac {1} {2^{n+1}} \\ &= \sum_{n=0}^{\infty} \frac {\phi_X(t)^n} {2^{n+1}} \\ &= \frac {1/2} {1 - \phi_X(t)/2} \\ &= \frac {1} {2 - \phi_X(t)} \end{align}$$
where line $1$ using the definition of characteristic function, line $2$ using the definition of $Z$, line $3$ using the law of total expectation, line $4$ using the independence of $A$ and $Y_i$ and the convention of the random sum, line $5$ using the basic property of exponential function, line $6$ using the independence of $Y_i$, line $7$ using the definition of characteristic function of $Y_i$ and they are identically distributed with the identical CF $\phi_X$, and the remaining lines are just some algebra to simplify the expression.
I have not study those complex analysis so if there is any holes in the above steps in regarding complex number, please help to fill them. I guess this should work.