How to express the probability $P(X \text{ mod } 2 = 0)$ that an integer-valued $X$ is even using characteristic function? Suppose $X$ is an integer-valued random variable. My question is, how can I express the probability $P(X \text{ mod } 2 = 0)$ that $X$ is even knowing the characteristic function $\phi_X(t)$? Can I do it without using a general inversion formula by summing up $P(X=2k)$ over $k \in \mathbb{Z}$? Maybe, I could use some sort of simpler argument and end up with some nice expression?
I almost don't have any idea. I tried to use $\phi_X(2t) = \sum\limits_{k \in \mathbb{Z}} e^{2ikt}P(X=k)$, but it just doesn't work. Do I miss something trivial here?
Actually, this is only a part of the problem that I'm trying to solve. Another part is the following:

Let $\{X_n\}_{n=1}^{+\infty}$ is a sequence of i.i.d random variables, and $P(X_1 \text{ mod } 2 = 0) = p \in (0, 1)$. Let $Y_n = X_1 + X_2 + ... + X_n$. Prove that $P(Y_n \text{ mod } 2 = 0) = 1/2$.

And I want to solve this problem using characteristic functions. I suspect, for this I would need to find $\phi_{Y_n}(t) = (\phi_{Z_1}(t))^n$ with $Z_i = X_i \text{ mod } 2$.
 A: You're making this way too complicated. $Y_n$ is even if and only if an even number of $\{X_1, \dots, X_n\}$ are odd. So
$$P(Y_n \text{ is even}) = \sum_{k \text{ even}} \binom{n}{k} (1 - p)^kp^{n-k}.$$
So one strategy is to compare this sum to the expansion of $(p + q)^n$ where $q = 1 - p$ (so of course $p + q = 0$).
For instance, one can show that
$$\sum_{k \text{ even}} \binom{n}{k} = \sum_{k \text{ odd}} \binom{n}{k},$$
by looking at the binomial expansion of $(1 - 1)^n$.
A: Ok let's try something else. We'll start with the probability generating function
$$f_X(q) = \sum_{k \in \mathbf Z} \mathrm{P}(X = k) q^k. $$
So the characteristic function is where $q = e^{it}$.
Notice:

*

*$f_{Y_n} = f_X^n$


*$\displaystyle \tfrac12(f_X(q) + f_X(-q)) = \sum_{k \text{ even}} \mathrm{P}(X = k) q^k$
This is called "bisecting" the series.
So the probability that $Y$ is even is
$$\mathrm{P}(Y \text{ is even}) = \frac{f_X(1)^n + f_X(-1)^n}{2} = \frac12 + \frac{f_X(-1)^n}{2}.$$
You can write this in terms of the characteristic function by choosing appropriate values for $q = e^{it}$ (namely $t = 0, t = \pi$).
Also notice that
$$\mathrm{P}(X \text{ is even}) = p = \frac12 + \frac{f_X(-1)}{2}. $$
So combining these gives the limit of $1/2$.
