Let $f_n$ be a set of independent, identically distributed random variables such that $\mu_{f_n} (1) = p$ and $\mu_{f_n} (-1) = 1-p.$
I have to calculate the characteristic function and distribution of $f_1 + f_2 + \cdots + f_n$.
My attempts:
Since the $f_i$ are independent, we can multiply their individual characteristic functions to get the overall characteristic function. Moreover, since they all have the same distribution, the individual characteristic functions are the same.
$$\phi_{f_n} (t) = \int_{\mathbb{R}} e^{it f_n} \mu.$$
Since $f_n = 1$ with probability $p$ and $-1$ with probability $1-p$, this becomes: $$p \cdot e^{it} + (1-p) \cdot e^{-it}.$$ Thus, the characteristic function of $f_1 + f_2 + \cdots + f_n$ is $$ \left(p \cdot e^{it} + (1-p) \cdot e^{-it} \right)^n.$$
Now, to get the distribution, we have to take the inverse Fourier Transform: $$f(t) = \frac{1}{2\pi} \int_{\mathbb{R}} e^{-itx} \left(p \cdot e^{it} + (1-p) \cdot e^{-it} \right)^n dx.$$
However, I cannot figure out how to simplify this any further, and am not sure if this approach/solution is even correct.
Am I on the right track?
Your computation of the characteristic function is correct.
The inversion formula is valid if the characteristic function is integrable, which is not the case here.
Moreover, $f_1+\dots+f_n$ takes integer values hence this random variable cannot have a density.
In order to find the distribution of $f_1+\dots+f_n$, notice that $(f_n+1)/2$ has a Bernoulli distribution.