We know that the normal form of Bernoulli distribution is, X = 1 with probability p and X = 0 with probability 1 - p. But what is the p.m.f of Bernoulli distribution if $X \in {-1, 1}$?
My approach is, the probability equals p if X = 1 and $p^{-1} (1-p)^2$ if X = -1. Is it correct?
So, to find the mean, using the definition and yields something like $p - \frac{(1-p)^2}{p}$. It seems strange, I am wondering if it is correct or not.
As @Ian points out in the comments, any random variable with 2-point support is just a scaled and shifted Bernoulli random variable. That is, if $X$ has 2 point support $\{a,b\}$ and assigns mass $1-p$ to $a$ and mass $p$ to $b$, then we can write
$$X=a+(b-a)B$$
where $B$ is a standard Bernoulli random variable. From here, the mean and variance can be computed (the mean should be immediate to begin with from the definition):
$$E[X]=a+(b-a)E[B]=a+(b-a)p=a(1-p)+bp\\ V(X)=(b-a)^2V(B)=(b-a)^2p(1-p).$$