Let $X_i$ be iid random variables.
It equals to $1$ with probability $(1+p)/2$ and equals to $-1$ with probability $(1-p)/2$.
My question is: What is the probability that $\sum_{i=1}^nX_i>0$?
I tried the following deduction, but I'm very unsure about the correctness, since I plugged this result into my original problem, it seems incorrect. Could someone help to see whether I did this in the right way or if not, where could be the issue?
Note that $\frac{X_i+1}{2}:=B_i$ is Bernoulli$(p)$ (it equals to $1$ with probability $p$). Thus $P(\sum_{i=1}^nX_i>0)=P(\sum B_i>\frac{n}{2})$. And $\sum B_i$ obeys Binomial distribution with parameter $n,p$. Moreover, $\mathbb{E}(\sum_{i=1}^nB_i)=np$.
Now I split the problem into two cases:
(1) $\mathbb{E}(\sum_{i=1}^nB_i)<\frac{n}{2}$, i.e. $p<\frac{1}{2}$ and
(2) $\mathbb{E}(\sum_{i=1}^nB_i)>\frac{n}{2}$, i.e. $p>\frac{1}{2}$
To solve the case (1).
$$P(\sum B_i>\frac{n}{2})=P\Big(\big(\sum B_i\big)-np>\frac{n}{2}-np\Big)\geq \frac{1}{(n+1)^2}\exp\{-nD(\frac{\frac{n+1}{2}-np}{n+1}\|p)\}$$
Here I used the estimation here
To solve the case (2).
$$P(\sum B_i>\frac{n}{2})=1-P(\sum B_i<\frac{n}{2})$$
We compute the upper bound of $P(\sum B_i<\frac{n}{2})$.
$$P(\sum B_i<\frac{n}{2})=P(\sum B_i-\mathbb{E}(\sum B_i)\leq \frac{n}{2}-\mathbb{E}(\sum B_i))\leq \exp\{-np\}(\frac{enp}{t})^t$$ where $t=\frac{n}{2}-np$ where I used Chernoff’s inequality for lower tails of Bernoulli sum. See Exercise 2.3.2.
Thus $$P(\sum B_i>\frac{n}{2})=1-P(\sum B_i<\frac{n}{2})\geq 1- \exp\{-np\}(\frac{enp}{t})^t$$ where $t=\frac{n}{2}-np$
