# Confidence interval for parameter of discrete distribution for large sample size

$$X$$ is a random variable that can take on $$(-a, 0, a), a > 0$$ values with corresponding probabilities $$(\frac{1}{2}p, 1 - p, \frac{1}{2}p), 0 < p < 1$$. Find confidence interval for $$p$$ with confidence level of $$1 - \alpha$$, assuming sample size, $$n$$, is large.

I'm stuck on this exercise. Because the sample size is large I assume that in the end I can approximate it with normal distribution to get the confidence intervals. However how to get the parameters of that normal distribution?

I can find some estimate of $$p$$ using method of moments. Then I get that $$p = \frac{S^{2}_n}{a^2}$$, can I use it as a mean for this normal distribution? What about variance? I'm not really sure how to move further and would appreciate some hints.

We have an i.i.d set of $$X_1,X_2,\ldots,X_n$$ as described about. Let $$S^-$$, $$S^0$$, and $$S^+$$ be sets of indices such that, for $$i\in S^-$$, $$X_i=-a$$, and likewise for the other two sets, and let corresponding $$c^-$$, $$c^0$$, and $$c^+$$, be the respective counts of the number of times $$-a$$, $$0$$, and $$+a$$, appear in the sample.
First, see that $$c^0$$ is a sufficient statistic for $$p$$.
\begin{align} f(x_1,\ldots,x_n) &= \prod_{i=1}^{n}f(x_i) \\ &= \prod_{i \in S^{-}} f(x_i) \times \prod_{i \in S^{0}} f(x_i) \times \prod_{i \in S^{+}} f(x_i) \\ &= \prod_{i \in S^{-}} \frac{1}{2}p \times \prod_{i \in S^{0}} (1-p) \times \prod_{i \in S^{+}} \frac{1}{2}p \\ &= \left(\frac{1}{2}p\right)^{c^-} \times \left(1-p\right)^{c^0} \times \left(\frac{1}{2}p\right)^{c^+} \\ &= \left(\frac{1}{2}p\right)^{c^-+c^+} \times \left(1-p\right)^{c^0} \\ &= \left(\frac{1}{2}p\right)^{n-c^0} \left(1-p\right)^{c^0} \\ \end{align}
That fact that $$c^0$$ is a sufficient statistic tells us that we can estimate $$p$$ from the number of times zero appears in the experiment. Each trial gives a zero with probability $$1-p$$ and we have $$n$$ trials, so $$c^0 \sim Binom(1-p,n)$$ $$c^0$$ will have an expectation of $$n(1-p)$$ and a variance of $$np(1-p)$$, so we can estimate $$p$$ by $$\hat p = 1-\frac{c^0}{n}$$ and we have the usual Normal approximation confidence interval as $$\hat p \pm Z_\alpha \sqrt{\hat p (1-\hat p)/n}$$
Define a new random variable $$Y=\frac{1}{a}|X|$$. Then $$Y$$ is of Bernoulli distribution with the probability $$p$$ for $$Y=1$$ and the probability $$(1-p)$$ for $$Y=0$$. For Bernoulli distribution with a mean equal to $$p$$, the variance equals $$p(1-p)$$. Thus, when the sample size $$n$$ is sufficiently large, the variable $$Z=\frac{n\bar{y}-np}{\sqrt{np(1-p)}}$$ well approximates the standard normal distribution, where $$\bar{y}=\frac{1}{n}\sum_{i=1}^ny_i$$ is the sample mean of $$y$$. The rest calculation is straightforward to find the confidence interval of $$p$$ with multiple optional methods.