1
$\begingroup$

$X$ 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.

$\endgroup$
4
$\begingroup$

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 $$ \hat p \pm Z_\alpha \sqrt{\hat p (1-\hat p)/n} $$

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.