# asymptotic and exact confidence interval

Assume we toss a thumbtack 300 times. After every time, we note $1$ if it lands point up or $0$ if it lands point down. In summary, we get 124 times $1$.

So we know that the number of the rounds with outcome $1$ is $\sim Bin(300,p)$ with unknown parameter $p$. Furthermore, $\mathcal X:=\{0,1\}^{300}$, $\Theta:=[0,1]$, $p \in \Theta$ (is this correct?)

Define the term 'asymptotic and exact confidence interval' for the niveau $1-\alpha>0$. Give a $95\%$ confidence interval for the probability that the thumb will land point up.

I do have the formal definitions of the asymptotic and exact confidence interval, but I don't really understand it. Could anyone explain it to me referring to this specific example?

The definitions are:

$Definition$

Let $(\mathbb P_\theta)_{\theta \in \Theta}$ be a statistical model with $\Theta \subset \mathbb R^n$ on the sample space $\mathcal X$. A reel parameter is a mapping $\gamma: \Theta \to \mathbb R$. An interval-valued mapping $$I:\mathcal X \to \mathcal P(\mathbb R), I(x)=[U(x),O(x)]$$ with the statistics $U,O: \mathcal X \to \mathbb R$ with $U \le O$ is called an interval estimation for the parameter $\gamma$

$Definition$

The coverage probability of an interval estimation $I$ for a parameter $\gamma$ is the mapping $$\theta \to \mathbb P_\theta (\{x \in \mathcal X: \gamma(\theta) \in I(x)\}), \theta \in \Theta$$ A confidence niveau of an interval estimation is the minimal coverage probability $$\inf_{\theta \in \Theta} \mathbb P_\theta(\gamma(\theta) \in I(x))$$

$Definition$

An inverval estimation $I$ is called (exact) confidence interval for the confidence niveau $1-\alpha$ (for a fixed $\alpha \in [0,1]$), if $$\forall \theta \in \Theta: \mathbb P_\theta(\gamma(\theta)\in I(x)) \ge 1-\alpha$$

$Definition$

For all $n \ge n_0$ let $I_n$ be an inverval estimation on $\mathcal X^n$. A sequence $(I_n)_{n \ge 1}$ of interval estimators is called asymptotic confidence interval for the confidence niveau $1-\alpha$, if $$\forall \theta \in \Theta: {\lim \inf}_{n\to\infty}P^{\otimes n}_\theta(\{x\in\mathcal X^n:\gamma(\theta)\in I_n(x)\}) \ge 1-\alpha$$

• Can you add the definitions you have to your question? That way it'll help to answer you with the notation / concepts you are familiar with. – owen88 Feb 18 '18 at 10:59
• Just done, thanks for the suggestion! – newbie Feb 18 '18 at 12:25

In your context, you are looking to define a confidence interval for the parameter $p$ associated with a Bernoulli distribution (i.e. the true probability $p$ that a thumbtack will land point up).

Fortunately, due to the relationship between Bernoulli and Binomial variables, as you observe this is equivalent to finding a confidence interval for the parameter $p$ of a $\text{Bin}(n,p)$ distribution (with $n=300$ in your instance), based on the observed outcome ($X = 124$, in your instance).

For a Binomial distribution, there is one standard example of an exact (1-$\alpha$) confidence interval, called the Clopper-Pearson interval. This has a rather messy formula, and is given by

$$I_{\alpha} = \bigg( {\textstyle B\left(\frac{\alpha}{2}; X; n - X + 1\right), \, B\left(1-\frac{\alpha}{2}; X+1; n - X \right)} \bigg) ,$$

here $B(r\,;v,w)$ denotes the percentile function of a Beta distribution with shape parameters $v,w$. For you, I'd imagine what this function is doesn't matter. In your particular instance the interval is at $\alpha = 0.05$ (i.e. a 95% confidence interval)

$$I_\alpha = \big( {\textstyle B\left(0.975; 125, 177\right), B\left(0.025; 124, 176\right)} \big) = (0.3570, 0.4714).$$ This is an exact confidence interval: which means that it is guaranteed that at least 95% of the time the true parameter will lie within the interval you calculate.

As an example of a asymptotic confidence interval we can use the standard Normal approximation to the binomial distribution, and the associated confidence interval. Denoting $\hat p = X/n$, this interval is given by

$$J_\alpha = \hat p \pm \Phi^{-1}\left(1 - \frac{\alpha}{2} \right) \sqrt{\frac{\hat p (1 - \hat p)}{n} },$$ where $\Phi$ is the cumulative distribution of a normal distribution. In your example this gives the interval $$J_\alpha = (0.3576,0.4691).$$

The difference with this interval is that we cannot say for certain that 95% of the time the result will lie in this interval. In particular when $n$ is small this will not be true, but as $n$ gets large it becomes increasingly close to being true that 95% of observations would fall in the interval. To see why this formula doesn't work for small $n$, suppose that we know $p = 1/2$, and suppose we make one throw, $n=1$. If this lands point up then the interval we would obtain (from the above formula) would be $J_\alpha = [1,1]$, whilst if it didn't land point up it would be $J_\alpha = [0,0]$. In either case, the probability that the true value falls in the interval $J_\alpha$ is clearly $0$ (since $p = 1/2$). i.e. the answer does not fall into the interval 95% of the time.