0
$\begingroup$

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?)

Now there is the task:

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 $$\forall \theta \in \Theta: {\lim \inf}_{n\to\infty}P^{\otimes n}_\theta(\{x\in\mathcal X^n:\gamma(\theta)\in I_n(x)\}) \ge 1-\alpha$$

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – owen88 Feb 18 '18 at 10:59
  • $\begingroup$ Just done, thanks for the suggestion! $\endgroup$ – newbie Feb 18 '18 at 12:25
2
$\begingroup$

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.

$\endgroup$

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.