0
$\begingroup$

Let $\theta$ be the proportion of people who are ready to quit smoking within 6 months. Let's say we perform a survey in $2017$ with a $n$ volunteers who ask people this question until they obtain yes as an answer. Then an appropriate statistical model for this problem will be the geometric distribution.

Now suppose that we perform the survey again in $2018$ expecting that the proportion of $\theta$ obtained in $2017$ remains the same. Then we assume the following prior distribution $\text{Beta}(k\theta,k(1-\theta))$ where $k$ is some constant, say equal to $5.$ Then I want to understand why using this prior distribution makes sense for this situation.

I read that the beta distribution is the conjugate prior to the geometric distribution but the parameters here are quite different. So maybe there is something else that I am missing.

I also wanted to know if the following is true: the proportion $p$ for the survey is now a random variable such that $$p\sim \text{Beta}(k\theta,k(1-\theta)).$$ Then $\mathbb{P}[p=x|\theta] =\frac{1}{\beta(k\theta,k-k\theta)}\cdot x^{k\theta-1}(1-x)^{k(1-\theta)-1}.$

$\endgroup$
  • 1
    $\begingroup$ The beta distribution is the conjugate prior to the binomial distribution as well as to the negative binomial distribution, of which the geometric is a special case. $\endgroup$ – jbowman Jan 1 at 23:53
1
$\begingroup$

Let θ be the proportion of people who are ready to quit smoking (...) We perform a survey in 2017 with n volunteers who ask people this question until they obtain yes as an answer. Then an appropriate statistical model for this problem will be the geometric distribution.

This is not correct: if there are $n$ volunteers, the correct model is a Negative Binomial, $X\sim\text{Neg}(n,\theta)$, and I do not understand how this observation is used in the sequel.

...perform the survey again in 2018 expecting that the proportion θ obtained in 2017 remains the same. Then assume the prior distribution Beta(kθ,k(1−θ)) where k is some constant, say 5. Then I want to understand why using this prior distribution makes sense for this situation.

This is 200% confusing because, since "the proportion θ remains the same", the unobserved and hence unknown $\theta$ should be the parameter modelled by a prior, rather than appearing in the prior for another parameter (which one?).

At last, $\mathbb{P}[p=x|\theta]=0$ since the Beta distribution is absolutely continuous wrt the Lebesgue measure. The function $$\frac{1}{\beta(k\theta,k-k\theta)}\cdot x^{k\theta-1}(1-x)^{k(1-\theta)-1}$$ is the density of the Beta distribution.

$\endgroup$
  • $\begingroup$ If $X$ is a random vector of samples such that $X = (X_1,X_2,\cdots,X_n)$ then each $X_i \sim \text{Geo}(\theta).$ I don't see why $X_i\sim \text{Neg}(n,\theta)?$ Furthermore it is not certain the proportion in $2018$ is the same as that in $2017.$ We are modelling the unknown parameter $p$ (in $2018$) with beta distribution whose parameters use the proportion obtained in $2017.$ $\endgroup$ – SuperMario Jan 3 at 17:28
  • $\begingroup$ (a) use sufficiency as $\sum_i X_i\sim \text{Neg}(n,\theta)$ and (b) this is not a fully Bayesian approach to the problem $\endgroup$ – Xi'an Jan 3 at 17:39
  • $\begingroup$ Agreed. But do you seen any reason to have a prior distribution $\text{Beta}(k\theta,k(1-\theta))$ with its specific parameters? $\endgroup$ – SuperMario Jan 3 at 18:23
  • $\begingroup$ The mean of the $\text{Beta}(k\theta,k(1-\theta))$ distribution is $\theta$. And $k$ sets the imprecision about this value. $\endgroup$ – Xi'an Jan 3 at 18:58

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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