Given the i.i.d. random variables $X_{1}, X_{2},...,X_{n}$ with frequenty function: $f(x|p)=(x-1)(1-p)^{x-2}p^2$, with $ x =2,3,...$

Pick as a priori density $f_{P}(p) \propto p(1-p)1_{(0,1]}(p)$.

Give the posterior estimate for $p$ based on this prior.

I find Bayesian estimation very hard to grasp, can someone help me out on this one?


Outline and Hints: According to the applicable version of Bayes' Theorem, you have $$f(p\,|\,\mathbf{x}) \propto f(p) \times f(\mathsf{x}\,|\, p).$$ Here $f(p\,|\,\mathbf{x})$ represents the posterior distribution of $p$ given data, $f(\mathsf{x}\,|\, p)$ is the likelihood function of the data, and $f(p)$ is the PDF of the prior distribution [in your case $\mathsf{Beta}(2,2)].$

With your prior and data consisting of $x$ successes in $n$ binomial trials, one can see that the posterior distribution would be $\mathsf{Beta}(2+x,\; 2+n-x).$ [This is easy to see looking at the kernel of the likelihood because the beta prior and binomial likelihood are 'conjugate' (mathematically compatible).]

You don't say what kind of Bayesian estimate you want. For point estimates, the mode and mean (and even the median) of the posterior distribution have been used. For a Bayesian 95% probability interval estimate one method is to find quantiles .025 and .975 of the posterior distribution. (Something like the qbeta function in R can be used to do the numerical integration, if necessary.)

Addendum per comment: I answered for binomial data. But your likelihood function is $\prod_{i=1}^n f(x_i|p)$, for the given $f(x|p).$ Then you would get somewhat different beta parameters for the posterior. My guess is that the answer is the mode of the posterior distribution, but I don't have the full context of what you've been studying. In meetings next few days so can't quickly go further into it. If you can't get it on your own, then I suggest you 'unaccept' my 'outline and hints' and maybe someone else will help.

  • $\begingroup$ Well the answer is $(2n-1)/(n\bar{X}_{n}-2)$, could you give me some hints on the method to get there? (and thank you for your previous answer) $\endgroup$ – Keep_On_Cruising Oct 2 '17 at 12:53
  • $\begingroup$ Please see addendum to my 'answer'. $\endgroup$ – BruceET Oct 2 '17 at 15:30
  • $\begingroup$ @O.Kro I think the answer is $\frac{2n+1}{n\bar{X}_n+2}$. $\endgroup$ – Sayan Oct 2 '17 at 17:20

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.