# Estimate Posterior with: Binomial Likelihood with 1/θ Prior

I am trying to estimate the posterior distribution using Bayes theorem. The following information is given:

\begin{align} \newcommand{\prior}{{\rm prior}} \newcommand{\likelihood }{{\rm likelihood }} \newcommand{\posterior}{{\rm posterior}} \posterior&∝\likelihood \times \prior \\ \likelihood(x|θ)&∝ θ^k\times(1-θ)^{n-k} \\ \prior(θ)&∝1/θ \end{align}

Here I interested in estimating the posterior distribution of θ, with Binomial likelihood(k successes of the n trials).

By multiplying the likelihood and prior I get the following:

Question: The posterior distribution looks like Beta distribution: $θ^{\alpha-1}(1-θ)^{\beta-1}$. It is quite clear that $\alpha=k$ , but I am struggling to figure out how to transform power $n-k$ to look like $\beta_{new}={\rm something} - 1$. Could anyone point me to the right direction?

• Is your data discrete or continuous? Is the parameter a proportion. I am thinking you may be talking about a situation wher you have a beta posterior and a binomial likelihood. But why does a 1/theta prior enter into it? Is that even a proper prior? – Michael R. Chernick Jan 6 '17 at 23:00
• $\beta = n-k+1$. – whuber Jan 6 '17 at 23:11
• This prior should not be used because the posterior is not defined when $X=n$, which has a positive probability to occur under the Binomial model. – Xi'an Jan 7 '17 at 8:36
• @MichaelChernick, the data is discrete. The random variable is the numer of patience with no side effects. I am a bit comfused myself about the form of the prior, but that was given by the assignment: p(theta) ∝ 1/theta. – Artur Jan 7 '17 at 9:37
• @Xi'an indeed it looks wierd, but that was given by the assignment: p(theta) ∝ 1/theta – Artur Jan 7 '17 at 9:39

Maybe I'm missing something, but isn't the prior, $p(\theta) \propto \theta^{-1}$ merely a $\mbox{beta}(a=0,b=1)$ distribution? (If $a=0$ is uncomfortable, then set $a=\epsilon$ for small $\epsilon$.) Therefore the usual beta-Bernoulli update rule should apply. That is, with a Bernoulli likelihood function (which this is, for $N$ independent trials), and a conjugate beta prior (for which the present problem is a special case), the posterior is again a beta distribution: $\mbox{beta}(a+k,b+N-k)$. Here we start with $a=\epsilon$ and $b=1$, so the posterior is $\mbox{beta}(\epsilon+k,1+N-k)$.
• @Artur: as indicated in my earlier comment, you cannot use this improper prior because the posterior is not defined when $X=n$. – Xi'an Jan 7 '17 at 14:47