# How can I plug in the value of parameter found by Maximum a Posterior?

Suppose I have 1 heads and 4 tails from 5 coin tosses. To find out the probability of 1 heads and 4 tails in my coin toss experiments, I decided to use Binomial Probability Mass Function for the calculation of the probability on the current observations.

I used Maximum a Posterior with the Beta prior $$(\alpha=5, \beta=5)$$, instead of Maximum Likelihood Estimation, to estimate the parameter value of $$\theta$$ and I got $$0.3846$$ from Maximum a Posterior. Now, I have the parameter value of $$\theta$$ and I want to find out the probability that I would observe 1 heads and 4 tails, which one of the following equations should I plug in the estimate parameter value, $$0.3846$$:

1. $${n}\choose{k}$$ $$\Pi^n_{i=1} \theta^{x_i} (1-\theta)^{(1-x_i)} \theta^{(\alpha-1)} (1-\theta)^{(\beta-1)}$$

The above equation is considering the prior probability for the calculation of the probability for my experiment. Or should I just plug the value of $$\theta$$ into the Binomial Probability Mass Function?

1. $${n}\choose{k}$$ $$\Pi^n_{i=1} \theta^{x_i} (1-\theta)^{(1-x_i)}$$

I know this sounds very naive but I just want to make sure I am thinking correctly.

You have data $$X \sim \mathcal{B}(\theta)$$, and you observe $$(X_1,\dots,X_n)$$, $$X_o = \sum X_i$$, which means that for $$k\in \{0,\dots,n\}$$,

$$\mathbb P(X_o = k \mid \theta) = \binom{n}{k} \theta^k(1-\theta)^{n-k}$$

From a bayesian perspective you also have a prior distribution on $$\theta$$ which is a Beta distribution $$p(\alpha,\beta)$$.
Then you get the posterior distribution of $$\theta$$, \begin{align*} p(\theta \mid X_o) &\propto p(\alpha,\beta) \mathbb P(X_o \mid \theta) \\ &\propto p(\alpha,\beta) \theta^{X_o} (1-\theta)^{n-X_o}, \end{align*} and you can derive the maximum a posteriori,

$$\hat \theta= \arg\max_{\theta} p( \theta \mid X_o)$$

Now if you want to compute the probability of getting 1 head out of 5 (new and independent) coin tosses, given that the probability of head is $$\hat \theta$$, you can simply put $$\hat \theta$$ in the likelihood $$\mathbb P(X \mid \theta)$$:

$$\mathbb P(X=1 \mid \hat \theta ) = \binom{5}{1} \hat \theta(1-\hat \theta)^4$$

But from a Bayesian point of view it may be more relevant to consider all possible values of $$\theta$$, and not just a point estimate like $$\hat \theta$$, and rather compute

$$\mathbb P(X= 1 \mid X_o) = \int_\Theta P(X = 1 \mid \theta) p(\theta \mid X_o)d\theta$$

• Thank you for the answer and I am trying to understand the last equation but it does not come straight. Do you mean a prior probability distribution by $p(\theta | X_0)$ and do you mean a likelihood by $P(X=1 | \theta)$? I don't know If you can stretch the last equation a bit further, I would really appreciate. Jan 23 '20 at 13:21
• The last equation is the sum of $\mathbb P (X \mid \theta)$ over all values of $\theta$ weighted by $p(\theta \mid X_o)$ which is your current knowledge about the distribution of $\theta$. This weigthed sum is called the posterior predictive distribution. See stats.stackexchange.com/…. Jan 23 '20 at 13:25
• Thank you very much!! Jan 23 '20 at 13:29