Bayesian – Posterior for Discrete Prior from Normal Sample with Known Variance

bayesianposteriorr

Given a sample $(1,2,5,2.5,3,7)$, and known $\sigma^2 = 4$ with possible values for prior mean of $(0,0.5,1,1.5)$ all equally likely, find the posterior distribution and the posterior probability that $\mu\leq 1$

So in R I have:

> sample = c(1,2,5,2.5,3,7)
> m = c(0,0.5,1,1.5)
> prior = (rep(1/4,4))
> lilklihood = dnorm(mean(sample), mean = m, sd = 2/sqrt(length(sample)))
> posterior = prior*liklihood/sum(prior*liklihood)
> posterior
[1] 0.002021756 0.021735920 0.160607934 0.815634390

So my posterior distribution just is this vector correct? The thing I find is that all my values are actually less than 1, so is posterior probability that $\mu\leq 1$ just 1?

Best Answer

The resulting posterior is the posterior probability associated with four possible values for the mean; therefore; $$ \begin{aligned} &P(\mu = 0 \vert \textrm{data}) &&= 0.002021756,\\ &P(\mu = 0.5 \vert \textrm{data}) &&= 0.021735920,\\ &P(\mu = 1 \vert \textrm{data}) &&= 0.160607934,\\ &P(\mu = 1.5 \vert \textrm{data}) &&= 0.815634390,\\ \end{aligned} $$ therefore $P(\mu \leq 1 \vert \textrm{data})$ is given by $$ \begin{aligned} P(\mu \leq 1 \vert \textrm{data}) &= P(\mu = 0 \vert \textrm{data}) + P(\mu = 0.5 \vert \textrm{data}) + P(\mu = 1 \vert \textrm{data})\\ &= 0.18436561\\ &\approx 18\%. \end{aligned} $$

Related Solutions

Solved – Bayesian analysis with histogram prior. Why draw simulations from the posterior

To answer your subquestion: How to do the following more elegantly?

post.vector <- vector()
post.vector[1] <- sum(post[p < 0.1])
post.vector[2] <- sum(post[p > 0.1 & p <= 0.2])
post.vector[3] <- sum(post[p > 0.2 & p <= 0.3])
post.vector[4] <- sum(post[p > 0.3 & p <= 0.4])
post.vector[5] <- sum(post[p > 0.4 & p <= 0.5])
post.vector[6] <- sum(post[p > 0.5 & p <= 0.6])
post.vector[7] <- sum(post[p > 0.6 & p <= 0.7])
post.vector[8] <- sum(post[p > 0.7 & p <= 0.8])
post.vector[9] <- sum(post[p > 0.8 & p <= 0.9])
post.vector[10] <- sum(post[p > 0.9 & p <= 1])

The easiest way to do it using base R is:

group <- cut(p, breaks=seq(0,1,0.1), include.lowest = T)
post.vector.alt <- aggregate(post, FUN=sum, by=list(group))

Note that the breaks go from 0 to 1. This yields:

     Group.1            x
1    [0,0.1] 3.030528e-13
2  (0.1,0.2] 1.251849e-08
3  (0.2,0.3] 6.385088e-06
4  (0.3,0.4] 6.732672e-04
5  (0.4,0.5] 2.376448e-01
6  (0.5,0.6] 7.372805e-01
7  (0.6,0.7] 2.158296e-02
8  (0.7,0.8] 2.691182e-03
9  (0.8,0.9] 1.205200e-04
10   (0.9,1] 3.345072e-07

And we have:

> all.equal (post.vector.alt$x, post.vector)
[1] TRUE

Solved – Why is it necessary to sample from the posterior distribution if we already KNOW the posterior distribution

This question has likely been considered already on this forum.

When you state that you "have the posterior distribution", what exactly do you mean? "Having" a function of $\theta$ that I know is proportional to the posterior, namely$$\pi(\theta|x) \propto \pi(\theta) \times f(x|\theta)$$for instance the completely artificial target$$\pi(\theta|x)\propto\exp\{-||\theta-x||^2-||\theta+x||^4-||\theta-2x||^6\},\ \ x,\theta\in\mathbb{R}^{18},$$does not tell me what is

the posterior expectation of a function of $\theta$, e.g., $\mathbb{E}[\mathfrak{h}(\theta)|x]$, posterior mean that operates as a Bayesian estimator under standard losses;
the optimal decision under an arbitrary utility function, decision that minimises the expected posterior loss;
a 90% or 95% range of uncertainty on the parameter(s), a sub-vector of the parameter(s), or a function of the parameter(s), aka HPD region$$\{h=\mathfrak{h}(\theta);\ \pi^\mathfrak{h}(h)\ge \underline{h}\}$$
the most likely model to choose between setting some components of the parameter(s) to specific values versus keeping them unknown (and random).

These are only examples of many usages of the posterior distribution. In all cases but the most simple ones, I cannot provide the answers by staring at the posterior distribution density and do need to proceed through numerical resolutions like Monte Carlo and Markov chain Monte Carlo methods.

Best Answer

Related Solutions

Solved – Bayesian analysis with histogram prior. Why draw simulations from the posterior

Solved – Why is it necessary to sample from the posterior distribution if we already KNOW the posterior distribution

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