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The random variable $X∼Binomial(n= 2,θ)$ and $θ∼Uniform(0,1)$. Calculate the posterior probability $Π(θ <0.5|X= 2)$.

Solution:

$θ|X∼Beta (1 +x,1 +n−x)$

$θ|X= 2∼Beta (1 + 2,1 + 2−2) = Beta(3,1) $ $i.e. \pi(\theta | X= 2) = 3 \theta^2$

$Π(θ <0.5|X= 2) =\int_{0}^{0.5}3θ^2dθ =1/8$.


Why is $θ|X∼Beta (1 +x,1 +n−x)$? I don't understand how they came up with that.

Also I don't understand how $i.e. \pi(\theta | X= 2) = 3 \theta^2$ for beta$(3, 1)$.

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  • $\begingroup$ Seem the prior distribution (unmentioned in your question) may have been $\theta \sim UNIF(0,1) \equiv BETA(1,1).$ Then with data $n = x = 2,$ you'd have posterior dist'n $BETA(3,1).$ Result is from multiplying Prior $\times$ Likelihood to get Posterior according to Bayes' Thm. $\endgroup$ – BruceET Apr 12 at 0:33
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Let $\text{Unif}(0,1)$ be a prior distribution on $\theta$, $p(\theta)=1$ on $(0,1)$. Thus, the prior is just a constant (equal 1).

The likelihood for a single data point is the binomial pmf $p(x|\theta)=$ $n \choose x$ $\theta^{x}(1-\theta)^{n-x}$.

Then, the posterior distribution, $p(\theta| \text{data}=x)$ is proportional to a multiple of prior and likelihood:

$$ p(\theta| x) \propto \theta^{x}(1-\theta)^{n-x}. $$

Note that we treat the RHS as a function of $\theta$ and drop the multiplying constants, such as $n \choose x$.

Now, you need to recognize the RHS as a form of a density (of $\theta$) and it is a Beta density with parameters $x+1$ and $n-x+1$ (see the definition to convince yourself). In particular, for $n=2, x=2$, $p(\theta|x)=3\theta^2$.

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