Say I have a coin and I don't know what is the probability of getting heads. So I set $\ p(H) = \theta $ where $\ \theta \sim U(0,1) $

Suppose I have flipped the coin once and observed one head so according to this formula:

$$\ (1)\ \ \ f( \theta | y) = \frac{f(y|\theta)\cdot f(\theta)}{\int_{-\infty}^{\infty} f(y | \theta ) \cdot f(\theta) \ d\theta} = \frac{likelihood \cdot prior}{normalizing \ constant} $$

the probability of heads (as shown by the teacher) will be

$$\ (2) \ \ \ f(\theta \ | \ y = 1) = \frac{\theta^1(1-\theta)^{1-y}}{\int_0^1\theta^1(1-\theta)^{1-y}d\theta} = \frac{\theta}{\int_0^1 \theta d\theta} = \frac{\theta}{1/2} = 2\theta$$

now suppose I have flipped again and observed another head, what will be the posterior probability of getting head?

If I understand correctly then $\ (y_1 | \theta) \sim Bernulli(\theta) $ and I flipped the coin more than once then $\ (y | \theta) \sim Binomial(n, \theta) $ but what exactly is "normalizing constant" and why did the teacher just dropped $\ f(\theta) $ from his calculations in $\ (2) $ ?


The prior here is uniformly distributed over $[0,1]$. Therefore the probability density function (pdf) is $f(\theta)=1$ for $\theta\in[0,1]$, and $f(\theta)=0$ for $\theta\not\in[0,1]$. So the integral over $\mathbb R$ turned into the integral over $[0,1]$ where $f(\theta)=1$.

For two coin flips the probability that we observe two heads $y_1=1,y_2=1$ under fixed value of $\theta$ is $$f(y_1=1,y_2=1\mid \theta)=\mathbb P(\text{two heads}\mid \theta)=\theta^2$$ and normalizing constant is $$ \int_{-\infty}^{\infty} f(y_1=1,y_2=1 | \theta ) \cdot f(\theta) \ d\theta = \int_0^1\theta^2\cdot 1\,d\theta = \frac13. $$ So for $\theta\in[0,1]$, pdf of posterior distribution is $$ f(\theta \mid y_1=1,y_2=1) = \frac{\theta^2}{1/3}=3\theta^2. $$

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  • $\begingroup$ Thanks for your answer. Can you please explain what is normalizing constant? Also, does it mean that $\ f(\theta) $ is just an indicator so if $\ \theta \in [0,1] $ it is $\ 1 $ and else it's $\ 0 $ ? Why do we need it? It's not like $\ \theta$ could be anything else than $\ [0,1] $ ?? $\endgroup$ – Yarden Gur Jan 1 at 15:32
  • $\begingroup$ The function $f(\theta|y)=\theta^2$ is not a valid density on $[0,1]$ since it does not integrates to $1$. Therefore normalizing constant $c$ is needed s.t. $f(\theta|y)=\frac{\theta^2}{c}$ integrates to 1. And this constant is exactly $\int_0^1 \theta^2 d\theta=1/3$. With this constant $f(\theta|y)=3\theta^2$ is the valid density: $\int_0^1 3\theta^2 d\theta=1$. Yes, density of uniform distribution is $\mathbb 1_{[0,1]}$. $\endgroup$ – NCh Jan 1 at 16:36

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