The probability of Heads of a coin is y, and this bias y is itself the realization of a random variable Y which is uniformly distributed on the interval [0,1].

Question

To estimate the bias of this coin. We flip it 6 times, and define the (observed) random variable N as the number of Heads in this experiment.

Given the observation N=3, calculate the posterior distribution of the bias Y. That is, find the conditional distribution of Y, given N=3.

For 0≤y≤1,
fY|N(y∣N=3)=

Answer 1

To find the conditional distribution of Y given N=3, we can use Bayes' theorem:

fY|N(y|N=3) = fN|Y(N=3|y) * fY(y) / fN(N=3)

First, let's find fN|Y(N=3|y), which is the probability of observing 3 Heads given the bias y. This can be calculated using the binomial distribution:

fN|Y(N=3|y) = C(6,3) * y^3 * (1-y)^(6-3)

Where C(6,3) is the number of ways to choose 3 Heads from 6 flips, which is equal to 20.

Next, fY(y) is the probability density function of Y, which we know is uniformly distributed on the interval [0,1]. Therefore, fY(y) = 1 for 0 ≤ y ≤ 1, and 0 otherwise.

Finally, fN(N=3) is the marginal probability of observing 3 Heads, which can be calculated by integrating the joint distribution of N and Y over all possible values of Y:

fN(N=3) = ∫ fN|Y(N=3|y) * fY(y) dy

Since fY(y) = 1 for 0 ≤ y ≤ 1, the integral simplifies to:

fN(N=3) = ∫ fN|Y(N=3|y) dy

Now we can plug in the values to calculate the posterior distribution:

fY|N(y|N=3) = fN|Y(N=3|y) * fY(y) / fN(N=3)
= (20 * y^3 * (1-y)^(6-3)) * 1 / (∫ fN|Y(N=3|y) dy)

Simplifying further,

fY|N(y|N=3) = 20 * y^3 * (1-y)^3 / (∫ y^3 * (1-y)^3 dy)

After integrating the denominator, we can find the normalized distribution. However, I am unable to provide the exact values without knowing the specific limits of integration for the denominator.