Question

In Example $5$e, what is the conditional probability that the ith coin was selected given that the first n trials all result in heads?

Short answer

$P(C_i\mid F_n)=\frac{i^n}{\sum _{j=0}^k{j}^n}\text{, the beginning formula is given in the example.}$

The denominator is the power sum, for which there are no explicit formulae.

Step-by-step solution

Given Information

The conditional probability that the ith coin was selected given that the first n trials all result in heads

Given:

Example $5$_e

Events F_n,C_i

i=$0,1,...K$

Explanation 

Probabilities:

$P\left(C_i\right)=\frac{1}{k+1},i=0,1,\dots ,k$

$P(H\mid C_i)=\frac{i}{k}$

Events of flipping heads in different flips are independent given C_i, Therefore:

$P(F_n\mid C_i)=\prod _{i=1}^{n}P(H\mid C_i)={\left(\frac{i}{k}\right)}^n$

$\text{Calculate}P(C_i\mid F_n)$

From the example

$P(C_i\mid F_n)=\frac{P\left(C_i{F}_n\right)}{P\left(F_n\right)}=\frac{P(F_n\mid C_i)P\left(C_i\right)}{\sum _{j=0}^{k}P(F_n\mid C_j)P\left(C_j\right)}=\frac{{\left(\frac{i}{k}\right)}^n\cdot \frac{1}{k+1}}{\sum _{j=0}^k{\left(\frac{j}{k}\right)}^n\cdot \frac{1}{k+1}}$

$\text{Reducing the fraction by factor}\frac{1}{k^n}\cdot \frac{1}{k+1}$

$P(C_i\mid F_n)=\frac{i^n}{\sum _{j=0}^k{j}^n}$

The number below is a power sum, and it cant be calculated by a simple explicit formula, some of the variations.

$S_n\left(k\right)=\sum _{j=0}^k{j}^n=\zeta (-n)-\zeta (-n,k+1)$

$\text{Where}\zeta \text{is the Reimann zeta function},f(n,a)$

Final Answer

$P(C_i\mid F_n)=\frac{i^n}{\sum _{j=0}^k{j}^n}\text{, the beginning formula is given in the example.}$

The denominator is the power sum, for which there are no explicit formulae.