- **Advanced Statistics**
(*http://mymathforum.com/advanced-statistics/*)

- - **Bayesian hypothesis testing**
(*http://mymathforum.com/advanced-statistics/345614-bayesian-hypothesis-testing.html*)

Bayesian hypothesis testingSuppose you are flipping a coin. You are considering the following hypothesis: The coin is weigthed so it comes up head 1/4 of the time and tail 3/4 of the time. Assume that the prior probability is 1/3. You now flip the coin ten times and get seven heads. What are the probabilities of this hypothesis given the result of the coin flips? Definition of Bayesian Prior P(F = t | X = k) = P(X = k | F = t ) * P(F = t) / P(X = k) P(F = t) is called the prior. It expresses our preconceptions about the relative likeihood of different values of the fraction of the population that yields head *before* we even see the poll result. I am not sure how to approach this problem using Bayesian hypothesis testing. Any tips would be greatly appreciated. Thanks. |

$P\left[\left . p_H = \dfrac 1 4 \right| \text{observe 7 heads in 10 tries}\right] = $ $\dfrac{P\left[\text{observe 7 heads in 10 tries } | p_H = \dfrac 1 4\right]P\left[p_H = \dfrac 1 4\right]}{P[\text{observe 7 heads in 10 tries}]} = $ $\dfrac{\dbinom{10}{7}\left(\dfrac 1 4\right)^7\left(\dfrac 3 4\right)^3\cdot \dfrac 1 3}{\dbinom{10}{7}\left(\dfrac 1 4\right)^7\left(\dfrac 3 4\right)^3 + \dbinom{10}{7}\left(\dfrac 1 2\right)^{10}} = \dfrac{9}{1051}$ |

one last question....Should the following be considered? $\displaystyle [\text{observe 7 heads in 10 tries } \cup not( p_H = \dfrac 1 4 ) ] = \dbinom{10}{7}\left( 1 - \left(\dfrac 1 4\right)^7\left(\dfrac 3 4\right)^3 \right) $ instead of $\displaystyle \dbinom{10}{7}\left(\dfrac 1 2\right)^{10} $ |

Quote:
$P[\text{observe 7/10 heads}] = \displaystyle \int_0^1 \dbinom{10}{7}p^7(1-p)^3~dp = \dfrac{1}{11}$ |

All times are GMT -8. The time now is 02:16 AM. |

Copyright © 2019 My Math Forum. All rights reserved.