My Math Forum Bayesian hypothesis testing

 January 12th, 2019, 10:21 AM #1 Senior Member   Joined: Jan 2017 From: Toronto Posts: 209 Thanks: 3 Bayesian hypothesis testing Suppose 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.
 January 12th, 2019, 10:10 PM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 2,494 Thanks: 1369 $P\left[\left . p_H = \dfrac 1 4 \right| \text{observe 7 heads in 10 tries}\right] =$ $\dfrac{P\left[\text{observe 7 heads in 10 tries } | p_H = \dfrac 1 4\right]P\left[p_H = \dfrac 1 4\right]}{P[\text{observe 7 heads in 10 tries}]} =$ $\dfrac{\dbinom{10}{7}\left(\dfrac 1 4\right)^7\left(\dfrac 3 4\right)^3\cdot \dfrac 1 3}{\dbinom{10}{7}\left(\dfrac 1 4\right)^7\left(\dfrac 3 4\right)^3 + \dbinom{10}{7}\left(\dfrac 1 2\right)^{10}} = \dfrac{9}{1051}$ Thanks from zollen
 January 13th, 2019, 12:38 AM #3 Senior Member   Joined: Jan 2017 From: Toronto Posts: 209 Thanks: 3 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 $\displaystyle \dbinom{10}{7}\left(\dfrac 1 2\right)^{10}$
January 13th, 2019, 07:44 AM   #4
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Quote:
 Originally Posted by zollen 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 $\displaystyle \dbinom{10}{7}\left(\dfrac 1 2\right)^{10}$
well... if $p_H$ can be anything between 0 and 1 then

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

Last edited by romsek; January 13th, 2019 at 07:46 AM.

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