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 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: 2,574 Thanks: 1422 $P\left[\left . p_H = \dfrac 1 4 \right| \text{observe 7 heads in 10 tries}\right] =$ $\dfrac{P\left[\text{observe 7 heads in 10 tries } | p_H = \dfrac 1 4\right]P\left[p_H = \dfrac 1 4\right]}{P[\text{observe 7 heads in 10 tries}]} =$ $\dfrac{\dbinom{10}{7}\left(\dfrac 1 4\right)^7\left(\dfrac 3 4\right)^3\cdot \dfrac 1 3}{\dbinom{10}{7}\left(\dfrac 1 4\right)^7\left(\dfrac 3 4\right)^3 + \dbinom{10}{7}\left(\dfrac 1 2\right)^{10}} = \dfrac{9}{1051}$ 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. Tags bayesian, hypothesis, testing Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post redsaints2010 Probability and Statistics 1 May 23rd, 2017 02:13 AM redsaints2010 Advanced Statistics 2 May 1st, 2017 01:35 PM kstudent Advanced Statistics 3 April 12th, 2015 11:31 AM tsl182forever8 Advanced Statistics 2 April 27th, 2012 01:53 PM

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