My Math Forum  

Go Back   My Math Forum > College Math Forum > Advanced Statistics

Advanced Statistics Advanced Probability and Statistics Math Forum


Thanks Tree2Thanks
  • 1 Post By romsek
  • 1 Post By romsek
Reply
 
LinkBack Thread Tools Display Modes
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.
zollen is offline  
 
January 12th, 2019, 10:10 PM   #2
Senior Member
 
romsek's Avatar
 
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
romsek is offline  
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}
$
zollen is offline  
January 13th, 2019, 07:44 AM   #4
Senior Member
 
romsek's Avatar
 
Joined: Sep 2015
From: USA

Posts: 2,494
Thanks: 1369

Quote:
Originally Posted by zollen View Post
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

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

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

  My Math Forum > College Math Forum > Advanced Statistics

Tags
bayesian, hypothesis, testing



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
Hypothesis Testing Help redsaints2010 Probability and Statistics 1 May 23rd, 2017 02:13 AM
Hypothesis Testing Help redsaints2010 Advanced Statistics 2 May 1st, 2017 01:35 PM
Hypothesis testing kstudent Advanced Statistics 3 April 12th, 2015 11:31 AM
hypothesis testing tsl182forever8 Advanced Statistics 2 April 27th, 2012 01:53 PM





Copyright © 2019 My Math Forum. All rights reserved.