Re: Bayesian estimation of Expectation in Bernoulli problem
- From: isabellesup@xxxxxxxxxxx
- Date: Thu, 23 Aug 2007 10:12:48 -0700
On Aug 23, 10:49 am, quebecs...@xxxxxxx wrote:
Hi everyone,
I would be interested in a hint, a pointer or some help with the
following problem.
Consider two coins, C1 and C2, with the following characteristics:
Pr(heads| C1) = 0.6 and Pr(heads| C2) = 0.4.
Choose one of the coins at random and imagine spinning it repeatedly.
Given that the first two spins from the chosen coin are tails, what is
the expectation of the number of additional spins until a head shows
up?
Below I have indicated my work so far.
Many thanks
Let N be the number of additional spins until a head shows up
We have N|Pr(heads)=Geometric(Pr(heads))
and E{N|Pr(heads)}=1/Pr(heads)
E{N}=E[E{N|Pr(heads)}]=E[1/Pr(heads)]
Now I need to find Pr(heads)
That is where I am stuck. I do not know how to write the fact that the
coin is chosen at random.
What I have is
Pr(heads|first 2 spins are tails, C1)= ( (1-p)^2 *0.6 ) /((1-p)^2
*0.6+(1-p)^2 *.4)
Thanks for your help.
Hello quebecstat,
Lets see if I can help.
It looks like the prior for p is
P{p=0.6}=1/2
P{p=0.4}=1/2
So that the posterior of p is
Pr(p=0.6|first 2 spins are tails)
= ( (1-0.6)^2 *.5 ) /
((1-.6)^2 *.5+(1-.4)^2 *.5)
=( .4^2 *.5 ) /
(.4^2 *.5+.6^2 *.5)
=.308
Pr(p=0.4|first 2 spins are tails)
= ( .6^2 *.5 ) /
(.4^2 *.5+.6^2 *.5)
=.692
Now this should help you compute E[1/Pr(heads)]. Although I have to
say, I am not too sure how to compute E[1/Pr(heads)].
cheers
.
- References:
- Bayesian estimation of Expectation in Bernoulli problem
- From: quebecstat
- Bayesian estimation of Expectation in Bernoulli problem
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