1. ## Posterior distribution

Question:
Let the unknown probability that a basketball player makes a shot successfully be Θ. Suppose your prior on Θ is uniform on [0,1] and that she then makes two shots in a row. Assume that the outcomes of the two shots are independent.
What is the posterior density of Θ?
What would you estimate the probability that she makes a third shot to be?

My attempt:
If I let X be the event that she made two shots then I have the posterior as
f(Θ|X) = f(X,Θ)/f(X)
Given that making a shot is binomial,
f(X,Θ) = (n C x)*Θ^x*(1-Θ)^n-x
where (n C x) is the binomial coefficient, n choose x.
Then f(Θ|X) = f(X,Θ)/f(X)
If I integrate out the marginal distribution from f(X|Θ) over the interval [0,1], I get
f(X) = 1/(n+1)

Hence,
f(Θ|X) = [(n+1)*n!/x!(n-x)!]*Θ^x*(1-Θ)^n-x
= [Γ(n+2)/Γ(x+1)*Γ(n-x+1)]*Θ^x*(1-Θ)^n-x
which is beta(x-1,n-x+1)

Does this look correct?

Then for the second question, would I just take the expected value of the beta with these parameters? I.e. if y ~ beta(1,1), E[y] = 0.5?
Here, I got the parameters from the fact that we now need 1 success in 1 try.

Thanks in advance for any input you may have.

2. Originally Posted by speedBoots
Question:
Let the unknown probability that a basketball player makes a shot successfully be Θ. Suppose your prior on Θ is uniform on [0,1] and that she then makes two shots in a row. Assume that the outcomes of the two shots are independent.
What is the posterior density of Θ?
What would you estimate the probability that she makes a third shot to be?

My attempt:
If I let X be the event that she made two shots then I have the posterior as
f(Θ|X) = f(X,Θ)/f(X)
$p(\theta|x)=\frac{p(x|\theta)p(\theta)}{p(x)}$

where:

$p(x|\theta)=\frac{2!}{x!(2-x)!}\theta^x (1-\theta)^{2-x} ,\ x\in\{0,1,2\}$

$
p(\theta)=1 ,\ \theta \in [0,1]
$

and:

$
p(x)=\int_0^1 p(x|\theta)p(\theta)\; d\theta
$

where we are interested in the case $x=2$.

Then the probability of making the third shot will be:

$p_3=p(\theta|2)$

CB