Truncated prior distribution

Can anyone show me how to set out this problem so i can tackle it myself?

Suppose that a random sample of size n = 100 from an exponential distribution with parameter $\displaystyle \theta$ is about to be collected. It is thought that a Ga(2, 2) prior distribution for $\displaystyle \theta$ is appropriate but, since it is believed that values of $\displaystyle \theta$ exceeding 6 are impossible, the prior distribution is truncated above 6.

Calculate the prior probability that $\displaystyle \theta$ exceeds 6 assuming the truncated prior distribution. What is the probability when the prior distribution is not truncated?

Thanks!

I think what they intend for you to do is renormalize the gamma so that it has support (0, 6). So, you would want to find a k so that

$\displaystyle \displaystyle
\int_0 ^ 6 k x e^{x/2} \ dx = 1
$

and then take your prior to be $\displaystyle \pi(\theta) = k \theta e^{\theta / 2} I(0 < \theta < 6)$.

(Note: there are different parameterizations of the gamma. I am using the one where Gamma(a, b) implies that the mean of the distribution is ab.)

Ok so then i can do that integration and i find that k = $\displaystyle \frac{1}{8e^{3}+4}$. So is I the fisher matrix?

Also, i think actually i am more used to seeing the parameterization of the gamma where the mean is a/b.

So assumably this would change my integral?

$\displaystyle I$ is the indicator function. It is equal to 1 when its argument is true, and 0 otherwise.

And yes, the integral would change if you are using a different parameterization. I hate the one where the mean is a/b.

Ok so what do i need to do to form my integral in order to find k?

Just integrate the gamma density from 0 to 6 and normalize it so that it integrates to 1.

Ok sorry so i dont know which gamma density i am integrating....? because its got to be different to the 1 you suggested i guess? What was the difference between "the gamma density" and the one you suggested?

ok iv managed to find what k is and so i beleive iv got my prior distribution..... but i dont understand the question asks "Calculate the prior probability that $\displaystyle \theta$ exceeds 6 assuming the truncated prior distribution". This confuses me, as isnt the probability 0 because we have truncated it. i.e. i have a pdf for 0 < $\displaystyle \theta$< 6 and then i have 0, otherwise? Help!

Quote:

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Originally Posted by sirellwood

ok iv managed to find what k is and so i beleive iv got my prior distribution..... but i dont understand the question asks "Calculate the prior probability that $\displaystyle \theta$ exceeds 6 assuming the truncated prior distribution". This confuses me, as isnt the probability 0 because we have truncated it. i.e. i have a pdf for 0 < $\displaystyle \theta$< 6 and then i have 0, otherwise? Help!

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Yes, that is odd. They might have meant to say the probability that X exceeds 6.