Truncated prior distribution

**sirellwood** · March 23rd 2011, 05:20 AM

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 $\theta$ is about to be collected. It is thought that a Ga(2, 2) prior distribution for $\theta$ is appropriate but, since it is believed that values of $\theta$ exceeding 6 are impossible, the prior distribution is truncated above 6.

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

Thanks!

**theodds** · March 23rd 2011, 09:13 AM

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

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

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

(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.)

**sirellwood** · March 23rd 2011, 09:50 AM

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

**sirellwood** · March 23rd 2011, 09:54 AM

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?

**theodds** · March 23rd 2011, 10:09 AM

$I$ 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.

**sirellwood** · March 23rd 2011, 10:15 AM

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

**theodds** · March 23rd 2011, 10:21 AM

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

**sirellwood** · March 23rd 2011, 10:29 AM

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?

**sirellwood** · March 23rd 2011, 05:58 PM

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 $\theta$ 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 < $\theta$ < 6 and then i have 0, otherwise? Help!

**theodds** · March 23rd 2011, 07:15 PM

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 $\theta$ 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 < $\theta$ < 6 and then i have 0, otherwise? Help!

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

Thread: Truncated prior distribution

LinkBack

Thread Tools

Display

Truncated prior distribution

Similar Math Help Forum Discussions

partial differentiation variance truncated normal distribution

normal distribution prior and posterior distribution proof

distribution of an mle for the Truncated Exponential Distribution

Truncated exponential distribution problem.

Truncated and Non-truncated distribution

Search Tags