# Thread: Definite integrals for Bayesian statistics

1. ## Definite integrals for Bayesian statistics

Hi,

As far as I know this has an analytical solution but I'm not sure how to go about doing it. Just to clarify, I'm not looking for a numerical method.

Sorry, don't know how to use maths symbols here:

I need to find x and y, such that the integral of exp(-ax^2 - b), evaluated from -infinity to x, equals 0.025, and the same integral evaluated from y to infinity equals 0.0975.

a and b are known constants.

(The question is just to find the 95% confidence interval from a posterior distribution).

Can anyone help?

Thanks!

2. Originally Posted by bsk
Hi,

As far as I know this has an analytical solution but I'm not sure how to go about doing it. Just to clarify, I'm not looking for a numerical method.

Sorry, don't know how to use maths symbols here:

I need to find x and y, such that the integral of exp(-ax^2 - b), evaluated from -infinity to x, equals 0.025, and the same integral evaluated from y to infinity equals 0.0975.

a and b are known constants.

(The question is just to find the 95% confidence interval from a posterior distribution).

Can anyone help?
Thanks!
up to some factors which you can pick out for instance take the $\displaystyle e^{-b}$ out of the integrand and note that $\displaystyle e^{-ax^2}$ can be handled with the proper choice of $\displaystyle \sigma$, where $\displaystyle \sigma$ is the square root of the variance of a normally distributed r.v., use the table for the normal distribution for answer.

3. Thanks, I think I've got it.

4. just be sure u scale things properly in the end, for there is the $\displaystyle \frac1{|\sigma|\sqrt{2\pi}}$ to consider, i.e., if X=$\displaystyle \mu+\sigma{Z}$ is a normal r.v., then p(x), the probability density function of X is given by,
$\displaystyle p(x) = \frac1{|\sigma|\sqrt{2\pi}}exp(-\frac{(x-\mu)^2}{2\sigma^2})$

And thus $\displaystyle P(a\leq{X}\leq{b}) = \int_a^bp(x)dx$