Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

So the equation that I was trying to solve was:

$$I=\int\limits_{a}^\infty \frac{\exp(x)}{2\pi\lambda}\int\limits_0^\infty \frac{1}{\tau^2}\exp\left(-\frac{(x-\mu)^2}{2\tau^2}-\frac{(\log(\sigma/\tau))^2}{2\lambda^2}\right)d\tau dx$$.

Where $a\in\mathbb{R}$, $\mu\in\mathbb{R}$, $\lambda\in\mathbb{R}^+$ and $\sigma\in\mathbb{R}^+$.

Now for the entire integration-range the integrand is positive so if I change the order of integration, then I should get the same solution for this one. If I do that this gives:

$$I=\frac{1}{2\pi\lambda}\int\limits_{0}^\infty \frac{\exp(-\frac{(\log(\sigma/\tau))^2}{2\lambda^2})}{\tau^2}\int\limits_a^\infty \exp\left(-\frac{(x-\mu)^2}{2\tau^2}+x\right)dx d\tau$$.

The integral over x is doable and gives: $$\int\limits_a^\infty \exp\left(-\frac{(x-\mu)^2}{2\tau^2}+x\right)dx=\exp\left(\mu+\frac{\tau^2}{2}\right)\sqrt{\frac{\pi}{2}}\tau\left(1+\text{erf}\left(\frac{-a+\mu+\tau^2}{\sqrt{2}\tau}\right)\right).$$

Where erf($x$) denotes the error function. Now with the exponential going as $\exp(\tau^2)$ the integral over $\tau$ will diverge! (I also checked the limit $\tau\rightarrow\infty$ and this yielded that the integrand indeed went to $\infty$.

This seems strange to mee since I've done numerical integration of the distribution:

$$P(x) = \frac{1}{2\pi\lambda}\int\limits_0^\infty \frac{1}{\tau^2}\exp\left(-\frac{(x-\mu)^2}{2\tau^2}-\frac{(\log(\sigma/\tau))^2}{2\lambda^2}\right)d\tau$$, which is a mixture variance model, where the normal distribution is broadend by a lognormal distribution of the standarddeviation. And this one seemed to converge. I've also been able to calculate the integral over x for the case where $\lambda \rightarrow 0$ and this even gave me some analytical results! So the integral over $\tau$ should definetely be finite (for as far as I know) and the integration over x for $\exp(x)$ shouldn't blow up the intire thing because the $\exp(-x^2)$ should normally compensate for that. Does anyone know what I might be doing wrong ?

share|improve this question
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.