1
$\begingroup$

For a group of random variables $X_{i:\,0\leq i < k}$, I am interested in finding the liklihood $f(x\in X_i)=\text{P}(x|\text{model}$) given that $X_i \sim \mathcal{N}(\mu_i+\Theta_i, e^{\beta_0}(\mu_i+\Theta_i)^{\beta_1})$, $\Theta \sim \text{Exp}(\phi)$. Or, in other words, an exponentially modified gaussian with a mean-variance relationship modeled with two parameters $\beta_0,\beta_1$ s.t. $\log\text{Var}(X) = \beta_0+\beta_1\log\mathbb{E}(X)$. This gives the marginal liklihood:$$f(x) = \frac{1}{\phi\sqrt{2\pi}}\int_0^\infty e^{-\frac{1}{2}\beta_0}(\mu_i+\Theta_i)^{-\frac{1}{2}\beta_1}\exp\left(-\frac{(x-(\mu_i+\Theta_i))^2}{e^{\beta_0}(\mu_i+\Theta_i)^{\beta_1}}-\frac{1}{\phi}\Theta_i \right)d\Theta_i $$ Currently I use numerical integration, but I am wondering if there is a canonical form that could speed up the integration (such as in terms of hypergeometric functions)? Any suggestions, including suggestions on algorithms that avoid integration all together, are welcome.

EDIT: While writing this, I found this post about the integral of a polynomial exponential, which I adapted to my particular rational function to yield: $$\frac{1}{\phi\sqrt{2\pi}}e^{-\frac{\mu_i}{\phi}}\sum_k^\infty \left[\sum_{\substack{p \\2n+m=k}}^\infty\frac{\phi^N\Gamma(N)}{(-2)^{n+p}e^{(n+m+p)\beta_0}n!m!p!}\right]x^k$$ where $N=m+2p+1-(n+m+p+\frac{1}{2})\beta_1$. Not sure if this actually converges, or if it is any better than numerical integration, though if the series converges for at least $|x| < 1,\phi < 1,$ and is rapid enough, it might be. It would also be nice to have a closed form for the coefficients. They seem to converge only when $\beta_1 > 1$, even though the integral should converge for all values of $\beta_1$. Using $2-\beta_1$ as the highest power when evaluating the integral might fix that.

$\endgroup$
1
$\begingroup$

I found a solution in the Gauss-Laguerre quadrature, which significantly sped up the numerical integration when compared to the standard integrate function in R.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.