# Log Sum Exponential Trick On Weibull Mixture [duplicate]

I am trying to evaluate the log-likelihood of a mixture of weibull distributions and am running into problems with the numerical aspect. In short, I have $$M$$ mixtures and want to evaluate: $$\log \left[ \sum_{m=1}^M \pi_m \frac{\beta_m}{\alpha_m} \left( \frac{x}{\alpha_m}\right)^{\beta_m -1} e^{-\left( \frac{x}{\alpha_m}\right)^{\beta_m}} \right]$$

with $$\pi_m$$ being the mixing probability, $$\alpha_m$$ and $$\beta_m$$ being parameters for distribution $$m$$ and $$x$$ being some data-point we've collected.

My problem comes from evaluating the exponential term here, where I often just get 0 on my computer due to the power being very negative. I have been reading and found the 'log-sum-exponential trick' at various sources, but it is unclear to me if this is valid when we have coefficients in-front of the exponential. Is there a known numerically stable way to evaluate this, and if the log-sum-exponential trick is indeed valid, can someone show how this works with the coefficients?

• if all the terms in front of each exponential are positive you can do $ae^x = e^{x + \log a}$ to turn it into an actual log-sum-exp – jld Sep 1 '20 at 15:17
• All the coefficients in front of the exponential are also exponentials (of their logs, of course!) For an example of this see the function Y in my post at stats.stackexchange.com/a/485474/919: it implements a product and sum of Weibull distribution values. – whuber Sep 1 '20 at 15:18
• Thanks whuber (and @jld but I can't tag both) with this I've expressed the entire function as standard form log-sum-exp and applied it from there. – Kieran108 Sep 1 '20 at 16:40
• great, glad this was helpful! – jld Sep 1 '20 at 18:34