Sampling with exponential tilting from a Weibull distribution

Question

Given the Weibull distribution: $$F(x)=1-e^{-\theta x^\beta}$$ Is there a way to sample from the tilted distribution? First we find probability density: $$f(x)=\beta x^{\beta-1}\theta e^{-\theta x^\beta}$$ The weibull distribution has a MGF, so there is no problem there. Is there a method to sample from the tilted distribution with density: $$f_c=e^{cx}\beta x^{\beta-1}\theta e^{-\theta x^\beta}/\phi(c),\text{where $\phi$ denotes the MGF }$$ I attempted to use generalized rejection sampling, but could not find a function that I could prove was dominating $f_c(x)$. Any ideas?

Answer 1

For the analysis, set $\theta=1$ (without loss of generality). For $\beta<1$ the MGF is infinite for any $c>0$, so for these $\beta$ exponential tilting cannot work (assuming the intention is to tilt toward larger values).

For $\beta>1$, rejection sampling from a Weibull proposal with $1< \beta_0 <\beta$ would work in the right tail; there, the ratio of $f_c(x; \beta)/f(x; \beta_0)$ is bounded, as can be seen from the log (omitting terms not depending on $x$): $$ (\beta-\beta_0) \log x + cx-x^{\beta_0} \left (x^{\beta-\beta_0}-1 \right ); $$ for actual implementation a lot of details remain. However, near zero the ratio is unbounded---it seems this can only be solved by splitting the domain and using a different proposal on the lower part.

All in all, there seem to be many difficulties, the source of which seems to be that the Weibull is not a member of the exponential family of distributions, so tilting results in a non-Weibull, non exponential-family distribution. I don't know what your purpose is, but perhaps this is useful: if $Z$ is a standard exponential random variable, then $X=(Z/\theta)^{1/\beta}$ has a Weibull($\theta$, $\beta$) distribution; this expression is amenable to producing larger values, by changing the parameters.