I'm trying to implement Gibbs Sampler for the distribution: $$\pi(x,y)=e^{-10(x^2-y)^2-(y-1/4)^4}$$ So, like the first step, I need to find: $$\phi(t) = \int_{-\infty}^{t} e^{-10(x^2-y)^2-(y-0.25)^4} dx$$ After that, find $u\sim U(0,1)$ and $x_1 = \phi^{-1}(u)$.

But my problem is that $\phi(t)$ can not be calculated analytically, because it is equivalent to: $$\phi(t) = c \int_{-\infty}^t e^{-10(x^2-y)^2}dx$$

What can I do? I'm getting something wrong?

  • $\begingroup$ Are these questions for some class? $\endgroup$ – Glen_b May 11 '15 at 0:36

Gibbs sampling requires the full conditionals associated with$$π(x,y)\propto\exp\left\{ −10(x^2−y)^2−(y−1/4)^4 \right\}$$namely $$π(x|y)\propto\exp\left\{ −10(x^2−y)^2\right\}$$ and $$π(y|x)\propto\exp\left\{ −10(2yx^2−y^2)−(y−1/4)^4 \right\}$$ both of which are non-standard distributions.

Two approaches (at least) are possible:

  1. Working hard enough to figure out accept-reject algorithms for the two conditionals. This may be arduous as for instance the conditional in $X$ is bimodal;
  2. Using Metropolis-within-Gibbs which means constructing a random walk proposal in both $x$ and $y$ and accepting the moves with a Metropolis-Hastings probability.
  • $\begingroup$ I'm interested in the first approach, but I do not know how to find (aproximate?) the value of $x_1 = \phi^{-1}(u)$. On the other hand, the second approach is not exactly Gibbs Sampler, right? Thanks for your help. $\endgroup$ – Hiperion May 10 '15 at 22:17

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