This is from a Bayesian problem I'm working on. I have worked out \begin{align} f(y_1,...,y_T|\varphi)=f(y_1|\varphi)f(y_2|y_1,\varphi)...f(y_T|y_1,y_2,...,y_{T-1},\varphi), \end{align} and all terms in the equation above are known.

Now let $X=(x_1,...,x_T)$ with $x_i=y_i$ if $y_i>0$ and $x_i=0$ if $y_i\le0$. How do I calculate $f(x_i,...,x_T|\varphi)$?

This is a bit like the tobit model but $f(y_i)$ is unknown in my case, I only have their conditional density..

Many thanks!

  • 1
    $\begingroup$ This question is posted simultaneously on math.SE as well. $\endgroup$ Feb 11, 2014 at 6:21
  • $\begingroup$ Take T=1. How do you calculate $f(x_1|\varphi)$ from $f(y_1|\varphi)$? (from X you can calculate Y. Reverse case, you require some assumption.) $\endgroup$
    – vinux
    Feb 11, 2014 at 10:12
  • $\begingroup$ Umm I don't think you can calculate Y from X. Because for $X=0$ you only know $Y\le0$ but don't know the exact values.. Could you please be more specific? $\endgroup$
    – dynamic89
    Feb 11, 2014 at 10:27
  • $\begingroup$ You are right.. it is in the other way. $\endgroup$
    – vinux
    Feb 11, 2014 at 12:29
  • $\begingroup$ Can you define the notation? $\endgroup$
    – AdamO
    Jun 5, 2014 at 21:03

1 Answer 1


Let $c(x|y)$ be your censoring function. Then $$ f(x_1,...,x_T) = \int_{y_1} \cdots \int_{y_T} f(y_1,...,y_T) \prod_{i=1}^T c(x_i|y_i) dy_i $$ Note that if any $x_i > 0$ then $c(x_i|y_i)$ forces $y_i = x_i$ and the integral over that $y_i$ disappears. You are only left with integrals over the $y$'s where $x_i=0$.


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