I have this density $f(y|z) = \kappa*\exp(-\kappa y) / (1 - \exp(-\kappa z))$, where $\kappa$ is some known value and $0 < y < \kappa$. I get the distribution by integrating with respect to $y$. I want to sample $y$'s from the resulting distribution and I have tried Markov chain Monte Carlo to do this task. As I am not a statistician and I am reading on my own, I was wondering if there is a better or a simpler way to sample from the resulting distribution. Thank you for any assistance you can provide.
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$\begingroup$ If you want to sample $y$, you want to integrate out $z$. This is simple calculus which gives $k C e^{-ky}$, assuming the integral $C$ is finite. Now, you can draw from $e^{-ky}$. $\endgroup$– horaceTCommented Jun 20, 2016 at 23:58
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1$\begingroup$ 1. Your stated conditional density doesn't integrate to 1. $\:$ 2. Conditional on $z$, the denominator is a scaling constant, so you simply have a truncated exponential on $(0,\kappa)$ (as you state it); a variety of standard RNG methods will for for that. $\endgroup$– Glen_bCommented Jun 21, 2016 at 1:01
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The density can be integrated with respect to $y$ and then inverted, which allows the inverse transform sampling method https://en.wikipedia.org/wiki/Inverse_transform_sampling to be used to generate random numbers from the distribution having the density $f(y|z)$.
First of all, note that in order to make $f(y|z)$ a proper density, i.e., integrating to 1 over its domain, the needed condition is $0 \le y \le z$, not $0 \lt y \lt \kappa$.
The cumulative distribution $$F(y|z) = -(exp(\kappa*z)-1-exp(-\kappa*(y-z))+exp(-\kappa*y))/((exp(\kappa*z)-1)*(-1+exp(-\kappa*z)))$$ for $0 \le y \le z$. Note that $F(0|z) = 0$ and $F(z|z) = 1$.
This can be inverted, resulting in $$ln(1/(U-exp(\kappa*z)*(U-1)))/\kappa+z$$ as the formula to generate a random number from a distribution having density $f(y|z)$, where $U$ is a random number drawn from a $Uniform[0,1]$ random number generator. For each random number to be drawn from the distribution having density $f(y|z)$, a single value of $U$ is drawn, and this single value is used in both locations in which it appears in the formula.
As can be seen, $U$ values of $0$ to $1$ produce random numbers via the formula ranging from $0$ to $z$.
answered Jun 21, 2016 at 0:48
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$\begingroup$ Thank you very much for your reply. I will redo it on my own to see if I got it. Thank you. $\endgroup$– MariaCommented Jun 21, 2016 at 11:48
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$\begingroup$ Can I please ask another question? If I want to sample from the conditional distribution of $(Y_{1}+Y_{2}) | (z_{1},z_{2})$ how would I do it? Is it enough to sample $Y_{1}|z_{1}$ and $Y_{2}|z_{2}$ using the inversted formula and then add them? $\endgroup$– MariaCommented Jun 23, 2016 at 14:19
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$\begingroup$ I don't know the context of what you are doing. Is there a meaning to $(Y_1+Y_2)|(z_1,z_2)$ other than as $Y_1|z_1 + Y_2|z_2$? $\endgroup$ Commented Jun 23, 2016 at 15:13
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$\begingroup$ I am trying to understand a bit more. With your answer to my question I am able now to sample $y$'s from the conditional distribution of $y|z$. Now I want to go further and I ask myself. If I sample from $y_{1}|z_{1}$ and name these $W_{1}$ and $y_{2}|z_{2}$ and name these $W_{2}$, how can I get the distribution of the sum $W_{1}+W_{2}$ . I understand that this is a conditional distribution since the $W$'s come from conditional distributions. My previous notation perhaps is not written correctly but I explain now what I am after. Thank you. $\endgroup$– MariaCommented Jun 23, 2016 at 15:24
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$\begingroup$ it sounds like you can draw $W_1$ and $W_2$ values separately (independently) and add them as you propose. But as I wrote before, I still don't know your bigger context (any dependency between $Y$ values for different $z$ values?), or some other kind of dependency going on, in which case we need to know more than $f(y|z)$ to define the problem.. $\endgroup$ Commented Jun 23, 2016 at 15:37