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2017 Aug 08
0
Latin hypercube sampling from a non-uniform distribution
> However, my variable is simulated from the cumulative distribution function > of the Poisson distribution. Then I am afraid I don't know what you're trying to achieve. Or why. However, the principle holds; write a function that maps [0,1] to the 'pattern' you want, do that and apply it to the result from randomLHS. It happens that for generating data that follow a given
2017 Aug 08
1
Latin hypercube sampling from a non-uniform distribution
...in England 2991879. Registered office: Queens Road, Teddington, Middlesex, TW11 0LY, UK -------------- next part -------------- A non-text attachment was scrubbed... Name: Rplot.png Type: image/png Size: 5579 bytes Desc: Rplot.png URL: <https://stat.ethz.ch/pipermail/r-help/attachments/20170808/6252fb45/attachment.png>
2017 Aug 07
2
Latin hypercube sampling from a non-uniform distribution
Thanks for your answer. However, my variable is simulated from the cumulative distribution function of the Poisson distribution. So, the pattern obtained from the function "qpois" is not the same as the observed pattern (i.e., obtained from the function "ppois") set.seed(5) mortality_probability <- round(ppois(seq(0, 7, by = 1), lambda = 0.9), 2)