R help - Aug 2017 - Latin hypercube sampling from a non-uniform distribution

Hello,

I am performing a sensitivity analysis using a Latin Hypercube sampling.
However, I have difficulty to draw a Hypercube sample for one variable. I?ve
generated this variable from a Poisson distribution as follows:

set.seed(5)
mortality_probability <- round(ppois(seq(0, 7, by = 1), lambda = 0.9), 2)
barplot(mortality_probability, names.arg = seq(0, 7, by = 1), xlab = "Age
class", ylab = "Probability")

How can I draw a Hypercube sample for the variable ?mortality_probability? so
that this variable exhibits the same pattern as the observed distribution?

Here is a reproducible code to draw Hypercube samples (my sensitivity analysis
includes several parameters and the variables ?var1? and ?var2? follow a uniform
distribution):

library(lhs)
set.seed(1)
parm <- c("var1", "var2",
"mortality_probability")
X <- randomLHS(100, length(parm))

Any suggestions would be much welcome.

Thanks for your time
Marine


	[[alternative HTML version deleted]]

> How can I draw a Hypercube sample for the variable  mortality_probability 
so
> that this variable exhibits the same pattern as the observed distribution?
One simple way is to use the uniform random output of randomLHS as input to the
quantile function for your desired distribution(s).

For example:

q <- randomLHS(1000, 3)
colnames(q) <- c("A", "B", "mort")
q[, "mort"] <- qpois(q[,"mort"], 1.5)


S Ellison






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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)
barplot(mortality_probability, names.arg = seq(0, 7, by = 1), xlab = "Age
class", ylab = "Probability")

library(lhs)
set.seed(1)
parm <- c("var1", "var2",
"mortality_probability")
X <- randomLHS(100, length(parm))
colnames(X) <- c("var1", "var2",
"mortality_probability")
X[, "mortality_probability"] <- qpois(X[,
"mortality_probability"], 0.9)
hist(X[, "mortality_probability"])


Thanks for your time

Marine




________________________________
De : S Ellison <S.Ellison at LGCGroup.com>
Envoy? : lundi 7 ao?t 2017 14:36
? : Marine Regis; r-help at r-project.org
Objet : RE: Latin hypercube sampling from a non-uniform distribution
> How can I draw a Hypercube sample for the variable  mortality_probability 
so
> that this variable exhibits the same pattern as the observed distribution?
One simple way is to use the uniform random output of randomLHS as input to the
quantile function for your desired distribution(s).

For example:

q <- randomLHS(1000, 3)
colnames(q) <- c("A", "B", "mort")
q[, "mort"] <- qpois(q[,"mort"], 1.5)


S Ellison






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