Timeline for Logarithmic sampling in Monte Carlo instead of linear
Current License: CC BY-SA 3.0
10 events
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| Jan 8, 2018 at 17:30 | comment | added | jbowman | ... or you could just do something like increment each successive one by one, so, for example, instead of $[30, 30, 30, ... ]$ you get $[30, 31, 32, ...]$, at the lower end. | |
| Jan 8, 2018 at 17:28 | comment | added | user929304 | no, they are integer. But in any case, you're right it's easiest to just remove the duplicates. Thanks again. | |
| Jan 8, 2018 at 1:14 | comment | added | jbowman | In your application, do those values need to be integers? I made them so because I had to generate an integer number of random values, but maybe you don't. | |
| Jan 7, 2018 at 21:34 | comment | added | jbowman | Well, I'd just eliminate the duplicate values myself, leaving you with one copy of each value. | |
| Jan 7, 2018 at 21:09 | comment | added | jbowman |
That looks right to me, with the caveat that the maximum i you have is 29, so your maximum x[i] will be exp(29*logmax/30), not exp(logmax). This is because you need n+1 steps to get both the endpoints of the interval [lower, upper], so i <= 30 would do the job. Otherwise, if you want 30 values, you need interval = logmax/29;.
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| Jan 7, 2018 at 20:35 | comment | added | user929304 |
I am still a bit struggling in my attempt. I m writing the routine in c++, in your case the max value is 10000, and to sample you re linearly picking values between 0 to log(10000), and then you exponentiate the values so they range from 1 to 10000 again but logarithmically separated now. Right? My way of implementing your seq function has been to set logmax = log(10000);, interval=logmax/30; and then for (int i=0; i<30; i++){ x[i]=round(exp(i*interval)); } is this a correct way of recreating basically what you had shown in R? Many thanks for your help in advance
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| Dec 28, 2017 at 18:07 | comment | added | jbowman |
Yes, they are in R. seq basically just creates a vector of equally-spaced numbers starting with from, ending with to, and with length.out entries in it. So, seq(from=0, to=1.5, length.out=4) would create (0, 0.5, 1.0, 1.5). To add some possibly-useful information, rpareto creates random numbers ~ Pareto, with the first parameter (x[i]) being the number of them created - so you could think of that function call as being x[i] executions of the underlying simulation, with mean(.) as being the function that aggregates the x[i] simulation results.
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| Dec 28, 2017 at 17:32 | vote | accept | user929304 | ||
| Dec 28, 2017 at 17:32 | comment | added | user929304 | Dear jbowman, this is exactly what I was looking for, thank you very much for taking the time to write this up. Are the sample codes in R? What does seq do here? | |
| Dec 23, 2017 at 19:20 | history | answered | jbowman | CC BY-SA 3.0 |