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I'm writing a Monte Carlo simulation. I'm using a Do loop to create the different simulations, iterating over different SeedRandom[] parameters in my pseudorandom number generator. I'm trying to formulate it as a function that returns a list of correlations that I can then manipulate.

I expect there is a very simple solution, but I have to admit that I'm pretty new to Mathematica so I'm having problems. The Mathematica documentation seems to conveniently dodge this issue whenever it comes up! This might be a little complicated, so I'll explain what my code does first, then I'll describe the problem precisely.

First I make a list with the parameters of each of my random variables (I want them to have different means):

varparams = Table[i, {i, 0.1, 0.9, 0.1}];

Then I define my random variables:

var[i_]:=RandomChoice[{varparams[[i]],1-varparams[[i]]}->{1,0},10]

(I'm generating [for argument's sake] a set of 10 values for each run of the simulation).

Now, I define the function that iterates over different SeedRandom[] values, so as to generate different data sets. I want to calculate a list of correlations:

corrlist = {};
corrcheck[i_, j_, n_] := Do[
  SeedRandom[k];
  AppendTo[corrlist, Correlation[var[i], var[j]]];
  , {k, n}]

(n is the number of different runs of the simulation).

Now this sort of works, in the sense that I can run corrcheck[i,j,n] and then work with the corrlist that is produced, but what I want is for corrcheck[i,j,n] itself to be a list, so that I can make tables and other manipulations of corrcheck for different values of i and j. I hope this makes sense. To reiterate: I want the function corrcheck, which consists of a Do loop, to produce a unique list for each set of i,j,n, so that I can do things like:

Table[corrcheck[i,j,100],{i,10},{j,10}]

and manipulations thereof.

The answer has to be obvious. I just can't see it!

Thank you very much in advance.

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  • $\begingroup$ Something like corrcheck[i_, j_, n_] := Table[BlockRandom[SeedRandom[k]; Correlation[var[i], var[j]]], {k, n}] then? $\endgroup$ – J. M.'s ennui♦ Sep 27 '12 at 14:10
  • $\begingroup$ Wow. I thought it would be easy, but that's really obvious now that I think about it. Thanks! I don't know why I had the idea that a Do[] loop was necessary. $\endgroup$ – Ooku Sep 27 '12 at 14:13
  • $\begingroup$ I'll probably accept that as an answer, if you care to submit it - I can't imagine a simpler solution. $\endgroup$ – Ooku Sep 27 '12 at 14:16
  • $\begingroup$ Tell me if I understood correctly: you want corrcheck[i_, j_, n_] to build a list with n simulations of correlation (10 samples each) between 2 independent bernoullis with weights defined by i and j, and you want that function to always return the same output for the same i,j,n ? $\endgroup$ – Rojo Sep 27 '12 at 14:16
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    $\begingroup$ @Ooku Do loops are not necessarily bad; in fact, their syntax is identical to Table - a functional stalwort. When a loop is necessary, Do is typically the way to go. The biggest issue with your code is the AppendTo command; look into useing Reap and Sow, rather than Append or AppendTo. $\endgroup$ – Mark McClure Sep 27 '12 at 14:45
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This seems to be what you want:

corrcheck[i_, j_, n_, opts___] := Table[BlockRandom[
   SeedRandom[k, FilterRules[{opts}, Method]];
   Correlation[var[i], var[j]]], {k, n}]

The additional optional argument allows you to change the algorithm used behind the pseudorandom number generator; you can thus do either a call like corrcheck[7, 10, 100] if you're happy with the defaults, or corrcheck[7, 10, 100, Method -> "MersenneTwister"] if you want to fool around with the Mersenne Twister.

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