Avoiding cycles when projecting with a growth rate

Question

I have a set of simulated growth rates, say, 8 time periods (and 5 simulated growth paths),

r <- matrix(rnorm(40,0.05,0.01),5,8) r [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [1,] 0.04229559 0.02846659 0.04948458 0.06144443 0.05657848 0.05782358 0.05545835 0.04090866 [2,] 0.06270360 0.06045967 0.04213729 0.05413941 0.06291148 0.05382643 0.05844549 0.03824342 [3,] 0.07846056 0.05503713 0.06800700 0.05888937 0.05759237 0.03789024 0.05250413 0.05011601 [4,] 0.04248757 0.04632404 0.04199074 0.04542522 0.03473972 0.04129197 0.06614095 0.06024244 [5,] 0.04382759 0.03555406 0.06630673 0.06019894 0.05057905 0.06336362 0.04954486 0.05092946 

Then I want to use these bets using x_{t+1} = x_{t} (1+r_{t}) . I can do this using a for loop,

 x.fn<-function(x,rr){ xx<-cbind(x,rr) for(i in 1:ncol(rr)){ xx[,i+1]<-xx[,i]*(1+rr[,i]) } xx } x.fn(x=100, rr=r) x [1,] 100 104.4835 110.8389 116.7353 122.0573 128.6441 136.0210 142.7797 146.9829 [2,] 100 106.1199 111.8381 115.9955 120.3976 125.9980 132.5384 138.1477 142.3214 [3,] 100 103.5990 106.6866 111.5629 117.5999 124.9613 131.6552 137.2966 142.9163 [4,] 100 105.7215 109.7624 113.8707 120.5216 128.8663 136.1915 143.3542 150.7466 [5,] 100 104.1080 109.6981 112.7072 118.6346 126.1118 130.1564 135.8039 142.0652 

Is it possible to speed up / avoid using a for loop with a function like apply ?

Answer 1

What you think is the r + 1 cumulative product . For my "data"

 > r <- matrix(rnorm(40,0.05,0.01),5,8) > r [,1] [,2] [,3] [,4] [,5] [,6] [1,] 0.04305611 0.03545166 0.05882694 0.03910892 0.04796011 0.05631498 [2,] 0.05623084 0.03747785 0.04809927 0.05644677 0.06747468 0.05405979 [3,] 0.04389437 0.05353846 0.04600529 0.04363427 0.05399780 0.04260270 [4,] 0.04644610 0.05471288 0.06279882 0.02195831 0.05971777 0.05024525 [5,] 0.04426485 0.03294009 0.04325665 0.06006569 0.07416615 0.06585176 [,7] [,8] [1,] 0.02724187 0.03515898 [2,] 0.04285792 0.04220358 [3,] 0.04302487 0.06662048 [4,] 0.04948629 0.04256084 [5,] 0.04659738 0.03395883 

here's how to get it:

 > t(apply(r+1, 1, cumprod)) [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [1,] 1.043056 1.080034 1.143569 1.188293 1.245284 1.315412 1.351246 1.398755 [2,] 1.056231 1.095816 1.148524 1.213355 1.295225 1.365245 1.423756 1.483844 [3,] 1.043894 1.099783 1.150379 1.200575 1.265403 1.319313 1.376076 1.467751 [4,] 1.046446 1.103700 1.173011 1.198769 1.270356 1.334186 1.400210 1.459804 [5,] 1.044265 1.078663 1.125322 1.192916 1.281390 1.365771 1.429413 1.477954 

(Don't ask me why t() needed.)

Then all this is multiplied by the original size:

 > t(apply(r+1, 1, cumprod)) * 100 [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [1,] 104.3056 108.0034 114.3569 118.8293 124.5284 131.5412 135.1246 139.8755 [2,] 105.6231 109.5816 114.8524 121.3355 129.5225 136.5245 142.3756 148.3844 [3,] 104.3894 109.9783 115.0379 120.0575 126.5403 131.9313 137.6076 146.7751 [4,] 104.6446 110.3700 117.3011 119.8769 127.0356 133.4186 140.0210 145.9804 [5,] 104.4265 107.8663 112.5322 119.2916 128.1390 136.5771 142.9413 147.7954 

Oh, and remember to also add a column for the initial size:

 > cbind(1, t(apply(r+1, 1, cumprod))) * 100 [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [1,] 100 104.3056 108.0034 114.3569 118.8293 124.5284 131.5412 135.1246 [2,] 100 105.6231 109.5816 114.8524 121.3355 129.5225 136.5245 142.3756 [3,] 100 104.3894 109.9783 115.0379 120.0575 126.5403 131.9313 137.6076 [4,] 100 104.6446 110.3700 117.3011 119.8769 127.0356 133.4186 140.0210 [5,] 100 104.4265 107.8663 112.5322 119.2916 128.1390 136.5771 142.9413 [,9] [1,] 139.8755 [2,] 148.3844 [3,] 146.7751 [4,] 145.9804 [5,] 147.7954 

Answer 2

Your task depends on the previous value. This is similar to the operation of a recursive function. But I'm not sure if there will be acceleration. Here's a version that doesn't use per-se recursive functions, but calculates values ​​independently (using recursive principles):

 my_fun <- function(x, rr, idx) { i <- 1 xx <- rep(x, nrow(rr)) while( i <= idx) { xx <- xx * (1 + rr[, i]) i <- i + 1 } xx } apply(as.matrix(0:ncol(r), ncol=1), 1, function(ix) my_fun(100, r, ix)) 

If you have too many columns, this function can be used to compute all columns in parallel. I do not see other advantages of using this function over yours. But it might be better to check by comparing it.