I am trying to determine the weights of 9 metrics which will return the highest accuracy ratio. Since they are weights, the values need to sum to 1 and lie between 0 & 1. I am currently using the optim function, but do to constraints, I think I need to switch to constrOptim. I was wondering the best way to do this. Below I have included the code i am currently using. x.matrix is 20,000 by 9 matrix of values ranked between 1-10.
pars<-c(w1=(1/9),w2=(1/9),w3=(1/9),w4=(1/9),w5=(1/9),w6=(1/9),w7=(1/9),w8=(1/9),w9=(1/9))
OptPars<-function(pars){(-(rcorr.cens(x.matrix%*%pars),f)["Dxy"])}
opt<-optim(pars,OptPars)
Say you have values x on the range (-Inf, Inf) and you need values p in the range [0,1] that sum to 1, you can do the following transformation
p <- exp(x)/sum(exp(x))
If you do that translation in your optimization function and do the same transformation on the best set of parameters, you should get what you want.
Related
I want to generate some data which correspond to a quantile function. But the data need a min and a max value.
set.seed(30)
a1<-950 ; a2<-0; a3<-2.48; a4<-1.92
invcdf<-function (x)(a1+a2*a3*((-log(x))^(1/a4)))/(a3*((-log(x))^(1/a4))+1)
t<-invcdf(runif(2000,min=80,max=800))
When I use min and max in the runif function NaN's are produced.
How can I improve this code to avoid NaN's? I can't change the parameters.
Since you don't explain what exactly you are trying to do (which distribution are you trying to sample?), all I can do is interpret this as an attempt to generate random variable according to some distribution using its inverse CDF function. Because I don't know which it is, I can't comment on whether your implementation of it is correct.
However, when you use this method, you should know that the CDF function takes values between 0 and 1, as it is a cumulative density, starting at 0, and going to 1 in some limit.
The inverse of that function then only makes sense if you feed it values between 0 and 1, and that is where a possible error lies. runif(2000,min=80,max=800) generates random values between 80 and 800, way outside the (0,1) interval.
If you instead do this:
t <- invcdf(runif(2000))
We do get results (which happen to lie between 80 and 800 mostly):
Objective function to be maximized : pos%*%mu where pos is the weights row vector and mu is the column vector of mean returns of d stocks
Constraints: 1) ones%*%pos = 1 where ones is a row vector of 1's of size 1*d (d is the number of stocks)
2) pos%*%cov%*%t(pos) = rb^2 # where cov is the covariance matrix of size d*d and rb is risk budget which is the free parameter whose values will be changed to draw the efficient frontier
I want to write a code for this optimization problem in R but I can't think of any function or library for help.
PS: solve.QP in library quadprog has been used to minimize covariance subject to a target return . Can this function be also used to maximize return subject to a risk budget ? How should I specify the Dmat matrix and dvec vector for this problem ?
EDIT :
library(quadprog)
mu <- matrix(c(0.01,0.02,0.03),3,1)
cov # predefined covariance matrix of size 3*3
pos <- matrix(c(1/3,1/3,1/3),1,3) # random weights vector
edr <- pos%*%mu # expected daily return on portfolio
m1 <- matrix(1,1,3) # constraint no.1 ( sum of weights = 1 )
m2 <- pos%*%cov # constraint no.2
Amat <- rbind(m1,m2)
bvec <- matrix(c(1,0.1),2,1)
solve.QP(Dmat= ,dvec= ,Amat=Amat,bvec=bvec,meq=2)
How should I specify Dmat and dvec ? I want to optimize over pos
Also, I think I have not specified constraint no.2 correctly. It should make the variance of portfolio equal to the risk budget.
(Disclaimer: There may be a better way to do this in R. I am by no means an expert in anything related to R, and I'm making a few assumptions about how R is doing things, notably that you're using an interior-point method. Also, there is likely an R package for what you're trying to do, but I don't know what it is or how to use it.)
Minimising risk subject to a target return is a linearly-constrained problem with a quadratic objective, looking like this:
min x^T Q x
subject to sum x_i = 1
sum ret_i x_i >= target
(and x >= 0 if you want to be long-only).
Maximising return subject to a risk budget is quadratically-constrained, however; it looks like this:
max ret^T x
subject to sum x_i = 1
x^T Q x <= riskbudget
(and maybe x >= 0).
Convex quadratic terms in the objective impose less of a computational cost in an interior-point method compared to introducing a convex quadratic constraint. With a quadratic objective term, the Q matrix just shows up in the augmented system. With a convex quadratic constraint, you need to optimise over a more complicated cone containing a second-order cone factor and you need to be careful about how you solve the linear systems that arise.
I would suggest you use the risk-minimisation formulation repeatedly, doing a binary search on the target parameter until you've found a portfolio approximately maximising return subject to your risk budget. I am suggesting this approach because it is likely sufficient for your needs.
If you really want to solve your problem directly, I would suggest using an interface Todd, Toh, and Tutuncu's SDPT3. This really is overkill; SDPT3 permits you to formulate and solve symmetric cone programs of your choosing. I would also note that portfolio optimisation problems are particularly special cases of symmetric cone programs; other approaches exist that are reportedly very successful. Unfortunately, I'm not studied up on them.
I am using the OptimalCutpoints package in R to find the optimal cutoff point from ROC curve. The criterion for finding the optimal threshold is maximizing Youden's index:
J = sensitivity + specificity - 1
I am trying to do the same in matlab with the function perfcurve. I run perfcurve with the default criteria for two axis, the FPR in x-coordinates and TPR in y-coordinates. The perfcurve returns a matrix with thresholds and chooses one of them according to the criteria.
The problem is that the optimal threshold that matlab gives, is not the same as in R. However, the optimal threshold according to R is included in the threshold matrix that matlab returns.
How can I replicate the results that R returns with the ones in matlab? I am suspecting that the criteria are not correctly set in matlab for Youden's index.
If you look at the documentation for perfcurve (specifically the OPTROCPT row), you would see that the formula that matlab uses to find the best threshold is quite different, and includes a cost matrix in the optimality criterion.
If you want to replicate what is done in R exactly, use the X and Y return values to compute the Youden index for each threshold, and then choose the best (see how to find max and it's index in array in matlab for some idea how to do it).
I'm new to R and need a little help with a simple optimization.
I want to apply a functional transformation to a variable (sales_revenue) over time (24 month forecast values 1 to 24). Basically I want to push sales revenue for products from later months into earlier month.
The functional transformations on t time is:
trans=D+(t/(A+B*t+C*t^2))
I will then want to solve:
1) sales_revenue=sales_revenue*trans
where total_sales_revenue=1,000,000 (or within +/- 2.5%)
total_sales_revenue is the sum of all sales_revenue over the 24 months forecast.
If trans has too many parameters I can fix most of them if required and leave B free to estimate.
I think the approach should be fix all parameters except B, differentiate function (1) (not sure what ti diff by) and solve for a non zero minima (use constraints to make sure its the right minima and no-zero, run optimization on that function with the constraint that the total sum of sales_revenue*trans will be equal (or close to) 1,000,000.
#user2138362, did you mean "1) sales_revenue=total_sales_revenue*trans"?
I'm supposing your parameters A, C and D are fixed, and you want to find B such that the distance between your observed values and your predicted values is minimized.
Let's say your time is in months. So we can write a function to give you the squared distance:
dist <- function(B)
{
t <- 1:length(sales_revenue)
total_sales_revenue <- sum(sales_revenue)
predicted <- total_sales_revenue * (D+(t/(A+B*t+C*t^2)))
sum((sales_revenue-predicted)^2)
}
I'm also using the squared euclidean distance as a measure of distance. Make the appropriate changes if that is not the case.
Now, dist is the function you have to minimize. You can use optim, as pointed out by #iTech. But even at the minimum of dist it probably won't be zero, as you have many (24) observations. But you can get the best fit, plot it, and see if it's nice.
I'm trying to generate random numbers with a multivariate skew normal distribution using the rmsn command from the sn package in R. I would like, ideally, to be able to get three columns of numbers with a specified variances and covariances, while having one column strongly skewed. But I'm struggling to achieve both goals simultaneously.
The post at skew normal distribution was related and useful (and the source of some of the code below), but hasn't completely clarified the issue for me.
I've been trying:
a <- c(5, 0, 0) # set shape parameter
s <- diag(3) # create variance-covariance matrix
w <- sqrt(1/(1-((2*(a^2)/(1 + a^2))/pi))) # determine scale parameter to get sd of 1
xi <- w*a/sqrt(1 + a^2)*sqrt(2/pi) # determine location parameter to get mean of 0
apply(rmsn(n=1000, xi=c(xi), Omega=s, alpha=a), 2, sd)
colMeans(rmsn(n=1000, xi=c(xi), Omega=s, alpha=a))
The columns means and SDs are correct for the second and third columns (which have no skew) but not the first (which does). Can anyone clarify where my code above, or my thinking, has gone wrong? I may be misunderstanding how to use rmsn, or the output. Any assistance would be appreciated.
The location is not the mean (except when there is no skew). From the documentation:
Notice that the location vector ‘xi’ does not represent the mean
vector of the distribution (which in fact may not even exist if ‘df <=
1’), and similarly ‘Omega’ is not the covariance matrix of the
distribution
And you may want to replace Omega=s with Omega=w.
And this is supposed to be a variance matrix: there should be no square root.