Introduction

Consider a simple linear regression problem where it is desired to estimate a set of parameters using a least squares criterion.

We generate some synthetic data where we know the model completely, that is

\[ Y = X\beta + \epsilon \]

where \(Y\) is a \(100\times 1\) vector, \(X\) is a \(100\times 10\) matrix, \(\beta = [-4,\ldots ,-1, 0, 1, \ldots, 5]\) is a \(10\times 1\) vector, and \(\epsilon \sim N(0, 1)\).

set.seed(123)

n <- 100
p <- 10
beta <- -4:5   # beta is just -4 through 5.

X <- matrix(rnorm(n * p), nrow=n)
colnames(X) <- paste0("beta_", beta)
Y <- X %*% beta + rnorm(n)

Given the data \(X\) and \(Y\), we can estimate the \(\beta\) vector using lm function in R that fits a standard regression model.

ls.model <- lm(Y ~ 0 + X)   # There is no intercept in our model above
m <- matrix(coef(ls.model), ncol = 1)
rownames(m) <- paste0("$\\beta_{", 1:p, "}$")
library(kableExtra)
knitr::kable(m, format = "html") %>%
    kable_styling("striped") %>%
    column_spec(1:2, background = "#ececec")
\(\beta_{1}\) -3.9196886
\(\beta_{2}\) -3.0117048
\(\beta_{3}\) -2.1248242
\(\beta_{4}\) -0.8666048
\(\beta_{5}\) 0.0914658
\(\beta_{6}\) 0.9490454
\(\beta_{7}\) 2.0764700
\(\beta_{8}\) 3.1272275
\(\beta_{9}\) 3.9609565
\(\beta_{10}\) 5.1348845

These are the least-squares estimates and can be seen to be reasonably close to the original \(\beta\) values -4 through 5.

The CVXR formulation

The CVXR formulation states the above as an optimization problem:

\[ \begin{array}{ll} \underset{\beta}{\mbox{minimize}} & \|y - X\beta\|_2^2, \end{array} \] which directly translates into a problem that CVXR can solve as shown in the steps below.

  • Step 0. Load the CVXR library
suppressWarnings(library(CVXR, warn.conflicts=FALSE))
  • Step 1. Define the variable to be estimated
betaHat <- Variable(p)
  • Step 2. Define the objective to be optimized
objective <- Minimize(sum((Y - X %*% betaHat)^2))

Notice how the objective is specified using functions such as sum, *%* and ^, that are familiar to R users despite that fact that betaHat is no ordinary R expression but a CVXR expression.

  • Step 3. Create a problem to solve
problem <- Problem(objective)
  • Step 4. Solve it!
result <- solve(problem)
  • Step 5. Extract solution and objective value
## Objective value: 97.847586

We can indeed satisfy ourselves that the results we get matches that from lm.

m <- cbind(result$getValue(betaHat), coef(ls.model))
colnames(m) <- c("CVXR est.", "lm est.")
rownames(m) <- paste0("$\\beta_{", 1:p, "}$")
knitr::kable(m, format = "html") %>%
    kable_styling("striped") %>%
    column_spec(1:3, background = "#ececec")
CVXR est. lm est.
\(\beta_{1}\) -3.9196887 -3.9196886
\(\beta_{2}\) -3.0117041 -3.0117048
\(\beta_{3}\) -2.1248257 -2.1248242
\(\beta_{4}\) -0.8666045 -0.8666048
\(\beta_{5}\) 0.0914653 0.0914658
\(\beta_{6}\) 0.9490453 0.9490454
\(\beta_{7}\) 2.0764693 2.0764700
\(\beta_{8}\) 3.1272271 3.1272275
\(\beta_{9}\) 3.9609564 3.9609565
\(\beta_{10}\) 5.1348848 5.1348845

Wait a minute! What have we gained?

On the surface, it appears that we have replaced one call to lm with at least five or six lines of new R code. On top of that, the code actually runs slower, and so it is not clear what was really achieved.

So suppose we knew that the \(\beta\)s were nonnegative and we wish to take this fact into account. This is nonnegative least squares regression and lm would no longer do the job.

In CVXR, the modified problem merely requires the addition of a constraint to the problem definition.

problem <- Problem(objective, constraints = list(betaHat >= 0))
result <- solve(problem)
m <- matrix(result$getValue(betaHat), ncol = 1)
rownames(m) <- paste0("$\\beta_{", 1:p, "}$")
knitr::kable(m, format = "html") %>%
    kable_styling("striped") %>%
    column_spec(1:2, background = "#ececec")
\(\beta_{1}\) 0.0000000
\(\beta_{2}\) 0.0000000
\(\beta_{3}\) 0.0000000
\(\beta_{4}\) 0.0000000
\(\beta_{5}\) 1.2374544
\(\beta_{6}\) 0.6234659
\(\beta_{7}\) 2.1230714
\(\beta_{8}\) 2.8035606
\(\beta_{9}\) 4.4448008
\(\beta_{10}\) 5.2073465

We can verify once again that these values are comparable to those obtained from another R package, say nnls.

library(nnls)
nnls.fit <- nnls(X, Y)$x
m <- cbind(result$getValue(betaHat), nnls.fit)
colnames(m) <- c("CVXR est.", "nnls est.")
rownames(m) <- paste0("$\\beta_{", 1:p, "}$")
knitr::kable(m, format = "html") %>%
    kable_styling("striped") %>%
    column_spec(1:3, background = "#ececec")
CVXR est. nnls est.
\(\beta_{1}\) 0.0000000 0.0000000
\(\beta_{2}\) 0.0000000 0.0000000
\(\beta_{3}\) 0.0000000 0.0000000
\(\beta_{4}\) 0.0000000 0.0000000
\(\beta_{5}\) 1.2374544 1.2374488
\(\beta_{6}\) 0.6234659 0.6234665
\(\beta_{7}\) 2.1230714 2.1230663
\(\beta_{8}\) 2.8035606 2.8035640
\(\beta_{9}\) 4.4448008 4.4448016
\(\beta_{10}\) 5.2073465 5.2073521

Okay that was cool, but…

As you no doubt noticed, we have done nothing that other R packages could not do.

So now suppose further, for some extraneous reason, that the sum of \(\beta_2\) and \(\beta_3\) is known to be negative and but all other \(\beta\)s are positive.

It is clear that this problem would not fit into any standard package. But in CVXR, this is easily done by adding a few constraints.

To express the fact that \(\beta_2 + \beta_3\) is negative, we construct a row matrix with zeros everywhere, except in positions 2 and 3 (for \(\beta_2\) and \(\beta_3\) respectively).

A <- matrix(c(0, 1, 1, rep(0, 7)), nrow = 1)
colnames(A) <- paste0("$\\beta_{", 1:p, "}$")
knitr::kable(A, format = "html") %>%
    kable_styling("striped") %>%
    column_spec(1:10, background = "#ececec")
\(\beta_{1}\) \(\beta_{2}\) \(\beta_{3}\) \(\beta_{4}\) \(\beta_{5}\) \(\beta_{6}\) \(\beta_{7}\) \(\beta_{8}\) \(\beta_{9}\) \(\beta_{10}\)
0 1 1 0 0 0 0 0 0 0

The sum constraint is nothing but \[ A\beta < 0 \]

which we express in R as

constraint1 <- A %*% betaHat < 0

NOTE: The above constraint can also be expressed simply as

constraint1 <- betaHat[2] + betaHat[3] < 0

but it is easier working with matrices in general with CVXR.

For the positivity for rest of the variables, we construct a \(10\times 10\) matrix \(A\) to have 1’s along the diagonal everywhere except rows 2 and 3 and zeros everywhere.

B <- diag(c(1, 0, 0, rep(1, 7)))
colnames(B) <- rownames(B) <- paste0("$\\beta_{", 1:p, "}$")
knitr::kable(B, format = "html") %>%
    kable_styling("striped") %>%
    column_spec(1:11, background = "#ececec")
\(\beta_{1}\) \(\beta_{2}\) \(\beta_{3}\) \(\beta_{4}\) \(\beta_{5}\) \(\beta_{6}\) \(\beta_{7}\) \(\beta_{8}\) \(\beta_{9}\) \(\beta_{10}\)
\(\beta_{1}\) 1 0 0 0 0 0 0 0 0 0
\(\beta_{2}\) 0 0 0 0 0 0 0 0 0 0
\(\beta_{3}\) 0 0 0 0 0 0 0 0 0 0
\(\beta_{4}\) 0 0 0 1 0 0 0 0 0 0
\(\beta_{5}\) 0 0 0 0 1 0 0 0 0 0
\(\beta_{6}\) 0 0 0 0 0 1 0 0 0 0
\(\beta_{7}\) 0 0 0 0 0 0 1 0 0 0
\(\beta_{8}\) 0 0 0 0 0 0 0 1 0 0
\(\beta_{9}\) 0 0 0 0 0 0 0 0 1 0
\(\beta_{10}\) 0 0 0 0 0 0 0 0 0 1

The constraint for positivity is \[ B\beta > 0 \]

which we express in R as

constraint2 <- B %*% betaHat > 0

Now we are ready to solve the problem just as before.

problem <- Problem(objective, constraints = list(constraint1, constraint2))
result <- solve(problem)

And we can get the estimates of \(\beta\).

m <- matrix(result$getValue(betaHat), ncol = 1)
rownames(m) <- paste0("$\\beta_{", 1:p, "}$")
knitr::kable(m, format = "html") %>%
    kable_styling("striped") %>%
    column_spec(1:2, background = "#ececec")
\(\beta_{1}\) 0.0000000
\(\beta_{2}\) -2.8447019
\(\beta_{3}\) -1.7109799
\(\beta_{4}\) 0.0000000
\(\beta_{5}\) 0.6641321
\(\beta_{6}\) 1.1780936
\(\beta_{7}\) 2.3286068
\(\beta_{8}\) 2.4144816
\(\beta_{9}\) 4.2119206
\(\beta_{10}\) 4.9483132

This demonstrates the chief advantage of CVXR: flexibility. Users can quickly modify and re-solve a problem, making our package ideal for prototyping new statistical methods. Its syntax is simple and mathematically intuitive. Furthermore, CVXR combines seamlessly with native R code as well as several popular packages, allowing it to be incorporated easily into a larger analytical framework. The user is free to construct statistical estimators that are solutions to a convex optimization problem where there may not be a closed form solution or even an implementation. Such solutions can then be combined with resampling techniques like the bootstrap to estimate variability.

Further Reading

We hope we have whet your appetite. You may wish to read a longer introduction with more examples.

We also have a number of tutorial examples available to study and mimic.

Session Info

sessionInfo()
## R version 3.4.2 (2017-09-28)
## Platform: x86_64-apple-darwin15.6.0 (64-bit)
## Running under: macOS High Sierra 10.13.1
## 
## Matrix products: default
## BLAS: /Library/Frameworks/R.framework/Versions/3.4/Resources/lib/libRblas.0.dylib
## LAPACK: /Library/Frameworks/R.framework/Versions/3.4/Resources/lib/libRlapack.dylib
## 
## locale:
## [1] en_US.UTF-8/en_US.UTF-8/en_US.UTF-8/C/en_US.UTF-8/en_US.UTF-8
## 
## attached base packages:
## [1] methods   stats     graphics  grDevices datasets  utils     base     
## 
## other attached packages:
## [1] nnls_1.4         CVXR_0.94-4      kableExtra_0.6.0
## 
## loaded via a namespace (and not attached):
##  [1] gmp_0.5-13.1      Rcpp_0.12.13      compiler_3.4.2   
##  [4] plyr_1.8.4        highr_0.6         R.methodsS3_1.7.1
##  [7] R.utils_2.6.0     tools_3.4.2       digest_0.6.12    
## [10] bit_1.1-12        evaluate_0.10.1   tibble_1.3.4     
## [13] viridisLite_0.2.0 lattice_0.20-35   rlang_0.1.2      
## [16] Matrix_1.2-11     yaml_2.1.14       blogdown_0.1.7   
## [19] Rmpfr_0.6-1       ECOSolveR_0.3-2   httr_1.3.1       
## [22] stringr_1.2.0     xml2_1.1.1        knitr_1.17       
## [25] hms_0.3           rprojroot_1.2     bit64_0.9-7      
## [28] grid_3.4.2        R6_2.2.2          rmarkdown_1.6    
## [31] bookdown_0.5      readr_1.1.1       magrittr_1.5     
## [34] scs_1.1-1         backports_1.1.1   scales_0.5.0     
## [37] htmltools_0.3.6   rvest_0.3.2       colorspace_1.3-2 
## [40] stringi_1.1.5     munsell_0.4.3     R.oo_1.21.0

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