I have some complex-valued time-series data, $y \in \mathbb{C}^n$ - a signal with additive Gaussian white noise. The goal is to find the Fourier coefficients of this signal.

Ideally, you would just do a discrete Fourier transform, but in this case the data is non-uniformly sampled and every data point has some arbitrary complex amplitude (a "feature" of the measurement method) multiplied with it (prior to injection of noise). So, I want to solve for the underlying signal model via a linear model.

We have $y = A x + \epsilon$ where $A \in \mathbb{C}^{n \times p}$ is essentially the analogue of the DFT Matrix but also includes the complex amplitudes, $x \in \mathbb{C}^p$ are the Fourier coefficients we want to solve for, and $\epsilon \sim \mathcal{CN}(0, \sigma^2)$ is a complex normal noise term.

Due to non-uniform sampling, $A$ does not necessarily have full rank and is most definitely ill-conditioned (large ratio between highest and lowest singular values). So, we need to add regularization.

So, the problem I am trying to solve is:

$$ \min_x \|x\|^2_2 \,\, \text{ subject to } \|y - Ax\|^2_2 \le \delta \tag{1}\label{eq1} $$

This is a nice way to frame the problem for two reasons:

  1. Our knowledge of the noise term allows us to set $\delta \approx n \sigma^2$ - this is the expected RSS. This constraint prevents underfitting.

  2. $\|x\|^2_2$ has the interpretation of being the "total power" in the model. Finding the model with minimum power prevents overfitting.

These conditions should hit the right spot in the bias-variance tradeoff.

How do you go about solving this sort of problem?

Here's some of my work on the problem:

I don't know how to rigorously prove this statement, but I am fairly certain my problem is essentially equivalent to Ridge Regression.

$$ \hat{x}_\lambda = \underset{x}{\operatorname{argmin}} \, (\| y - A x \|^2_2 + \lambda \|x\|^2_2)$$

Is this true?

You can claim that as $\lambda$ decreases, $\delta$ increases (if you switch the inequality to an equality in (1)). Ideally, there's a nice way to numerically solve (1) preserving the interpretability of $\delta$.

An idea is to write the RSS as a function of $\lambda$,

$$ f(\lambda) = \| y - A \hat{x}_\lambda \|^2_2 $$

and solve for $\lambda$ in the equation: $f(\lambda) = \delta$.

This gives us:

$$ \| y - A \hat{x}_\lambda \|^2_2 = \delta \\ (y - A \hat{x}_\lambda)^* (y - A \hat{x}_\lambda) = \delta $$

where $X^*$ is the conjugate transpose of $X$. We know that,

$$ \hat{x}_\lambda = (A^* A + \lambda I)^{-1} A^* y $$

So, we have:

$$ (y - A (A^* A + \lambda I)^{-1} A^* y)^* \, (y - A (A^* A + \lambda I)^{-1} A^* y) = \delta $$

I am not sure how to approach this.


1 Answer 1


Your intuition about the connection between your problem and Ridge regression is correct.

Let's consider the least squares regression problem:

$$\begin{bmatrix} y &=& Ax &+& \epsilon \\ 0 &=& \lambda I_p x &+& e\end{bmatrix}$$

where we have "stacked" the two regression problems $y = Ax + \epsilon$ and $0 = \lambda I_p x + e$. This is equivalent to ridge regression: see @whuber's answer to How to derive the ridge regression solution?.

Clearly, if we set $\lambda = 0$, we have an unconstrained regression problem $y = Ax + \epsilon$, and, as $\lambda \to \infty$, the problem approaches the unconstrained regression problem $0 = I_p x + e$. The latter problem is evidently equivalent to $\min_x ||x||_2^2$.

Let $x_{\lambda}$ be the least squares solution to this problem for a given value of $\lambda$. Then we define:

$$ ||y - Ax_{\lambda}||_2^2 = \delta_{\lambda}$$

$x_{\lambda}$ is the solution to $\min_x ||x||_2^2 \; \text{s.t.} \; ||y - Ax||_2^2 \leq\delta_{\lambda}$.

One proof of this is by contradiction. Assume there is an $x'$ that achieves a smaller value of $||x||_2^2$ given the constraint. Then it would have achieved a smaller value of $||e||_2^2 \; (= \lambda^2||x||_2^2)$ in the original regression problem while achieving the same or smaller value of $||\epsilon||_2^2$. Since the least squares solution minimizes $||e||_2^2 + ||\epsilon||_2^2$, and $x'$ achieves a smaller value of one or both terms while not increasing the other, $x_{\lambda}$ cannot be the least squares solution. But it is by definition, therefore, no such $x'$ can exist.

How do we find $\lambda$ such that $\delta_{\lambda} = $ the desired $\delta$? A little thought will convince us that $\delta_{\lambda}$ is strictly monotonic (increasing) in $\lambda$, implying that this is a straightforward one-dimensional root finding problem, and can be solved using any number of off-the-shelf routines.

Here's some sample code in R, solving a simpler (non-complex) version of your problem:


# Original problem
A <- matrix(rnorm(100), 10, 10)
x_true <- rep(1, 10)
y <- (A %*% x_true + rnorm(10))[,1]

# Extension to Ridge formulation
yr <- c(y, rep(0, 10))

# Objective function for root-finding routine
foo <- function(lambda, delta) {
  Ar <- rbind(A, diag(lambda, 10))
  e <- residuals(lm(yr~Ar))[1:10]
  sum(e*e) - delta

# We handwave over bracketing lambda, but, due to monotonicity, it's easy to do
lambda <- uniroot(foo, lower=0.0001, upper=100, delta=5)$root

Ar <- rbind(A, diag(lambda, 10))
m1 <- lm(yr~Ar)

# Compare this to our target delta of 5
e <- residuals(m1)[1:10]
[1] 4.999904   
> coef(m1)  # constrained coefficients
(Intercept)         Ar1         Ar2         Ar3         Ar4         Ar5         Ar6 
 -0.9502549   0.5227828   1.1742053   0.7032605   0.5329104   0.7776386   0.5640878 
        Ar7         Ar8         Ar9        Ar10 
  0.6192965   1.2160958   0.8605507   0.3305862 

The unconstrained coefficients would, naturally, all equal $0$.

  • $\begingroup$ Why do say $\delta$ is monotonically increasing in $\lambda$? I thought it would be decreasing. $\endgroup$
    – XYZT
    Sep 28, 2021 at 22:03
  • $\begingroup$ The way I have it set up is that large $\lambda$ will force $x$ closer to $0$, which will naturally increase the errors associated with $y - Ax$. This happens because the second block of the regression will cause least squares to try to make $\lambda I_p x$ as close to zero as possible, so large $\lambda \to $ small $x$. $\endgroup$
    – jbowman
    Sep 29, 2021 at 1:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.