Supplement to “Causal Strategic Linear Regression”

Let’s walk through the steps of the algorithm, bounding the error that accumulates along the way.

In the first round we set $\omega=0$ in order to obtain an estimate for $E[{\omega^{*}}^{T}x]$.

Since $\omega^{*}$ is a unit vector, the variance of ${\omega^{*}}^{T}x$ is at most $\lambda_{max}$ plus a constant (from the $1-$subgaussian noise).

By Chebyshev’s inequality, this means that $O(\lambda_{max}\epsilon^{-2}d^{2})$ samples suffice for the empirical estimator of $E[{\omega^{*}}^{T}x]$ to have no more than $\frac{\epsilon}{4d}$ error with failure probability $\Omega(\frac{1}{2d})$. We call the output of this estimator $\hat{\mu}$ and let $\hat{\mu}_{d}$ be the r-dimensional vector with $\hat{\mu}$ in every coordinate.

Now we choose $\omega_{1}....\omega_{d}$ that form an orthonormal basis of the image of the diagonal matrix $V$. For each $\omega$ we observe the reward ${\omega^{*}}^{T}(x+G\omega)+\eta$, subtract out $\hat{\mu}$, and plug it into the empirical mean estimator. For each $\omega_{i}$, let $\hat{\nu}_{i}$ be the resulting coefficient. After $O(\epsilon^{-1}d\lambda_{max})$ samples, each coefficient has at most $\frac{\epsilon}{4d}$ error with failure probability at most $\frac{1}{2d}$. Since we have computed $d+1$ estimators, each one with failure probability at most $\frac{1}{2d}$, a union bound gives us a total failure probability that is sub-constant.

We can now bound the total squared $\ell_{2}$ error between said coefficients and $G^{T}\omega^{*}$ in the $\omega_{1}...\omega_{d}$ basis (noting that the choice of basis does not affect the magnitude of the error). We can break up the error into two components using the triangle inequality: the error due to $\hat{\mu}_{d}$ and the error in the subsequent rounds. Each coordinate of $\hat{\mu}_{d}$ has error of magnitude at most $\frac{\epsilon}{4d}$, so the total magnitude of the error in $\hat{\mu}_{d}$ is at most $\frac{\epsilon}{4}$. The same argument applies for the error in the coordinate estimates, leading to a total $\ell_{2}$ error of at most $\epsilon/2$.

Recall that $\hat{\omega}=\hat{\nu}/\|\hat{\nu}\|$. Let $\nu:=G^{T}\omega^{*}$. We can now bound the gap between the agent outcomes incentivized by $\hat{\omega}$ and by $\omega_{imp}=\nu/\nu$:

∎

where the last line follows because $Ma$ and $x$ are uncorrelated. ∎

The first step of the algorithm is for recovering an estimate of $\Sigma$ and $\mu$. Note that $n_{1}$ samples suffice to recover $\hat{\Sigma}$ and $\hat{\mu}$ such that:

• •

  $||\hat{\Sigma}-\Sigma||^{2}\leq\epsilon$

• •

  $||\mu-\hat{\mu}||^{2}\leq\frac{\lambda_{max}(G^{T}G)\epsilon}{d}$

The for loop recovers an estimate of $G$. Via Lemma 3, the samples suffice to ensure that the following two conditions hold:

• •

  $||\hat{G}-G||^{2}\leq\frac{\epsilon}{d||\mu||^{2}}$

• •

  $||\hat{G}-G||^{2}\leq\frac{\epsilon}{d^{2}||G||^{2}}$

Then the algorithm computes an estimate of the optimal $\omega$. Via Lemma 2, we have that the optimum guarantees the minimum eigenvalue of an approximate solution will be within a constant factor of the optimum.

This $\omega$ guarantees that $n_{3}$ samples suffice to ensure the recover of $\omega^{*}$ within squared $L^{2}$-distance of $O(\epsilon)$ in expectation.

Finally the expectations can be used with a Markov inequality to ensure the algorithm succeeds with (arbitrarily high) constant probability. ∎

In this section we derive a bound on the convergence of the least squares estimator when a fixed design matrix $X$ is used. Note this is exactly the case we encounter, since the choice of $\omega$ lets us affect the entries of the design matrix. This is a standard, textbook result and not a main contribution of the paper.

In order to state the result more formally we have to introduce some notation. The goal of the procedure is to recover $\hat{\omega}_{opt}$, when given tuples $(x_{i},\hat{\omega}_{opt}x_{i}+\eta)$ where $\eta$ is 1-subgaussian. We aim to characterize $||\omega-\hat{\omega}_{opt}||$ where $\omega$ is obtained from ordinary least squares. Let $X$ be the vector with the $x_{i}$’s in its columns. Let $\kappa_{min}$ be the minimum eigenvalue of $\frac{1}{n}X^{T}X$ (the second moment matrix).

Below all expectations are taken only over the random noise. We assume the second moment matrix is full rank.

This motivates our procedure for parameter recovery. We do so in a fashion that attempts to maximize $\kappa_{min}$. Note that it is the minimum eigenvalue that determines the convergence rate. This is due to the fact that little variation along a dimension makes it hard to disentangle the features’ effect on the outcome via $\hat{\omega}_{opt}$ from the constant-variance noise $\eta$. ∎

In this section we describe how to choose the value of $\omega$ that maximizes the value of $\kappa_{min}$ for the samples we obtain.

To do so, we examine the expectation of the second moment matrix and make several observations. Let $\Sigma$ denote the expected second moment matrix of $x$ (i.e. $\mathbb{E}[xx^{T}]$. We have:

1. 1.

   The minimum eigenvalue of the above expression is concave with respect to $\omega$. This follows due to the following: $x+G\omega$ is a linear operator, the minimum eigenvalue of a Gramian matrix $X^{T}X$ is concave with respect to $X$, and the expectation of a concave function is concave (boyd2004convex).

2. 2.

   Since the agent attempts to maximize their motion in the $\omega$ direction, we want to ensure that we move them toward toward the direction that maximizes the minimum eigenvalue of $\mathbb{E}[(x+G\omega)(x+G\omega)^{T}]$.

However, we do not operate with exact knowledge of $G$, etc. It turns out that even approximately solving this optimization problem with estimates for $G,\Sigma,\mu$ suffices for our purposes, as long as the $\omega$ we obtain from our optimization (using the estimates) results in a high value for the minimum eigenvalue of $\mathbb{E}[(x+G\omega)(x+G\omega)^{T}]$. Let $\hat{\omega}$ be the maximizing argument for the estimated optimization problem and let $\omega$ be the maximizing argument for the original optimization problem. Let $Q$ be the true maximized second moment matrix including gaming, and $\hat{Q}$ be the maximizing second moment matrix with gaming resulting from replacing the true $\Sigma,\mu,G$ with the estimates. In formal terms, we need to show the minimum eigenvalue of the following is large: $\mathbb{E}[(x+G\hat{\omega})(x+G\hat{\omega})^{T}]$. We note that when $y^{T}\hat{Q}y$ is within $\epsilon$ of $y^{T}Qy$ for all $y$ in the unit ball, the minimum eigenvalues may differ by at most $\epsilon$.

And now we bound the norm of $||\hat{\Sigma}-\Sigma||^{2}$ assuming the following:

1. 1.

   $||\hat{\Sigma}-\Sigma||^{2}\leq\epsilon$

2. 2.

   $||\mu-\hat{\mu}||^{2}\leq\frac{\lambda_{max}(G^{T}G)\epsilon}{d}$

3. 3.

   $||\hat{G}-G||^{2}\leq\frac{\epsilon}{d||\mu||^{2}}$

4. 4.

   $||\hat{G}-G||^{2}\leq\frac{\epsilon}{d^{2}||G||^{2}}$

We work out the bound below.

This means if we find an $\epsilon-$approximate solution to the system with the estimated values, we obtain a $\kappa_{min}$ within $6\epsilon$ of the optimal. ∎

Recall that when the decision-maker fixes $\omega$, it receives samples of the form $x+G\omega$. We note this can be used to recover the matrix $G$. In particular, we show how $d$ rounds, each with $O(\frac{d\mathrm{tr}(\Sigma)}{\epsilon})$ samples, suffices to recover the matrix to squared Frobenius norm $\epsilon$. Recall the procedure we propose simply chooses $\omega=e_{1},...e_{d}$, one-hot coordinate vectors in each round. We first bound the error in $\hat{G}$. coordinate-wise: $\mathbb{E}[||\hat{G}_{i,j}-G_{i,j}||^{2}]\leq\frac{\mathbb{E}[x_{i}^{2}]}{n}$. A union bound across coordinates shows that $O(\frac{d^{2}\mathrm{tr}(\Sigma)}{\epsilon})$ samples suffice to recover $G$ within squared Frobenius norm $\epsilon$. ∎