Some questions about the implementation of the Waudby-Smith–Ramdas (WSR) bound

[christian-cahig]

If I understand correctly, bounds.WSR_mu_plus() implements the Waudby-Smith–Ramdas (WSR) bound.

rcps/core/bounds.py

Lines 126 to 137 in b400457

	def WSR_mu_plus(x, delta, maxiters): # this one is different.
	n = x.shape[0]
	muhat = (np.cumsum(x) + 0.5) / (1 + np.array(range(1,n+1)))
	sigma2hat = (np.cumsum((x - muhat)**2) + 0.25) / (1 + np.array(range(1,n+1)))
	sigma2hat[1:] = sigma2hat[:-1]
	sigma2hat[0] = 0.25
	nu = np.minimum(np.sqrt(2 * np.log( 1 / delta ) / n / sigma2hat), 1)
	def _Kn(mu):
	return np.max(np.cumsum(np.log(1 - nu * (x - mu)))) + np.log(delta)
	if _Kn(1) < 0:
	return 1
	return brentq(_Kn, 1e-10, 1-1e-10, maxiter=maxiters)

I have some (seemingly trivial) questions regarding the said function. Any form of guidance is appreciated.

1. In the subroutine _Kn(), how does one get the + np.log(delta) term from Proposition 5?
2. Why do we check whether _Kn(1) is negative, and, if it is, subsequently return a WSR bound of 1?
3. When invoking scipy.optimize.brentq(), why do we use a “smaller” search interval (i.e., between 1e-10 and 1-1e-10) instead of between 0 and 1?

[aangelopoulos]

1. We're inverting the capital process _Kn, so we need to find when it equals to log(delta). Looking at Theorem 3, we know that this happens when the process equals 1/delta. take log of both sides and set them equal.

2. If _Kn(1) is negative that means that we can never achieve an error of delta, which means that our upper-bound is 1. (This is basically just clipping).

3. This was just for numerical stability. I don't recall at this point exactly why this was needed, but I think some warnings were raised if you plugged in 0 or 1.

[christian-cahig]

@aangelopoulos thank you very much for the explanation. It's a big help.