The Mathematics of Saturation & Diminishing Returns

Adstock answered when spend acts. Saturation answers how much each additional dollar buys. The first $10k of search spend captures the high-intent shoppers; the next $10k reaches people who were going to convert anyway, or no one at all. Formally, the response is a nonlinear, increasing function of (adstocked) spend whose slope — the value of the next dollar — falls as spend rises. That falling slope is the whole point: it is what makes a finite optimal budget exist. This page derives both saturation families the framework ships, corrects a docstring myth about which one is "S-shaped," and lets you bend every curve live. It mirrors nbs/math/math_02_saturation.ipynb, where each claim ends in an assert.

The two families have different shapes — getting that distinction right is the spine of this page:

1 · The core curve: \( f(x) = 1 - e^{-\lambda x} \)

This is what the default BayesianMMM actually builds, per channel (model/base.py):

sat_lam     = pm.Exponential(f"sat_lam_{channel}", lam=0.5)   # rate prior, mean = 2
exponent    = pt.clip(-sat_lam * x_adstocked, -20, 0)         # numerical clamp only
x_saturated = 1 - pt.exp(exponent)                            # == 1 - exp(-sat_lam * x)

The curve acts on the adstocked spend, and the single parameter \(\lambda\) (sat_lam) gets an \(\mathrm{Exponential}(0.5)\) prior (mean 2). The pt.clip(..., -20, 0) only guards exp from underflow at huge spend; mathematically the function is

Properties, derived

Anchored at the origin. \(f(0) = 1 - e^{0} = 0\). No spend, no response.

Bounded above by 1, never reached. Since \(e^{-\lambda x} > 0\) for all finite \(x\), \(f(x) < 1\) strictly and \(f(x) \to 1\) as \(x \to \infty\): the range is \([0, 1)\). This is the ceiling — an intrinsic maximum effect the channel can deliver no matter how much you spend.

Strictly increasing, with falling marginal returns from the very first dollar. Differentiate twice:

The marginal response \(f'(x)\) is maximal at the origin — \(f'(0) = \lambda\) — and strictly decreasing thereafter: the first dollar is the most valuable and every dollar after is worth less. Always diminishing, from the start.

Half-saturation point. Solve \(f(x_{1/2}) = \tfrac12\):

Larger \(\lambda\) ⇒ smaller \(x_{1/2}\) ⇒ the channel saturates sooner. The slope at the half-saturation point is a clean number: \(f'(x_{1/2}) = \lambda e^{-\ln 2} = \lambda/2\).

Computed live in your browser — the same math as the notebook. \(f(x) = 1 - e^{-\lambda x}\) for four saturation rates. Each dotted dropline lands exactly on \(x_{1/2} = \ln 2 / \lambda\), where the curve crosses ½; the dashed line at 1 is the ceiling, approached but never reached. Larger λ saturates sooner.

Computed live in your browser — the same math as the notebook. The marginal response \(f'(x) = \lambda e^{-\lambda x}\): each curve starts at its maximum \(f'(0) = \lambda\) (dots) and only falls. The squares mark \(f'(x_{1/2}) = \lambda/2\) — the slope at half-saturation. Diminishing returns from dollar one, with no warm-up region.

2 · The correction: concave, not S-shaped

Because \(f''(x) = -\lambda^2 e^{-\lambda x} < 0\) everywhere on \(x \ge 0\), the core curve has no inflection point. It is the classic diminishing-returns silhouette, not a sigmoid. (Algebraically, \(1 - e^{-\lambda x}\) is the CDF of an \(\mathrm{Exponential}(\lambda)\) random variable; an exponential CDF is concave, unlike the S-shaped normal or logistic CDFs.) The notebook encodes this as np.diff(slope) < 0 on a dense grid — the machine-checked statement that the curve never stops bending downward.

Where does a genuine S come from? Take the Hill function \(f(x) = x^n / (x^n + \kappa^n)\) and differentiate twice. The first derivative is (derived in section 3) \( f'(x) = n \kappa^n x^{n-1} / (x^n + \kappa^n)^2 \), and its sign change works out to

For \(n \le 1\) there is no positive root — the curve is concave throughout, just like the exponential. For \(n > 1\) the root \(x^{\star}\) is a real inflection point: convex below it (slow start), concave above it (bend to the ceiling). That — and only that — is an S-curve. The two shapes are not cosmetically different: below the inflection the truth says "spend does almost nothing," while every concave curve is at its steepest there.

Computed live in your browser — the same math as the notebook. Both curves are calibrated to hit half-saturation at the same spend (0.45 of max): the core exponential with \(\lambda = \ln 2 / 0.45 \approx 1.54\) versus Hill with \(\kappa = 0.45,\ n = 3\) (dark dashed — the series' "truth" styling, because this is exactly the true curve in the stress-test world below). The triangle marks the Hill inflection \(x^{\star} = \kappa\,(2/4)^{1/3} \approx 0.357\). The green curve has no such point anywhere: steepest at zero, bending down forever.

3 · The Hill alternative: the real S-curve

The other family — selectable per channel via SaturationConfig.hill() and used throughout the mmm_extensions path — is the Hill function:

It has two interpretable knobs:

• \(\kappa\) — the half-saturation point ("EC50"). Plug in \(x = \kappa\): \( f_{\text{Hill}}(\kappa) = \dfrac{\kappa^n}{\kappa^n + \kappa^n} = \tfrac12 \). So \(\kappa\) is, exactly and for every \(n\), the spend at which the channel hits half its ceiling. Unlike the exponential's \(\ln 2/\lambda\), here the half-sat location is the parameter (the notebook asserts \(f(\kappa) = \tfrac12\) at \(10^{-12}\) tolerance for every \((\kappa, n)\) pair it sweeps).
• \(n\) — the shape (Hill coefficient). \(n = 1\) gives a hyperbolic, concave curve — the Michaelis–Menten special case \(f(x) = x/(x + \kappa)\), diminishing from the first dollar with \(f'(0) = 1/\kappa\). \(n > 1\) gives the genuine S: slow start, inflection at \(x^\star\), then the bend to the ceiling. \(n < 1\) saturates extremely fast (infinitely steep at the origin).

Like the core curve, \(f_{\text{Hill}}(0) = 0\) and \(f_{\text{Hill}} \to 1\) as \(x \to \infty\) — range \([0, 1)\), and \(f\) itself is monotone increasing for every \(n\), S-shapes included.

Marginal returns: where the two families part ways

Differentiate Hill with the quotient rule:

Behaviour at the origin splits on \(n\):

• \(n < 1\): \(f'(x) \to \infty\) as \(x \to 0^{+}\) — infinitely steep start, then collapse.
• \(n = 1\): \(f'(0) = 1/\kappa\), finite, then strictly decreasing — hyperbolic, like the exponential in spirit.
• \(n > 1\): \(f'(0) = 0\). The next dollar is worth almost nothing at low spend; the marginal curve rises to a peak at the inflection \(x^{\star} = \kappa\left(\frac{n-1}{n+1}\right)^{1/n}\), then falls. This is the "you must clear a threshold before the channel works" story.

A clean checkpoint the notebook asserts exactly: the slope at half-saturation is

against the exponential's \(f'(\ln 2/\lambda) = \lambda/2\) at its half-sat. The notebook also checks both analytic derivatives against finite differences of the actual curve functions, and that the Hill marginal for \(n > 1\) is genuinely non-monotone with its peak at the predicted \(x^\star\).

When do thresholds matter? Whenever a channel needs critical mass before it works — awareness media that must hit reach thresholds, channels with auction dynamics, or any case where low weekly spend is genuinely wasted. Under a threshold, small budgets can be worse than none, and the optimal move is concentrate-or-exit — a qualitatively different optimization problem from "every dollar worth less than the last." Picking the saturation family is a real modeling decision, not a cosmetic one. One caution from math 01's equifinality theme: \(\kappa\) is only weakly identified from observational data (it trades off against adstock decay and \(\beta\)), so SaturationConfig.compute_kappa_bounds_from_data(x, percentiles=(0.1, 0.9)) anchors \(\kappa\)'s prior to observed-spend percentiles, keeping the curve's elbow inside the region the data actually cover.

Curve explorer: response and marginal returns, live

Pick a family and drag its parameters. The left panel is the response \(f(x)\); the right panel is the marginal return \(f'(x)\) — the value of the next dollar. Watch what happens to the marginal curve the moment Hill's \(n\) crosses 1.

Computed live in your browser — the same math as the notebook. The dot marks the half-saturation point (\(\ln 2/\lambda\) for logistic, exactly \(\kappa\) for Hill at every \(n\)); the square on the marginal panel marks the slope there (\(\lambda/2\), resp. \(n/4\kappa\)). With logistic selected, the marginal curve starts at its peak \(f'(0) = \lambda\) and only falls — no λ produces a slow start. Switch to Hill and push \(n\) above 1: \(f'(0)\) pins to zero and the marginal curve humps at the inflection \(x^\star\) (triangle) before diminishing. At \(n = 1\) (Michaelis–Menten) the hump vanishes and \(f'(0) = 1/\kappa\).

4 · Why normalize the input before saturating

\(\lambda\) is a rate with units of 1/spend: the product \(\lambda x\) in \(e^{-\lambda x}\) must be dimensionless. If channel A spends in the thousands and channel B in the millions, the same numeric \(\lambda\) means completely different saturation behaviour — and a shared prior on \(\lambda\) would be incoherent across channels. The fix is to normalize each channel's (adstocked) spend to a common scale: divide by its own maximum, so \(x \in [0, 1]\). Then:

• \(\lambda\) is comparable across channels — it measures saturation relative to that channel's own observed range, so one prior fits all.
• The half-saturation point \(x_{1/2} = \ln 2/\lambda\) lands at a fraction of the channel's max spend — an interpretable quantity ("this channel half-saturates at 35% of its peak weekly spend").

This is why the prior \(\lambda \sim \mathrm{Exponential}(0.5)\) (mean 2) is sensible only on normalized input: at \(\lambda = 2\) on \(x \in [0,1]\), the elbow \(x_{1/2} = \ln 2/2 \approx 0.347\) sits comfortably inside the observed range, where the data can actually inform the curve (asserted in the notebook). On raw spend it would be nonsense.

Computed live in your browser — the same math as the notebook's normalization argument. Left: one shared λ applied to two channels' raw spend — A in the thousands, B near millions. A rate that bends B sensibly leaves A trapped in its linear toe (\(f < 0.005\) over A's entire range — no curvature the data could ever identify); tune λ for A instead and B saturates within the first pixel. One number cannot serve two scales. Right: after dividing each channel by its own max, the two curves at \(\lambda = 2\) coincide exactly (the notebook asserts identity), and the shared half-saturation point sits at 0.347 of max spend.

5 · Saturation × coefficient: the contribution ceiling

A channel's contribution to the outcome is the saturated response scaled by its coefficient \(\beta\) (model/base.py: channel_contrib = beta * x_saturated):

Two consequences fall straight out of \(f \in [0, 1)\):

1. \(\beta\) is a pure vertical scale. It stretches the whole curve; the shape (where the elbow is, how fast it bends) is owned entirely by \(\lambda\) (or \(\kappa, n\)). Timing/shape and size are cleanly separated — which is also why they can trade off against adstock (the identifiability theme of math 01: long-carryover + weak-saturation ≈ short-carryover + strong-saturation).
2. A hard ceiling at \(\beta\). Since \(f(x) < 1\) strictly and \(f \to 1\), we get \(\beta f(x) < \beta\) for all finite \(x\), approaching \(\beta\) as \(x \to \infty\). So \(\beta\) is the channel's maximum achievable contribution — the most outcome you could ever buy from it at any budget. That cap is what makes "throw money at the best channel" a bounded, not unbounded, strategy.

Computed live in your browser — the same math as the notebook. \(\beta \cdot f(x)\) at \(\lambda = 2\) for four coefficients: every curve is the same shape stretched vertically (the notebook asserts contribution/β is identical across β), and each approaches — never reaches — its own dotted ceiling at \(\beta\).

End to end: spend → adstock → saturation → ×β

The full single-channel media response, composed exactly as the model does, in order:

One subtlety carried over from math 01: the framework's geometric_adstock is the un-normalized recurrence whose impulse weights sum to \(1/(1-\alpha) > 1\), so a steady per-period spend \(s\) settles to the amplified level \(s/(1-\alpha)\) before saturation sees it (the notebook asserts the steady state: \(s/(1-\alpha) = 2.50\) at \(s = 1, \alpha = 0.6\)). Pushing that steady level through the curve gives the static response curve a planner reads to choose a budget — contribution as a function of steady weekly spend. Its half-response level solves \(s_{1/2}/(1-\alpha) = \ln 2/\lambda\), i.e. \(s_{1/2} = (1-\alpha)\ln 2/\lambda\) — 0.139 at the notebook's defaults (\(\alpha = 0.6, \lambda = 2\)), and the notebook asserts the composed curve is monotone increasing, concave, and bounded by \(\beta\).

Computed live in your browser — the same math as the notebook. The planner's diminishing-returns view: steady contribution \(\beta\,f\!\left(\frac{s}{1-\alpha}\right)\) against steady weekly spend \(s\). The dot marks the half-response spend \(s_{1/2} = (1-\alpha)\ln 2/\lambda\) (0.139 at the defaults); the dashed line is the \(\beta\) ceiling. Raising α makes the channel reach its ceiling at lower weekly spend — carryover amplifies the level saturation sees — while β only stretches the axis. Adstock, saturation and β each bend this curve in a way the others can partly imitate: that is the identifiability trade-off experiments are for (math 05).

6 · What the framework supports today

As of 2026-06-10, the core BayesianMMM honors a per-channel SaturationConfig in-graph:

Recap

The two saturation families, side by side:

Next in the series: math 03 builds the baseline these media curves sit on top of (Fourier seasonality and the trend families). To see these defaults fitted to data for the first time, take the beginner route through workshop 03 · Your First MMM. For the measured cost of choosing the wrong saturation form — with every convergence diagnostic green — see stress 01 · Carryover & Shape.