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Notebook 2 — Many channels

SWC ENC 2026 · ephys-one module

A real patch holds not one channel but tens to thousands. Each is the random switch of Notebook 1, flipping independently of the others. The current you record is their sum. This notebook is about what that sum looks like — and, above all, how much it fluctuates, because the fluctuations are where the information is.

In this notebook you will:

1. Watch the macroscopic current emerge as N grows from 1 to 1000.
2. Run the ensemble experiment: repeat a voltage step, stack the sweeps, and compute the mean and variance across sweeps at each time point (isochrones).
3. Understand the two formulas the whole method rests on, $\langle I\rangle = N i p$ and $\sigma^2 = N i^2 p(1-p)$, from coin flips.
4. See why bigger patches look smoother: relative noise falls as $1/\sqrt{N}$.

import numpy as np
import matplotlib.pyplot as plt
import picopatch as pp

scheme = pp.two_state()                                   # the voltage-dependent channel of NB1
protocol = pp.step_protocol(v_hold=-80, v_step=65)        # 2 ms at -80, then 12 ms at +65

1. From one switch to a thousand

Put N channels in the patch and apply the same voltage step. At each instant every channel is open with the same probability $p(t)$ — the curve we met at the end of Notebook 1 — but each decides for itself. The current is $i$ times the number of channels open right now. Watch what happens to one sweep as N grows:

i_unit = 1.0
fig, axes = plt.subplots(5, 1, figsize=(10, 9), sharex=True)
for ax, N in zip(axes, [1, 3, 10, 100, 1000]):
    exp = pp.make_experiment(N=N, i=i_unit, scheme=scheme, protocol=protocol, n_sweeps=1, seed=N)
    ax.plot(exp.t_ms, exp.sweeps[0], color="k", lw=0.8, drawstyle="steps-post")
    ax.set_ylabel("pA"); ax.set_title(f"N = {N} channel{'s' if N > 1 else ''}", loc="left", fontsize=10)
    ax.axvline(2, color="tab:purple", ls=":", lw=1)
axes[-1].set_xlabel("time (ms)  (step to +65 mV at 2 ms)")
plt.tight_layout(); plt.show()

With one channel you see the switch. With three, a staircase: the current takes values 0, 1, 2, 3 pA as channels open and close. With a hundred the steps blur into a rough rise; with a thousand the trace is a smooth, textbook macroscopic current — but look closely at the plateau and it still wiggles. Those wiggles are a thousand channels flickering, and they shrink relative to the mean as N grows. Hold that thought for section 4.

2. The ensemble experiment: sweeps and isochrones

One sweep is one random draw. To get at the statistics we repeat the identical step many times — here 300 — and stack the recordings into a matrix of shape (n_sweeps, n_samples). Here is the stack, shown two ways: overlaid, and as an image with one row per sweep.

exp = pp.make_experiment(N=1000, i=1.0, scheme=scheme, protocol=protocol, n_sweeps=300, seed=0)
print("sweeps:", exp.sweeps.shape, "(n_sweeps, n_samples)")

pp.plotting.plot_sweeps(exp, n_show=30, isochrones_ms=[3.0, 12.0])
plt.show()
pp.plotting.plot_sweep_image(exp)
plt.show()

sweeps: (300, 700) (n_sweeps, n_samples)

An isochrone ("same time") is a vertical slice through the stack: the 300 current values recorded at one particular time after the step, one from each sweep. The two dashed lines above mark two of them. Along an isochrone the sweeps disagree with each other a little — let's look at the spread of values on each:

fig, axes = plt.subplots(1, 2, figsize=(11, 3.6))
for ax, tm in zip(axes, [3.0, 12.0]):
    k = int(np.argmin(np.abs(exp.t_ms - tm)))
    vals = exp.sweeps[:, k]
    ax.hist(vals, bins=25, color="tab:blue", alpha=0.7)
    ax.axvline(vals.mean(), color="k", lw=2, label=f"mean = {vals.mean():.0f} pA")
    ax.axvspan(vals.mean() - vals.std(), vals.mean() + vals.std(), color="tab:red", alpha=0.15,
               label=f"± 1 SD = {vals.std():.1f} pA")
    ax.set_title(f"isochrone at t = {tm:.0f} ms: the 300 values across sweeps")
    ax.set_xlabel("current (pA)"); ax.set_ylabel("sweeps"); ax.legend(fontsize=9)
plt.tight_layout(); plt.show()

Two numbers summarise each isochrone:

• the mean across sweeps, $\langle I \rangle(t)$ — the smooth macroscopic current;
• the variance across sweeps, $\sigma^2(t)$ — the average squared distance of the sweeps from that mean. Its square root is the standard deviation (SD), the typical size of the wiggle, in pA; the variance itself is in pA².

Notice in the histograms that the spread is larger at 3 ms (where about half the channels are open) than at 12 ms (where nearly all are) even though the mean is smaller. That is not an accident; it's the whole story, and section 3 explains it.

Exercise 1 (~4 min). Complete isochrone_stats: return the mean and the variance across sweeps at every time point (i.e. along axis 0). Use ddof=1 for the variance (the usual sample variance). Then plot both against time.

Check / unstuck. Mean should rise from 0 to ~960 pA; variance should peak around 250 pA² during the rise and settle near 40 pA² on the plateau. Stuck? pp.isochrone_stats(sweeps).

def isochrone_stats(sweeps):
    mean = sweeps.mean(axis=0)
    var = sweeps.var(axis=0, ddof=1)
    return mean, var

mean, var = isochrone_stats(exp.sweeps)
pp.plotting.plot_mean_variance(exp.t_ms, mean, var)
plt.show()
print(f"peak variance {var.max():.0f} pA²  at t = {exp.t_ms[np.argmax(var)]:.2f} ms;  plateau variance ~{var[-100:].mean():.0f} pA²")

peak variance 241 pA²  at t = 3.26 ms;  plateau variance ~38 pA²

The variance has a shape of its own: zero before the step, a bump during the rise, then down to a lower plateau. The bump is where the channels are most "undecided".

3. Why the variance looks like that: coin flips

Forget channels for a moment. Toss N = 10 coins, each landing heads with probability $p$, and count the heads. Do it many times. The count is random; what are its mean and variance?

• Mean: each coin contributes $p$ heads on average, so the mean count is $N p$.
• Variance: a single coin is a 0/1 variable with variance $p(1-p)$ — zero if the coin always lands the same way ($p = 0$ or $1$), largest when $p = 0.5$. Independent coins add their variances, so the variance of the count is $N p (1-p)$.

That count is called a binomial random variable. Let's check both claims by simulation, sweeping $p$ from 0 to 1:

rng = np.random.default_rng(1)
N_coins, n_rep = 10, 5000
ps = np.linspace(0.02, 0.98, 25)
mean_heads = []; var_heads = []
for p in ps:
    heads = (rng.random((n_rep, N_coins)) < p).sum(axis=1)       # number of heads in each of n_rep tosses
    mean_heads.append(heads.mean()); var_heads.append(heads.var(ddof=1))

fig, axes = plt.subplots(1, 3, figsize=(15, 3.8))
# the distribution of the count at p = 0.5
heads = (rng.random((n_rep, N_coins)) < 0.5).sum(axis=1)
axes[0].hist(heads, bins=np.arange(-0.5, N_coins + 1.5), density=True, color="tab:blue", alpha=0.7, label="simulated")
from scipy.stats import binom
axes[0].plot(np.arange(N_coins + 1), binom.pmf(np.arange(N_coins + 1), N_coins, 0.5), "ko", label="binomial")
axes[0].set_title("10 coins, p = 0.5: number of heads"); axes[0].set_xlabel("heads"); axes[0].legend()
axes[1].plot(ps, mean_heads, "o", color="tab:blue", label="simulated"); axes[1].plot(ps, N_coins * ps, "k-", label="N·p")
axes[1].set_title("mean number of heads"); axes[1].set_xlabel("p"); axes[1].legend()
axes[2].plot(ps, var_heads, "o", color="tab:blue", label="simulated"); axes[2].plot(ps, N_coins * ps * (1 - ps), "k-", label="N·p·(1−p)")
axes[2].set_title("variance of the number of heads"); axes[2].set_xlabel("p"); axes[2].legend()
plt.tight_layout(); plt.show()

The variance against $p$ is an upside-down parabola: nothing to fluctuate when the coins are all heads or all tails, maximal uncertainty at $p = 0.5$. Now replace "coin" with "channel", "heads" with "open", and multiply the count by the unitary current $i$ (which multiplies the mean by $i$ and the variance by $i^2$):

$$\boxed{\;\langle I \rangle = N\, i\, p \qquad\qquad \sigma^2 = N\, i^2\, p\,(1-p)\;}$$

These two lines are the entire theory. During the rise after the step, $p(t)$ climbs from 0 through 0.5 to 0.96, so the variance climbs to its peak and comes back down — exactly the bump you measured. On the plateau $p = 0.96$, so the variance is small but not zero: $1000 \times 1 \times 0.96 \times 0.04 \approx 38$ pA².

▸ Go deeper: the binomial mean and variance (optional)

Let $x_k \in \{0, 1\}$ indicate whether channel $k$ is open. Then $E[x_k] = p$ and, since $x_k^2 = x_k$, $\mathrm{Var}[x_k] = E[x_k^2] - E[x_k]^2 = p - p^2 = p(1-p)$. The current is $I = i \sum_k x_k$. Expectation is linear, so $E[I] = N i p$. For independent channels variances add, so $\mathrm{Var}[I] = i^2 \sum_k \mathrm{Var}[x_k] = N i^2 p(1-p)$.

Independence is the one real assumption. It fails if channels gate cooperatively, or if something common to all of them fluctuates (e.g. the clamp voltage) — and background instrument noise adds its own variance on top (Notebook 4).

Exercise 2 (~5 min). Use the ground truth to predict the mean and variance you measured. exp.truth holds N, i and p_open — the exact open probability $p(t)$ at every sample. Compute the predicted $N i p(t)$ and $N i^2 p(t)(1-p(t))$ and overlay them on your measured curves.

Check / unstuck. The predictions should run straight through the measured curves (the measured variance is noisier — it's estimated from 300 sweeps). Stuck? pp.plotting.plot_mean_variance(exp.t_ms, mean, var, truth=exp.truth) draws the same overlay.

N, i, p = exp.truth.N, exp.truth.i, exp.truth.p_open
pred_mean = N * i * p
pred_var = N * i**2 * p * (1 - p)

fig, axes = plt.subplots(2, 1, figsize=(10, 5), sharex=True)
axes[0].plot(exp.t_ms, mean, "k", label="measured mean"); axes[0].plot(exp.t_ms, pred_mean, "r--", label="N·i·p(t)")
axes[1].plot(exp.t_ms, var, color="tab:blue", lw=0.8, label="measured variance"); axes[1].plot(exp.t_ms, pred_var, "r--", label="N·i²·p(1−p)")
axes[0].set_ylabel("mean (pA)"); axes[1].set_ylabel("variance (pA²)"); axes[1].set_xlabel("time (ms)")
axes[0].legend(); axes[1].legend(); plt.tight_layout(); plt.show()

4. Why big patches look smooth: relative noise

Divide the standard deviation by the mean to get the relative noise, the wiggle as a fraction of the current:

$$\frac{\sigma}{\langle I\rangle} = \frac{\sqrt{N i^2 p (1-p)}}{N i p} = \sqrt{\frac{1-p}{N p}} \;\propto\; \frac{1}{\sqrt N}.$$

Ten times more channels → the current looks about three times smoother. That is why the N = 1000 trace in section 1 looked clean, and it is the first warning sign for the method: with very many channels the fluctuations become tiny compared with the current, and anything that limits how finely you can measure (Notebook 4) starts to bite.

Exercise 3 (~6 min). For N in [10, 30, 100, 300, 1000, 3000], simulate 100 sweeps, compute the relative noise $\sigma/\langle I\rangle$ on the plateau (average the last 200 samples of mean and SD), and plot it against N on log–log axes together with the prediction $\sqrt{(1-p)/(N p)}$ with $p = 0.96$.

Check / unstuck. The points should fall on a straight line of slope −1/2. Stuck? pp.isochrone_stats for the mean/variance; the rest is arithmetic.

Ns = np.array([10, 30, 100, 300, 1000, 3000])
rel = []
for N in Ns:
    e = pp.make_experiment(N=int(N), i=1.0, scheme=scheme, protocol=protocol, n_sweeps=100, seed=int(N))
    m, v = pp.isochrone_stats(e.sweeps)
    rel.append(np.sqrt(v[-200:]).mean() / m[-200:].mean())
rel = np.array(rel)

p_plateau = 0.96
plt.figure(figsize=(5.5, 4))
plt.loglog(Ns, rel, "o", ms=8, color="tab:blue", label="simulated")
plt.loglog(Ns, np.sqrt((1 - p_plateau) / (Ns * p_plateau)), "k-", label=r"$\sqrt{(1-p)/(Np)}$")
plt.xlabel("number of channels N"); plt.ylabel("relative noise  SD / mean"); plt.legend()
plt.title("relative noise falls as 1/√N"); plt.show()

Where we are. We can record an ensemble of sweeps and reduce it to two curves, mean and variance against time, and we know what both should be: $\langle I\rangle = N i p$ and $\sigma^2 = N i^2 p(1-p)$. The catch is that $p(t)$ is unknown and changing. Notebook 3 shows the trick that gets rid of it — and turns these two curves into N and i.

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