How sure? How many? How stable?

Standard errors only mean something at a maximum of the likelihood, and the notebook's 15 iterations had not reached one. So the run is finished first: same data, same random start, until the log-likelihood moves by less than 10⁻¹⁰ (170 iterations, ℓ = −415.366). The notebook's own estimates stay in the first column, unchanged. Components are ordered by mean (DR-002): component 1 is the low group, the notebook's romance lovers.

Two kinds of uncertainty sit beside each estimate. The Hessian standard error measures how sharply the log-likelihood falls away from its peak: the observed information (minus the matrix of second derivatives of ℓ at the maximum), inverted. It gives a Wald interval, $\hat\theta \pm 1.96\,\mathrm{SE}$. The parametric bootstrap simulates 1,000 new data sets of 200 ratings from the fitted mixture (seed 42), refits EM to each and reads the 2.5% and 97.5% points of the 1,000 estimates.

A 95% interval should contain the true value in 95% of repeated samples. That is checkable: simulate data sets from the notebook's true mixture (60% at 7.5 with σ 1.2, 40% at 4.0 with σ 1.5, n = 200), fit each by EM from a k-means++ start, build the intervals and count. 500 data sets per scenario for Wald intervals (seed 1), once with the model exactly and once clipped to 1 to 10 like the notebook; 200 for the percentile bootstrap (B = 200 each, seed 7), paired with the Wald intervals on the same data sets.

Neither reaches 95%. Wald intervals covered between 84% and 92% of the time; percentile-bootstrap intervals between 90% and 92%. On the same data sets the bootstrap covered more often for π₁, μ₁ and σ₁ (paired 95% intervals exclude 0); for μ₂ and σ₂ the difference is within simulation noise. Part of the gain comes from width: the bootstrap intervals are wider for 4 of the 5 parameters (median widths in the table). At n = 200 with groups this close, the log-likelihood is not yet the parabola the Wald interval assumes. Clipping, perhaps surprisingly, changes little.

The notebook assumed two groups. Fitting K = 1 to 4 components (variance floor σ ≥ 0.1, DR-003) and scoring each with AIC and BIC gives an uncomfortable answer: BIC prefers 3 components and AIC 4, and neither picks two. BIC's extra component is small and narrow: 4.0% of the ratings at μ = 9.96 with σ = 0.10, held at the floor. That is the pile of 7 ratings the notebook's clipping put at exactly 10.0, not another kind of viewer.

Each K gets the best of 60 ordinary starts (k-means++, Forgy and random; seed 4) and 30 pile starts, which put a narrow component on the 7 tied ratings at 10.0. The pile starts matter: ordinary starts spread their means over the bulk of the ratings with wide spreads, so none of them isolates seven identical values. For K = 3 and 4 only pile starts reached the best fit; without them the table would report lower maxima for those K.

The choice also depends on the floor. BIC picks K = 3 at σ ≥ 0.05 and 0.1, but K = 4 at σ ≥ 0.25, where 3 values of K are within 1.1 BIC points of each other. A narrow component on tied values gains likelihood as its σ shrinks, so the floor sets how much the pile is worth (DR-003).

200 notebook-style random starts (μ ~ U(3, 8), σ ~ U(0.5, 2), π = 0.5; seed 2025) on the notebook's 200 ratings, each run until |Δℓ| < 10⁻¹² with its whole log-likelihood trace kept. The ascent property holds in every run. The answer does depend on the start: most runs reach the maximum the notebook was heading for, and a few find the narrow-component maximum with a higher likelihood. “Keep the best of many starts” maximises the likelihood; here that means more starts make the implausible answer more likely, not less.