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advanced29 min studyLesson 963 in Physics

Maximum likelihood estimation in physical models

Estimate a lifetime with a normalized sampling density and show how truncation changes the likelihood.

Before you begin
On this page

The question for today

What you’ll understand

  • Distinguish the parameter from its data-dependent estimator.
  • Derive a lifetime estimate and separate exact moments from asymptotic errors.
  • Include selection normalization and boundary checks in a likelihood.

A physical model predicts a distribution of observations, not merely a curve through measured points. Maximum likelihood chooses the parameter making the observed data density largest within that model. It does not assign a posterior probability to the parameter. We will estimate an exponential lifetime from ideal independent decay times, then impose a lower acceptance threshold. The example is deliberately simple enough to differentiate by hand, while exposing selection and small-sample issues that also occur in complicated detector fits.

Normalize before maximizing

Let t be a decay time and tau a positive mean lifetime. Assume perfect time resolution, no background, constant acceptance over the full nonnegative time axis, and a fixed number n of independently observed decays. The density contains both an exponential shape and an inverse-lifetime normalization. The log likelihood is a sum because of independence. Dropping the normalization would make the objective increase toward arbitrarily long lifetimes and would destroy the inference, even though the plotted exponential shape might appear familiar.

The times have units, so a density carries reciprocal-time units; likelihood ratios at different parameter values remain dimensionless. Changing the time unit consistently changes a density normalization by a parameter-independent factor and leaves the estimated physical lifetime unchanged. By contrast, omitting a parameter-dependent factor changes the maximization. If the observed number of decays also carries parameter information, its count probability belongs in an extended likelihood rather than being conditioned away without explanation.

f(t∣τ)=τ−1e−t/τ 1t≥0,ℓ(τ)=−nln⁡τ−∑itiτ+const,τ^=1n∑itif(t\mid\tau)=\tau^{-1}e^{-t/\tau}\,\mathbf1_{t\ge0},\qquad \ell(\tau)=-n\ln\tau-\frac{\sum_i t_i}{\tau}+\mathrm{const},\qquad \widehat\tau=\frac1n\sum_i t_i
Tau and the observed times have time units. The logarithm is understood relative to a fixed time unit; its derivative and likelihood ratios are unit-consistent.↔ Scroll wide equations horizontally.

Estimator properties are repeated-sample statements

Differentiating gives the sample mean as the interior maximum when the time sum is positive. For ideal exponential observations, this estimator is exactly unbiased and has variance tau squared divided by n. Its sampling distribution is a scaled gamma distribution, which is skewed for small n. A standard error obtained by substituting the estimate into the variance is convenient, but a symmetric normal interval is an additional approximation. Exact pivot-based intervals can instead use the known gamma sampling law.

The reciprocal rate parameter lambda equals inverse tau, and maximum-likelihood invariance gives its estimate as inverse sample mean. Invariance does not preserve unbiasedness: for more than one event, the expected estimated rate is n divided by n minus one times the true rate. For one event that expectation diverges. These facts are compatible with the lifetime estimator being unbiased. Bias, variance, consistency and loss functions refer to specified parameters and cannot be transferred indiscriminately through nonlinear transformations.

E⁡[τ^]=τ,Var⁡(τ^)=τ2n,λ^=n∑iti,E⁡[λ^]=nn−1λ(n>1)\operatorname E[\widehat\tau]=\tau,\qquad \operatorname{Var}(\widehat\tau)=\frac{\tau^2}{n},\qquad \widehat\lambda=\frac{n}{\sum_i t_i},\qquad \operatorname E[\widehat\lambda]=\frac{n}{n-1}\lambda\quad(n>1)
The first two moments are exact for fixed-n independent uncensored exponential times. The plug-in standard error and Gaussian interval approximation have distinct status.↔ Scroll wide equations horizontally.

Selection changes the probability of being observed

Suppose only decays later than a known threshold are retained. The conditional density divides the original density by the probability of passing that selection. For an exponential this produces the same lifetime dependence in the excess time above threshold. Fitting the retained raw times as if selection were absent overestimates the lifetime. If acceptance varies continuously with time, its shape and parameter-dependent normalization must both be included; the simple shift argument is no longer sufficient.

Left truncation means earlier events are absent from the sample. Right censoring means an event is known to have survived until a recorded time without its decay being observed. The latter contributes a survival probability, not an ordinary observed-decay density. Resolution smearing, backgrounds and uncertain thresholds also require generative terms. Numerical optimization should check allowed parameter ranges, alternative maxima and identifiability, and model checking should test time-dependent residual structure. Convergence of an optimizer cannot validate the assumed observation process.

Work it through

Three ideal decay times

For fixed n = 3, independent uncensored decay times are 1, 2 and 3 ns. Estimate tau, calculate its plug-in standard error, and estimate the inverse lifetime.

  1. The sum is six ns, so maximizing the normalized likelihood gives τ^=6/3=2 ns\widehat\tau=6/3=2\,\mathrm{ns}. The positive sum yields an interior maximum.

  2. The exact estimator variance is tau squared over three. Substitution gives estimated variance 4/3 ns24/3\,\mathrm{ns^2} and standard error 2/3≈1.15 ns2/\sqrt3\approx1.15\,\mathrm{ns}.

  3. Inverting the estimate gives λ^=0.5 ns−1\widehat\lambda=0.5\,\mathrm{ns^{-1}}. This is the MLE of the rate but is not an unbiased rate estimator; with three observations its expectation is three halves the true rate.

The result

The lifetime MLE is 2 ns, its plug-in standard error is about 1.15 ns, and the rate MLE is 0.5 ns⁻¹.

The living notebook

Explore the model

Interactive lab

Interactive lab · optimization

Optimization behavior depends on the update rule

Reset and set BOTH Step size η = 0.5 and Initial position x₀ = 2. Predict x₁ = 1 and use Step once. For a fresh comparison, Reset and set η = 1 and x₀ = 2; predict x₁ = 0 and Step once.

Optimization behavior depends on the update ruleGradient descent on f(x) = x²/2: xₙ = (1 − η)ⁿx₀. Playback advances two discrete iterations per second. Values outside |x| ≤ 4 are explicitly reported. Current readings: Position x0: 2; Objective x²/2: 2; Iteration behavior: Stable step. All parameters are adjustable with the labeled controls below.−4−202402.174.356.528.7Parameter xf(x) = x²/2n = 0
ObjectiveDiscrete iterate
Position x0
2
Objective x²/2
2
Iteration behavior
Stable step
n = 0 · t = 0 s

Gradient descent on f(x) = x²/2: xₙ = (1 − η)ⁿx₀. Playback advances two discrete iterations per second. Values outside |x| ≤ 4 are explicitly reported. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forThe clock advances one gradient-descent iteration per half-second Step for the fixed objective x²/2. It does not evaluate a lifetime likelihood, model selection, parameter boundaries or optimizer uncertainty.

Pause & reason

Check your understanding

Which factor may be dropped when maximizing a likelihood over tau?

Your turn

Try it yourself

Only times satisfying t≥1 nst\ge1\,\mathrm{ns} are recorded, with no other acceptance or resolution effect. The two recorded times are 2 and 4 ns. Derive the conditional density and lifetime MLE, and compare with ignoring the threshold.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Divide by the exponential survival probability at the known threshold before differentiating.

Compare with the solution

The passing probability is e−tmin⁡/τe^{-t_{\min}/\tau}, so f(t∣t≥tmin⁡,τ)=τ−1e−(t−tmin⁡)/τf(t\mid t\ge t_{\min},\tau)=\tau^{-1}e^{-(t-t_{\min})/\tau} on the accepted domain. The MLE averages excess times: [(2−1)+(4−1)]/2=2 ns[(2-1)+(4-1)]/2=2\,\mathrm{ns}. Ignoring truncation instead averages the raw times and gives three ns. The corrected result relies on a known fixed threshold and ideal exponential observations; it does not account for additional censoring, background or uncertain timing calibration.

Keep the idea

Take it with you

  • Likelihood normalization carries physical information.
  • An unbiased lifetime estimator need not give an unbiased reciprocal estimator.
  • Truncation, censoring and detector response must be modeled separately.

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