advanced36 min studyLesson 670 in Mathematics

Change of measure and density martingales

Construct a new probability law from an integrable density and distinguish finite-time consistency from a terminal absolutely continuous change.

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What you’ll understand

  • Construct a closed density martingale from a terminal Radon–Nikodym variable.
  • Apply conditional Bayes with the correct positivity and integrability scope.
  • Identify why finite-time density martingales may fail to define a terminal absolutely continuous law.

Changing a probability law means changing the weights assigned to outcomes, not changing the outcomes themselves. A nonnegative integrable density with mean one defines a new probability measure, and conditioning that density along a filtration produces a martingale. This is the basis of likelihood-ratio calculations and conditional Bayes formulas. The subtlety appears over infinitely many observations: a nonnegative mean-one martingale can lose mass in its almost-sure limit, so finite-time consistent weights need not come from one integrable terminal density.

A terminal density produces a closed martingale

Let Z be a nonnegative integrable random variable of expectation one under a base probability P. Defining Q of an event as the P-expectation of Z times its indicator gives a probability measure absolutely continuous with respect to P. If Z is strictly positive P-almost surely, the two measures are equivalent, meaning they have the same null events. Nonnegativity alone gives only one direction of absolute continuity; events supported where the density vanishes can have positive P-probability and zero Q-probability.

For an increasing filtration define the density process by conditional expectation of Z. The tower property makes it a nonnegative P-martingale. On each time-n event, integrating against this conditional density agrees with integrating against Z, so the process is exactly the Radon–Nikodym density of the restricted measure at that time. Because one integrable variable closes the process, it is uniformly integrable. Its limit is the conditional expectation of Z given the eventually observed sigma-algebra, equal to Z when that sigma-algebra contains all its information.

Q(A)=EP[Z1A],Z≥0,EPZ=1,Zn=EP[Z∣Fn],dQdP∣Fn=ZnQ(A)=\mathbb E_P[Z\mathbf1_A],\quad Z\ge0,\quad\mathbb E_PZ=1,\qquad Z_n=\mathbb E_P[Z\mid\mathcal F_n],\quad \left.\frac{dQ}{dP}\right|_{\mathcal F_n}=Z_n
Z gives absolute continuity. Strict positivity almost surely additionally gives equivalence. Terminal closure makes the density martingale uniformly integrable.↔ Scroll wide equations horizontally.

Conditional Bayes divides by the current density

For a Q-integrable variable X, the product ZX is P-integrable, and conditional Bayes says its P-conditional expectation divided by the current density is the Q-conditional expectation of X. The formula holds on the set where that density is positive. Its complement is Q-null, so arbitrary values can be assigned there for a Q-version; insisting on a ratio everywhere would create an unnecessary division by zero. Strictly positive terminal density ensures positive current densities P-almost surely and avoids that ambiguity.

If X is measurable at a later filtration time m, the numerator may use the time-m density in place of the terminal density, by conditioning first at m. The ratio still divides by the density at the earlier conditioning time n. This order is crucial: the density process is a P-martingale, but a P-martingale need not remain a Q-martingale after the law changes. Conditional means must be recalculated under the new weights. A finite coin model makes every term in the formula directly verifiable.

EQ[X∣Fn]=EP[ZX∣Fn]Znon {Zn>0},X∈L1(Q)\mathbb E_Q[X\mid\mathcal F_n]=\frac{\mathbb E_P[ZX\mid\mathcal F_n]}{Z_n}\quad\text{on }\{Z_n>0\},\qquad X\in L^1(Q)
The equality is Q-almost sure, with arbitrary definition on the Q-null zero-density set. For an Fm-measurable variable and n ≤ m, the numerator can equivalently use ZmX.↔ Scroll wide equations horizontally.

Finite-horizon likelihoods can become singular globally

Conversely, a nonnegative mean-one martingale supplies a consistent density on each finite-time sigma-algebra. Without uniform integrability its almost-sure limit can have mean less than one, preventing that limit from serving as a probability density with the original normalization. Additional measure-extension arguments may still construct a law on the full sequence space, but that law need not be absolutely continuous with respect to P on the eventual sigma-algebra. Finite-horizon consistency and terminal absolute continuity are different assertions.

For independent fair tosses under P and independent tosses with heads probability three quarters under Q, every finite likelihood ratio is strictly positive. Yet the strong law gives different limiting heads frequencies under the two infinite laws, making them singular on the full sequence space. Under P the likelihood ratio's logarithm has a negative linear limiting slope, so the ratio tends to zero almost surely despite expectation one at every finite time. Positivity at each finite horizon therefore cannot replace uniform integrability when claiming one global terminal density.

Tilt two fair observations toward heads

Under P, two coins are independent and fair. Reweight their outcomes so Q makes them independent with heads probability 3/43/4. Compute the two-step and one-step densities, and use conditional Bayes to find the Q-mean of the second sign.

  1. Each heads contributes likelihood factor 3/23/2 and each tails factor 1/21/2. In path order HH,HT,TH,TT, the terminal density values are 9/4,3/4,3/4,1/49/4,3/4,3/4,1/4. Multiplying by the P-path mass one quarter gives Q-masses 9,3,3,19,3,3,1 divided by sixteen, summing to one.

  2. Conditioning on the first toss under P gives Z1=3/2Z_1=3/2 after heads and Z1=1/2Z_1=1/2 after tails. Their average is one, and each equals the average of its two terminal-density children.

  3. For second sign ξ2\xi_2, the numerator after first heads is (9/4−3/4)/2=3/4(9/4-3/4)/2=3/4, and after first tails it is (3/4−1/4)/2=1/4(3/4-1/4)/2=1/4. Dividing by 3/23/2 or 1/21/2 respectively yields the same Q-conditional mean 1/21/2.

The result

The density is positive on every finite outcome, so the finite measures are equivalent. The second fair P-increment becomes positively biased under Q, exactly as conditional Bayes predicts.

Explore the model

Interactive lab

Interactive lab · probability

Compare base and tilted terminal laws

Pause and Reset, set both p=0.5,n=2p=0.5,n=2, and predict count masses 1/4,1/2,1/41/4,1/2,1/4. Reset and set both p=0.75,n=2p=0.75,n=2, predicting masses 1/16,6/16,9/161/16,6/16,9/16 for zero, one, and two heads. Compare them with the likelihood-weighted finite tree. The instrument changes the selected exact probability law; it does not calculate conditional Radon–Nikodym derivatives or simulate an infinite measure extension. Control edits retain time; Reset clears both controls and clock.

Compare base and tilted terminal lawsThese are exact binomial probabilities for independent trials with one fixed success probability. Playback tours the distribution; it does not resample outcomes. Current readings: Mean · np: 5; Variance · np(1−p): 2.5; Total probability: 1. All parameters are adjustable with the labeled controls below.03581000.070.150.220.3Successes kP(X = k)k = 0 · P = 9.77e-4
Exact probabilityHighlighted outcomeMean np
Mean · np
5
Variance · np(1−p)
2.5
Total probability
1
t = 0 s

These are exact binomial probabilities for independent trials with one fixed success probability. Playback tours the distribution; it does not resample outcomes. 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 forLikelihood weights connect two probability laws, while conditional expectations must use the law specified in their subscript.

Check your understanding

What extra property of a nonnegative mean-one density Z ensures Q and P have the same null events?

Try it yourself

For two fair coins under P, use terminal density Z=21{first toss is H}Z=2\mathbf1_{\{\text{first toss is H}\}}. Find the Q-law of both tosses, identify the density process after one toss, and explain the zero-denominator set in conditional Bayes.

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

Give me a hint

Multiply each P-path probability by its terminal density, then determine which histories Q never observes.

Compare with the solution

The Q-masses of HH,HT,TH,TT are 1/2,1/2,0,01/2,1/2,0,0. Thus the first toss is certainly heads and the second remains fair. After one toss the density is already Z, equal to two on heads and zero on tails. Q is absolutely continuous with respect to P but not equivalent to it. Conditional Bayes gives the second-sign mean zero on first heads; on first tails the formal ratio is zero divided by zero, but that history is Q-null and a Q-version can be defined arbitrarily there.

Take it with you

  • An integrable mean-one terminal density creates a closed UI density martingale.
  • Strict positivity gives equivalence, while nonnegativity gives absolute continuity.
  • Finite-time likelihood consistency does not guarantee a global terminal density without tail control.

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