CONTENTS

Solution: Variance Is a Submartingale — Doob's Decomposition

Exercise: Variance Is a Submartingale — Doob's Decomposition

Part 1

By conditional Jensen on the convex function ϕ(x)=x2\phi(x) = x^2:

E[Mn+12∣Fn]≥(E[Mn+1∣Fn])2=Mn2,\mathbb{E}[M_{n+1}^2 \mid \mathcal{F}_n] \ge (\mathbb{E}[M_{n+1} \mid \mathcal{F}_n])^2 = M_n^2,
where the equality uses the martingale property E[Mn+1∣Fn]=Mn\mathbb{E}[M_{n+1} \mid \mathcal{F}_n] = M_n and the convexity of x2x^2 drives the inequality. Both are needed: convexity alone would give a bound but not identify the pointwise lower bound as Mn2M_n^2; the martingale property pins down that lower bound.

Part 2

An=∑k=1nE[Mk2−Mk−12∣Fk−1]A_n = \sum_{k=1}^n \mathbb{E}[M_k^2 - M_{k-1}^2 \mid \mathcal{F}_{k-1}]. Expand:

Mk2−Mk−12=(Mk−Mk−1)2+2Mk−1(Mk−Mk−1).M_k^2 - M_{k-1}^2 = (M_k - M_{k-1})^2 + 2M_{k-1}(M_k - M_{k-1}).

Take conditional expectation: the second term vanishes because Mk−1M_{k-1} is Fk−1\mathcal{F}_{k-1}-measurable and E[Mk−Mk−1∣Fk−1]=0\mathbb{E}[M_k - M_{k-1} \mid \mathcal{F}_{k-1}] = 0 (martingale property). So:

E[Mk2−Mk−12∣Fk−1]=E[(Mk−Mk−1)2∣Fk−1]=Var⁡(Mk−Mk−1∣Fk−1).\mathbb{E}[M_k^2 - M_{k-1}^2 \mid \mathcal{F}_{k-1}] = \mathbb{E}[(M_k - M_{k-1})^2 \mid \mathcal{F}_{k-1}] = \operatorname{Var}(M_k - M_{k-1} \mid \mathcal{F}_{k-1}).

(The last equality uses that the conditional mean of Mk−Mk−1M_k - M_{k-1} given Fk−1\mathcal{F}_{k-1} is zero, by the martingale property.)

Therefore:

An=⟨M⟩n:=∑k=1nVar⁡(Mk−Mk−1∣Fk−1).A_n = \langle M\rangle_n := \sum_{k=1}^n \operatorname{Var}(M_k - M_{k-1} \mid \mathcal{F}_{k-1}).

Part 3

Shown above: ⟨M⟩n=∑k=1nE[(Mk−Mk−1)2∣Fk−1]\langle M\rangle_n = \sum_{k=1}^n \mathbb{E}[(M_k - M_{k-1})^2 \mid \mathcal{F}_{k-1}].

Part 4

For the symmetric random walk, the increments are i.i.d. ±1\pm 1, independent of Fk−1\mathcal{F}_{k-1}, so:

Var⁡(Sk−Sk−1∣Fk−1)=Var⁡(Xk)=1.\operatorname{Var}(S_k - S_{k-1} \mid \mathcal{F}_{k-1}) = \operatorname{Var}(X_k) = 1.

Summing: ⟨S⟩n=n\langle S\rangle_n = n.

The martingale part is Yn=Sn2−nY_n = S_n^2 - n. A direct check:

E[Yn+1∣Fn]=E[Sn+12∣Fn]−(n+1)=(Sn2+1)−(n+1)=Yn.\mathbb{E}[Y_{n+1} \mid \mathcal{F}_n] = \mathbb{E}[S_{n+1}^2 \mid \mathcal{F}_n] - (n+1) = (S_n^2 + 1) - (n+1) = Y_n.

So Sn2−nS_n^2 - n is a martingale — the discrete analogue of the continuous-time fact that Wt2−tW_t^2 - t is a martingale, and the content of "Brownian motion has quadratic variation tt."

Takeaways

  • Any square-integrable martingale has two structural pieces: the martingale itself and its predictable quadratic variation ⟨M⟩n\langle M\rangle_n, which measures the cumulative conditional variance.
  • Doob's decomposition is the discrete precursor of the Doob-Meyer decomposition for continuous-time submartingales, itself the precursor of the stochastic-integral decomposition for semimartingales.
  • Mn2−⟨M⟩nM_n^2 - \langle M\rangle_n is always a martingale. This is the workhorse identity for computing variances of martingale integrals and for bounding martingale deviations via L2L^2 methods.
  • For a random walk ⟨S⟩n=n\langle S\rangle_n = n. The accumulation of variance at unit rate is the defining property — and it is what (dW)2=dt(dW)^2 = dt encodes in continuous time.
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Solution — Variance Is a Submartingale: Doob's Decomposition | q4quant.studio