Backward Martingale

Backword Martingale

Theorems

Def: A series of r.v of index $n\in\{-1,-2,\cdots\}$ is a Backward Martingale if

\[\mathbb{E}(X_n|\mathcal{F}_{n-1})=X_{n-1} \]

In short, it is martingale looking backward.

In some sense, a martingale is "divergent", and backward martingale is "convergent" if we look at its sigma algebra. The following theorems explain my idea.

Thm: $X_n$ convergence in a.s and $L_1$.

Pf: To prove a.s. Use Doob's upwards inequality.

Define

\[U_N=\text{The times that } X_n \text{ goes from below }a \text{ over }b \]

We have

\[\mathbb{E}(U_N)\leq \frac{1}{b-a}\mathbb{E}(X_0^{+}) \]

Note that these two sets

\[\{\limsup_n X_n \geq b \text{and} \liminf_n X_n \leq a\} \supset \{U_\infty =\infty \} \]

but we already bounded $U_\infty=\lim_N U_N$ by taking the limit in the Doob's upwards inequality. So this set could not happen. By the arbitary of $a,b$ we finish the proof.

The $L_1$ is symple, since it is UI.

Thm: $X_{-\infty}=\lim_{n \to -\infty}X_n,\ \mathcal{F}_{-\infty}=\cap\mathcal{F}_n$, then $X_{-\infty}=\mathbb{E}(X_0|\mathcal{F}_{\infty})$

It is obvious.

Thm: $\mathcal{F}_n \downarrow \mathcal{F}_{-\infty}$, $Y$ integrable, then $\mathbb{E}(Y|\mathcal{F}n)\to \mathbb{E}(Y|\mathcal{F}) a.s. and $L_1$

Pf First, it is a backward martingale, then convegence got. Use the second theorem, we know the form of limits.

Applications

Setings

Let

\[\Omega=\{(\omega_1,\omega_2,\cdots) \omega_i \in S \} \]

\[\mathcal{F}=\mathcal{S}\times \mathcal{S}\cdots \]

\[X_n(\omega)=\omega_n \]

In settings, we first give a probability space and make a product for infinately times. Then we constract a r.p. But in application, we first get r.p. then consider its probability space.

$\mathcal{E}_n$ is the sigma field generate by events invarient undet permutations in $S_n$ acting on its indexs. its limit is $\mathcal{E}$

Exmp: Simple random walk:

\[\xi_1,\xi_2,\cdots \]

S is canonical space and so is sigma algebra. It is easy to give $\mathcal{E}_n$ which is

\[\sigma(\cup_{t \in S_n}\{\xi_{t(1)}=i_1,\cdots,\xi_{t(n)}=i_n\}) \]

It hard to give $\mathcal{E}$,but we know the sets like this

\[\{\frac{\sum_{i=1}^n \xi_i}{n} \in A\} \]

is in it. And this gives ideas to the following applications.

Strong Law of Large Numbers (Durrett Example 4.7.4)

Let $\xi_1,\xi_2,\cdots$ be i.i.d. with $\mathbb{E}|\xi_i|<\infty$, $S_n=\xi_1+\cdots+\xi_n$, and

\[X_{-n}=\frac{S_n}{n},\qquad \mathcal{F}_{-n}=\sigma(S_n,S_{n+1},S_{n+2},\cdots)=\sigma(S_n,\xi_{n+1},\xi_{n+2},\cdots) \]

Pf (backward martingale check): To compute $\mathbb{E}(X_{-n}|\mathcal{F}_{-n-1})$, first note that if $j,k\leq n+1$, symmetry implies

\[\mathbb{E}(\xi_j|\mathcal{F}_{-n-1})=\mathbb{E}(\xi_k|\mathcal{F}_{-n-1}) \]

\[\mathbb{E}(\xi_{n+1}|\mathcal{F}_{-n-1})=\frac{1}{n+1}\sum_{k=1}^{n+1}\mathbb{E}(\xi_k|\mathcal{F}_{-n-1})=\frac{1}{n+1}\mathbb{E}(S_{n+1}|\mathcal{F}_{-n-1})=\frac{S_{n+1}}{n+1} \]

(The trick: $\mathcal{F}_{-n-1}$ only sees the first $n+1$ variables through their sum $S_{n+1}$, and conditioning on a symmetric function preserves the symmetry among $\xi_1,\cdots,\xi_{n+1}$.)

Since $X_{-n}=(S_{n+1}-\xi_{n+1})/n$, it follows that

\[\mathbb{E}(X_{-n}|\mathcal{F}_{-n-1})=\mathbb{E}(S_{n+1}/n|\mathcal{F}_{-n-1})-\mathbb{E}(\xi_{n+1}/n|\mathcal{F}_{-n-1})=\frac{S_{n+1}}{n}-\frac{S_{n+1}}{n(n+1)}=\frac{S_{n+1}}{n+1}=X_{-n-1} \]

so $X_{-n}$ is a backward martingale. Moreover

\[\mathbb{E}|X_{-n}|\leq \frac{1}{n}\sum_{i=1}^n\mathbb{E}|\xi_i|=\mathbb{E}|\xi_1|<\infty \]

By the first theorem above, $S_n/n\to X_{-\infty}$ a.s. and in $L_1$; by the second theorem,

\[\lim_{n\to\infty}\frac{S_n}{n}=\mathbb{E}(X_{-1}|\mathcal{F}_{-\infty})=\mathbb{E}(\xi_1|\mathcal{F}_{-\infty}) \]

Since $\mathcal{F}_{-n}\subset\mathcal{E}_n$ (each $S_n$ is invariant under permutations of the first $n$ indices), we have $\mathcal{F}_{-\infty}\subset\mathcal{E}$. The Hewitt–Savage 0-1 law (below) says $\mathcal{E}$ is trivial, hence

\[\lim_{n\to\infty}\frac{S_n}{n}=\mathbb{E}(\xi_1)\quad a.s.\text{ and in }L_1 \]

Ballot Theorem (Durrett Example 4.7.5)

Let $\{\xi_j,\ 1\leq j\leq n\}$ be i.i.d. non-negative integer valued r.v.'s, $S_k=\xi_1+\cdots+\xi_k$, and $G=\{S_j<j \text{ for } 1\leq j\leq n\}$. Then

\[\mathbb{P}(G|S_n)=\left(1-\frac{S_n}{n}\right)^+ \]

Remark (the name): in an election, candidate $B$ gets $\beta$ votes, $A$ gets $\alpha>\beta$ votes. Take $\xi_1,\cdots,\xi_n$ i.i.d. with $\mathbb{P}(\xi=0)=\mathbb{P}(\xi=2)=1/2$ ($0$ = vote for $A$, $2$ = vote for $B$), $n=\alpha+\beta$. Then $G=\{A\text{ leads }B\text{ throughout the counting}\}$ and

\[\mathbb{P}(G|B\text{ gets }\beta\text{ votes})=\left(1-\frac{2\beta}{\alpha+\beta}\right)^+=\frac{\alpha-\beta}{\alpha+\beta} \]

Pf: The result is trivial when $S_n\geq n$ (then $(1-S_n/n)^+=0$ and $G$ is impossible), so suppose $S_n<n$. The computations in the SLLN example show $X_{-j}=S_j/j$ is a backward martingale w.r.t. $\mathcal{F}_{-j}=\sigma(S_j,\cdots,S_n)$ for $1\leq j\leq n$, and note $\mathcal{F}_{-n}=\sigma(S_n,\cdots,S_n)=\sigma(S_n)$.

Let $T=\inf\{k\leq -1:\ k\geq -n,\ X_k\geq 1\}$ (i.e. the largest $j\leq n$ with $S_j\geq j$), and set $T=-1$ if the set is empty.

Claim: $X_T=1$ on $G^c$. Indeed, if $S_{j+1}<j+1$ then, since the $\xi_i$ are non-negative integer valued, $S_j\leq S_{j+1}\leq j$; by maximality of $j$, on $G^c$ we have $S_{j+1}<j+1$ for $j+1$ the time just after the crossing, hence $S_j\leq S_{j+1}\leq j$ together with $S_j\geq j$ forces $S_j=j$, i.e. $X_{-j}=1$.

On the other hand $G\subset\{T=-1\}$, and $S_1<1$ with $S_1$ a non-negative integer implies $S_1=0$, so $X_T=X_{-1}=S_1=0$ on $G$.

Hence on $\{S_n<n\}$, $X_T=\mathbf{1}_{G^c}$. By optional stopping for the (bounded) backward martingale, $\mathbb{E}(X_T|\mathcal{F}_{-n})=X_{-n}$. Since $\mathcal{F}_{-n}=\sigma(S_n)$,

\[\mathbb{P}(G^c|S_n)=\mathbb{E}(X_T|\mathcal{F}_{-n})=X_{-n}=\frac{S_n}{n}\quad\text{on }\{S_n<n\} \]

Subtracting from $1$ and recalling this was under $S_n<n$ gives $\mathbb{P}(G|S_n)=(1-S_n/n)^+$.

Hewitt–Savage 0-1 Law (Durrett Example 4.7.6)

Thm: If $X_1,X_2,\cdots$ are i.i.d. and $A\in\mathcal{E}$, then $\mathbb{P}(A)\in\{0,1\}$.

The key to the proof is:

Lemma: Suppose $X_1,X_2,\cdots$ are i.i.d. and

\[A_n(\varphi)=\frac{1}{(n)_k}\sum_{\mathbf{i}}\varphi(X_{i_1},\cdots,X_{i_k}) \]

where the sum is over all sequences of distinct integers $1\leq i_1,\cdots,i_k\leq n$, and $(n)_k=n(n-1)\cdots(n-k+1)$ is the number of such sequences. If $\varphi$ is bounded, then

\[A_n(\varphi)\to\mathbb{E}\varphi(X_1,\cdots,X_k)\ a.s. \]

Pf: $A_n(\varphi)\in\mathcal{E}_n$ (it is invariant under permutations of the first $n$ indices), so

\[A_n(\varphi)=\mathbb{E}(A_n(\varphi)|\mathcal{E}_n)=\frac{1}{(n)_k}\sum_{\mathbf{i}}\mathbb{E}(\varphi(X_{i_1},\cdots,X_{i_k})|\mathcal{E}_n)=\mathbb{E}(\varphi(X_1,\cdots,X_k)|\mathcal{E}_n) \]

since all the terms in the sum are the same (each injective tuple can be permuted to $(1,\cdots,k)$, and $\mathcal{E}_n$-measurable sets are invariant under such permutations). By the third theorem above with $\mathcal{F}_{-m}=\mathcal{E}_m$, $\mathcal{E}_n\downarrow\mathcal{E}$, so

\[\mathbb{E}(\varphi(X_1,\cdots,X_k)|\mathcal{E}_n)\to\mathbb{E}(\varphi(X_1,\cdots,X_k)|\mathcal{E}) \]

We want to show the limit is $\mathbb{E}\varphi(X_1,\cdots,X_k)$. The first step: there are $k(n-1)^{k-1}$ terms in $A_n(\varphi)$ involving $X_1$, and $\varphi$ is bounded, so (writing $\sum_{1\in\mathbf{i}}$ for the sum over tuples containing $1$)

\[\frac{1}{(n)_k}\sum_{1\in\mathbf{i}}\varphi(X_{i_1},\cdots,X_{i_k})\leq\frac{k(n-1)^{k-1}}{(n)_k}\sup\varphi\to 0 \]

Passing to the limit in $A_n(\varphi)=\mathbb{E}(\varphi(X_1,\cdots,X_k)|\mathcal{E}_n)$, this shows

\[\mathbb{E}(\varphi(X_1,\cdots,X_k)|\mathcal{E})\in\sigma(X_2,X_3,\cdots) \]

Repeating the argument for $2,3,\cdots,k$ shows $\mathbb{E}(\varphi(X_1,\cdots,X_k)|\mathcal{E})\in\sigma(X_{k+1},X_{k+2},\cdots)$. Intuitively, if the conditional expectation of a r.v. is independent of the r.v. then

\[(a)\quad \mathbb{E}(\varphi(X_1,\cdots,X_k)|\mathcal{E})=\mathbb{E}(\varphi(X_1,\cdots,X_k)) \]

To show this, we prove:

\[(b)\quad \text{If }\mathbb{E}X^2<\infty,\ \mathbb{E}(X|\mathcal{G})\in\mathcal{F},\ \text{and }X\text{ is independent of }\mathcal{F},\text{ then }\mathbb{E}(X|\mathcal{G})=\mathbb{E}X \]

Pf of (b): Let $Y=\mathbb{E}(X|\mathcal{G})$; since $Y\in\mathcal{F}$ and $X\perp\mathcal{F}$, $\mathbb{E}(XY)=\mathbb{E}X\,\mathbb{E}Y=(\mathbb{E}Y)^2$ (using $\mathbb{E}Y=\mathbb{E}X$). From the orthogonal projection property of conditional expectation, $\mathbb{E}((X-Y)Y)=0$, so $\mathbb{E}Y^2=\mathbb{E}(XY)=(\mathbb{E}Y)^2$ and $\mathrm{var}(Y)=0$. $\square$

$(a)$ holds for all bounded $\varphi$, so $\mathcal{E}$ is independent of $\mathcal{G}_k=\sigma(X_1,\cdots,X_k)$ for every $k$. Since $\cup_k\mathcal{G}_k$ is a $\pi$-system containing $\Omega$, the $\pi$-$\lambda$ theorem implies $\mathcal{E}$ is independent of $\sigma(\cup_k\mathcal{G}_k)\supset\mathcal{E}$. Then the usual 0-1 law punch line: if $A\in\mathcal{E}$, it is independent of itself, so

\[\mathbb{P}(A)=\mathbb{P}(A\cap A)=\mathbb{P}(A)\mathbb{P}(A),\qquad \text{i.e. } \mathbb{P}(A)\in\{0,1\} \]

de Finetti's Theorem (Durrett Example 4.7.8, Theorem 4.7.9)

Def: $X_1,X_2,\cdots$ is exchangeable if $(X_1,\cdots,X_n)$ and $(X_{\sigma(1)},\cdots,X_{\sigma(n)})$ have the same distribution for each $n$ and each permutation $\sigma$ of $\{1,\cdots,n\}$.

Thm (de Finetti): If $X_1,X_2,\cdots$ are exchangeable, then conditional on $\mathcal{E}$, $X_1,X_2,\cdots$ are independent and identically distributed.

Pf: Repeating the first calculation in the proof of the Lemma (this needs only exchangeability, not independence) shows that for any exchangeable sequence

\[A_n(\varphi)=\mathbb{E}(A_n(\varphi)|\mathcal{E}_n)=\mathbb{E}(\varphi(X_1,\cdots,X_k)|\mathcal{E}_n)\xrightarrow{n\to\infty}\mathbb{E}(\varphi(X_1,\cdots,X_k)|\mathcal{E})\tag{4.7.2} \]

This time, however, $\mathcal{E}$ may be nontrivial, so we cannot hope to show the limit is $\mathbb{E}\varphi$.

Let $f,g$ be bounded functions on $\mathbb{R}^{k-1}$ and $\mathbb{R}$ respectively, and let $I_{n,k}$ be the set of all sequences of distinct integers $1\leq i_1,\cdots,i_k\leq n$. Then

\[(n)_{k-1}A_n(f)\cdot nA_n(g)=\sum_{\mathbf{i}\in I_{n,k-1}}f(X_{i_1},\cdots,X_{i_{k-1}})\sum_m g(X_m)\\ =\sum_{\mathbf{i}\in I_{n,k}}f(X_{i_1},\cdots,X_{i_{k-1}})g(X_{i_k})+\sum_{\mathbf{i}\in I_{n,k-1}}\sum_{j=1}^{k-1}f(X_{i_1},\cdots,X_{i_{k-1}})g(X_{i_j}) \]

If we let $\varphi(x_1,\cdots,x_k)=f(x_1,\cdots,x_{k-1})g(x_k)$, note that $\frac{(n)_{k-1}n}{(n)_k}=\frac{n}{n-k+1}$ and $\frac{(n)_{k-1}}{(n)_k}=\frac{1}{n-k+1}$; then rearranging gives

\[A_n(\varphi)=\frac{n}{n-k+1}A_n(f)A_n(g)-\frac{1}{n-k+1}\sum_{j=1}^{k-1}A_n(\varphi_j) \]

where $\varphi_j(x_1,\cdots,x_{k-1})=f(x_1,\cdots,x_{k-1})g(x_j)$. Applying $(4.7.2)$ to $\varphi$, $f$, $g$ and all the $\varphi_j$, and letting $n\to\infty$ (so $\frac{n}{n-k+1}\to 1$ and the correction term vanishes):

\[\mathbb{E}(f(X_1,\cdots,X_{k-1})g(X_k)|\mathcal{E})=\mathbb{E}(f(X_1,\cdots,X_{k-1})|\mathcal{E})\,\mathbb{E}(g(X_k)|\mathcal{E}) \]

It follows by induction that

\[\mathbb{E}\left(\prod_{j=1}^k f_j(X_j)\Big|\mathcal{E}\right)=\prod_{j=1}^k\mathbb{E}(f_j(X_j)|\mathcal{E}) \]

which is exactly conditional independence given $\mathcal{E}$, with a conditional law that is the same for each $X_j$ (by exchangeability $\mathbb{E}(g(X_j)|\mathcal{E})$ does not depend on $j$). Hence conditional on $\mathcal{E}$, the $X_i$ are i.i.d. $\square$

Remark: when the $X_i$ take values in a nice space, there is a regular conditional distribution for $(X_1,X_2,\cdots)$ given $\mathcal{E}$, and the sequence can be represented as a mixture of i.i.d. sequences (Hewitt and Savage call the sequence presentable). For arbitrary measure spaces this can fail; see Dubins and Freedman (1979), Freedman (1980).

Exchangeable Sequences in $\{0,1\}$ (Durrett Theorem 4.7.10)

Thm: If $X_1,X_2,\cdots$ are exchangeable and take values in $\{0,1\}$, then there is a probability distribution $F$ on $[0,1]$ such that

\[\mathbb{P}(X_1=1,\cdots,X_k=1,X_{k+1}=0,\cdots,X_n=0)=\int_0^1\theta^k(1-\theta)^{n-k}\,dF(\theta) \]

Pf (via de Finetti): Let $\theta:=\mathbb{P}(X_1=1|\mathcal{E})$ and let $F$ be the distribution of $\theta$. By de Finetti's theorem, conditional on $\mathcal{E}$ the $X_i$ are i.i.d. Bernoulli with parameter $\theta$, so

\[\mathbb{P}(X_1=1,\cdots,X_k=1,X_{k+1}=0,\cdots,X_n=0)=\mathbb{E}\big[\theta^k(1-\theta)^{n-k}\big]=\int_0^1\theta^k(1-\theta)^{n-k}\,dF(\theta) \]

Why important: this is the simplest case of "exchangeable = mixture of i.i.d." — from the palatable assumption of symmetry one gets the powerful conclusion that the sequence is a mixture of i.i.d. sequences. Given $\theta$, $S_n|X_1,\cdots$ behaves like $\mathrm{Binomial}(n,\theta)$, so summing over all configurations with $S_n=k$:

\[\mathbb{P}(S_n=k)=\binom{n}{k}\int_0^1\theta^k(1-\theta)^{n-k}\,dF(\theta) \]

This is the foundation of Bayesian statistics: $F$ is the prior distribution of the unknown parameter $\theta$, and observing the data updates it to the posterior. E.g. with a uniform prior ($F$ = uniform on $[0,1]$), after observing $k$ ones in $n$ trials the predictive probability of a success at the next trial is Laplace's rule of succession:

\[\mathbb{P}(X_{n+1}=1|S_n=k)=\frac{\int_0^1\theta^{k+1}(1-\theta)^{n-k}d\theta}{\int_0^1\theta^k(1-\theta)^{n-k}d\theta}=\frac{k+1}{n+2} \]

(Laplace's famous sunrise problem: with a uniform prior, after the sun has risen on $n$ days, the probability it rises tomorrow is $\frac{n+1}{n+2}$.)

Remark (finite-$n$ version, Durrett Exercise 4.7.3): one can also prove directly from the definition of exchangeability that for $X_1,\cdots,X_n\in\{0,1\}$ exchangeable,

\[\mathbb{P}(X_1=1,\cdots,X_k=1|S_n=m)=\frac{\binom{n-k}{n-m}}{\binom{n}{m}} \]

(draw without replacement from an urn with $m$ ones and $n-m$ zeros). As $n\to\infty$ with $m/n\to\theta$, this converges to $\theta^k$, which recovers the mixture representation; this is the proof of Diaconis and Freedman (1980), with explicit bounds on the rate of convergence.

posted on 2026-09-24 20:13 ChillmanLee 阅读(9) 评论(0) 收藏举报