18 — Doob’s $L^p$ Inequality and Branching Processes
Doob’s $L^p$ Inequality (for $p>1$)
Let ${X_n, \mathcal{F}n}{n\ge 0}$ be a submartingale with $X_n \ge 0$ a.s.
Then for any $p>1$, \(E\!\left( \max_{0\le k\le n} X_k^p \right) \le E(X_n^p)\left(\frac{p}{p-1}\right)^p.\)
For $p=2$, \(\left(\frac{p}{p-1}\right)^p = \left(\frac{2}{1}\right)^2 = 4.\)
Durrett Probablity 4e - Theorem 5.4.5. $L^p$ convergence Theorem
If X_n is a martingale with $\sup\vert X_n\vert^p <\infty$ where $p>1$, then $x_n\to X$ a.s and in $L^p$.
Durrett Probability 4e - Theorem 5.4.2. Doob’s Inequality.
Let X_n be a submartingale, \(\bar{X}_n=\max_{0\le m\le n}X^+_m\) $\lambda>0$, and $A={\bar{X}_n\ge\lambda}$. Then \(\lambda P(A)\le EX_n1_A\le EX^+_n\)
Remarks
- For a submartingale that is not necessarily nonnegative, use $(X_n^+)_{n\ge 0}$.
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For an honest martingale, the inequality applies to $( X_n )_{n\ge 0}$.
Application: $L^p$ Convergence of a Martingale
Let ${X_n,\mathcal{F}_n}$ be a martingale with \(\sup_n E(|X_n|^p) < \infty.\) Then:
- $X_n \to X$ a.s.
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$E X_n - X ^p \to 0$.
Reasoning:
\(|X_n - X|^p \le 2^p \max_{1\le k\le n} |X_k|^p,\)
and the RHS is integrable by Doob’s inequality.
Thus the dominating function allows DCT.
If we want $E|X_n - X|\to 0$, we use:
\(\sup_n E(|X_n|^p)<\infty \quad\Rightarrow\quad \{X_n\}\text{ uniformly integrable}.\)
Branching Processes
Let ${\xi_{n,i}}$ be i.i.d. offspring counts with
\(P(\xi=0)>0,\qquad m = E(\xi) < \infty.\)
Define the Galton–Watson process: \(Z_0=1,\qquad Z_{n+1} = \sum_{i=1}^{Z_n} \xi_{n+1,i}.\)
Let $\mathcal{F}_n = \sigma(Z_1,\dots,Z_n)$.
Claim
\(X_n = \frac{Z_n}{m^n}\) is a martingale.
Check: \(E\left( \frac{Z_{n+1}}{m^{n+1}} \mid \mathcal{F}_n \right) = \frac{m\,Z_n}{m^{n+1}} = \frac{Z_n}{m^n} = X_n.\)
Extinction Behavior
Case 1: $m \le 1$
Then
\(Z_n \xrightarrow[n\to\infty]{a.s.} 0.\)
Proof idea:
${X_n}$ is a nonnegative supermartingale when $m\le 1$:
\(X_n = \frac{Z_n}{m^n} \text{ is decreasing.}\)
Thus by MGCT:
\(X_n \to X_\infty \quad\text{a.s.}\)
Since integers converge only if eventually constant and extinction is possible, we must have $Z_n\to 0$.
Thus extinction probability is: \(e = 1.\)
Case 2: $m>1$
Here extinction probability satisfies: \(e < 1,\) and is equal to the smallest positive solution of the fixed-point equation: \(e = \varphi(e), \qquad \varphi(s)=E(s^\xi)=\sum_{k=0}^\infty P(\xi=k)s^k.\)
Properties of generating function $\varphi$:
- Increasing, convex.
- $\varphi(1)=1$.
- $\varphi(0)=P(\xi=0)>0$.
- Slope $\varphi’(1)=E(\xi)=m>1$.
Graphically, the line $y=s$ intersects $\varphi(s)$ at $s=1$ and one more point in $[0,1)$. That point is extinction probability $e$.
Doob’s Inequality in the Supercritical Case
Assume now: \(\mathrm{Var}(\xi)=\sigma^2<\infty,\qquad m>1,\qquad P(\xi=0)>0.\)
Still: \(X_n = \frac{Z_n}{m^n}\) is a martingale. Then by MGCT: \(X_n \xrightarrow[n\to\infty]{a.s.} X.\)
Compute the quadratic variation term: \(E_{\mathcal{F}_{n-1}}\!\left[(X_n - X_{n-1})^2\right] = \mu^{-2n} E\big[ (Z_n - m Z_{n-1})^2 \mid \mathcal{F}_{n-1} \big] = \mu^{-2n} \sigma^2 Z_{n-1}.\)
Taking expectation: \(E(X_n - X_{n-1})^2 = \mu^{-2n} \sigma^2 E(Z_{n-1}) = \mu^{-2n} \sigma^2 m^{n-1} = \mu^{-(n+1)} \sigma^2.\)
Thus \(E(X_n^2) = 1 + \sigma^2 \sum_{k=1}^n \mu^{-(k+1)} \le 1+\sigma^2\sum_{k=1}^\infty \mu^{-(k+1)} <\infty.\)
Hence Doob’s $L^2$ inequality implies: \(E|X_n - X|^2 \to 0,\) and therefore: \(E(X)=1.\)
This is the key step in the Kesten–Stigum theorem.
Kesten–Stigum Theorem (optimal result)
Assume $m>1$. Then: \(\frac{Z_n}{m^n} \xrightarrow[n\to\infty]{a.s.} 0 \quad\text{with positive probability}\) iff \(E\big[\xi \log^{+}(\xi)\big] < \infty.\)
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