6 — Wald’s Equations and Hitting Times
This lecture builds on stopping times from Lectures 4–5 and introduces the first and second Wald equations, with applications to simple symmetric random walk hitting times.
1. Setup
We work on a filtered probability space
\[(\Omega, \mathcal F, P),\qquad \{\mathcal F_n\}_{n\ge1}.\]Assume:
- $T$ is a stopping time w.r.t. ${\mathcal F_n}$.
- ${X_n}_{n\ge1}$ are i.i.d. and adapted, meaning
$X_n \in \mathcal F_n$ for all $n$. - Define partial sums
\(S_0 = 0,\qquad S_n = \sum_{k=1}^n X_k .\)
Define the stopped sum:
\[S_T = \sum_{k=1}^T X_k .\]2. First Wald Equation
If
- $E\vert X_1\vert < \infty$
- $E(T) < \infty$
then
\[\boxed{E(S_T) = E(X_1)\, E(T)} .\]This deceptively simple identity is extremely powerful and used repeatedly in gambling, random walks, queueing theory, and renewal theory.
Durrett Probability 4e - Theorem 5.1.5 Wald’s Equation
Let $Χ_1, Χ_2, …$ i.i.d. with $\mathbb{E}\vert X_i\vert<\infty$. If $N$ is a stopping time with $\mathbb{E}N<\infty$ then $\mathbb{E}S_N=\mathbb{E}X_1\mathbb{E}N$.
3. Example: SSRW Hitting Times
Simple symmetric random walk (SSRW):
\[P(X_k = +1) = P(X_k = -1) = \frac12 .\]Define hitting times of two levels:
-
For $a>0$
\(T_a = \min\{n\ge1 : S_n = a\}.\) -
For $b<0$
\(T_b = \min\{n\ge1 : S_n = b\}.\)
Define the two-sided hitting time
\[T_{a,b} = T_a \wedge T_b.\]On page 1 of the PDF :contentReference[oaicite:1]{index=1} you can see the professor’s diagram showing the walk fluctuating until it hits either level.
3.1 Using Wald’s Equation
Since $E(X_1)=0$, Wald gives
\[E\left(S_{T_{a,b}}\right) = 0.\]But
\[S_{T_{a,b}} = \begin{cases} a & \text{if } T_a < T_b, \\ b & \text{if } T_b < T_a. \end{cases}\]Thus
\[a\,P(T_a<T_b)+ b\,P(T_b<T_a)=0,\]and using
$P(T_a<T_b)+P(T_b<T_a)=1$, we solve:
These are the classical gambler’s ruin probabilities.
3.2 Almost sure finiteness
Using the formula above:
\[P(T_a < T_b) = \frac{-b}{a-b} \xrightarrow[b\to -\infty]{} 1.\]Hence
\[P(T_a < \infty) = 1.\]So SSRW hits every positive integer almost surely.
(This matches the intuitive recurrence of 1-D SSRW.)
3.3 Does $E(T_a)$ exist?
Suppose for contradiction that $E(T_a) <\infty$.
Then Wald’s equation gives:
\[E(S_{T_a}) = E(X_1)E(T_a)=0.\]But $S_{T_a} = a$, so
\[E(S_{T_a}) = a.\]Contradiction unless $a=0$.
Thus
\[\boxed{E(T_a)=\infty}.\]This is written explicitly on page 1 of the PDF :contentReference[oaicite:2]{index=2}.
4. A Side Argument Showing $E(T_{a,b}) < \infty$
While $E(T_a)=\infty$ for a one-sided boundary, the two-sided hitting time $T_{a,b}$ does have finite expectation.
The argument (page 2) uses:
- Geometric trial structure
- Blocks of length $(a+\vert b\vert)$
- Success probability $2^{-(a+\vert b\vert)}$
Thus
\[E(T) = \frac{1}{2^{-(a+|b|)}} < \infty\]and
\[E(T_{a,b}) \le (a+|b|)\,E(T) <\infty.\]5. Second Wald Equation
Assume now:
- $E(X_1)=0$
- $E(X_1^2) = \sigma^2 < \infty$
- $E(T) < \infty$
Then
\[\boxed{E(S_T^2)= \sigma^2\,E(T)}.\]This appears in red ink on page 2 of the PDF :contentReference[oaicite:3]{index=3}.
Durrett Probability 4e - Theorem 4.1.6 Wald’s second Equation.
Let $X_1, X_2, …$ be i.i.d. with $\mathbb{E}X_n=0$ and $\mathbb{E}X^2_n=\sigma^2<\infty$. If $T$ is a stopping time with $\mathbb{T}<\infty$ then $\mathbb{E}S^2_T=\sigma^2\mathbb{E}T$.
Proof Sketch
The proof (pages 2–3) uses:
- Square-difference decomposition
\(S_{T\wedge(n+1)} - S_{T\wedge n} = X_{n+1}\,\mathbf 1_{\{T\ge n+1\}}\) - Independence and adaptedness
\(X_{n+k}\perp \mathbf 1_{\{T\ge n+1\}}\)
because $X_{n+k}$ is independent of $\mathcal F_{n+k-1}$. - No cross-terms survive (zero expectation).
Then
\[E\bigl(S_{T\wedge(n+1)} - S_{T\wedge n}\bigr)^2 = E(X_1^2)\, P(T\ge n+1) = \sigma^2\, P(T\ge n+1).\]Summing from $n=0$ to $\infty$:
\[E(S_T^2) = \sigma^2\sum_{n=0}^{\infty} P(T\ge n+1) = \sigma^2\,E(T).\]As shown on page 3 of the PDF :contentReference[oaicite:4]{index=4}, the partial sums form a Cauchy sequence in $L^2$, implying $S_{T\wedge n}\to S_T$ in $L^2$.
6. Extension when $E(X_1)\ne 0$
Let $\mu = E(X_1)$ and consider centered variables
\[Y_k = X_k - \mu,\qquad E(Y_k)=0.\]Then
\[S_T = \sum_{k=1}^T Y_k + \mu T.\]Thus
\[E(S_T^2) = \operatorname{Var}(S_T) + \mu^2 E(T^2) + 2\mu\, E\bigl[(S_T - \mu T)\,T\bigr].\]If additionally $T$ is independent of the increments (e.g., in renewal theory), then the cross-term vanishes:
\[E\bigl[(S_T - \mu T)\,T\bigr] = 0.\]This corresponds to the final notes on page 4 of the PDF :contentReference[oaicite:5]{index=5}.
Summary
- First Wald: $E(S_T)=E(X_1)E(T)$
- Second Wald: $E(S_T^2)=\sigma^2 E(T)$ when $E(X_1)=0$
- SSRW:
- Hits any level a.s., but
- One-sided hitting times have infinite expectation: $E(T_a)=\infty$
- Two-sided hitting times have finite expectation: $E(T_{a,b})<\infty$
- Centering trick handles the general case $E(X_1)\ne 0$.
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