2022 Probability Prelim Exam (PDF)
Problem 5 (verbatim)
Let ${X_k,\mathcal F_k}$ be a martingale sequence with $\mathbb E(X_k^2)<\infty$, $k=0,1,\dots$.
Let
\(A_n^X=\sum_{k=1}^n \mathbb E_{\mathcal F_{k-1}}\bigl(X_k-X_{k-1}\bigr)^2,
\qquad n=1,2,\dots\)
(The sequence ${A_n^X}$ is known as the predictable and increasing process associated with
${X_k,\mathcal F_k}$.)
Assume $A_1^X=C>0$.
a.
Let \(Y_n=\sum_{k=1}^n \frac{X_k-X_{k-1}}{A_k^X},\qquad n=1,2,\dots\)
(i) Prove that \(\mathbb E_{\mathcal F_{k-1}}(Y_k-Y_{k-1})^2 =\frac{A_k^X-A_{k-1}^X}{(A_k^X)^2} \le \frac{1}{C} \quad\text{a.s.}\)
(ii) Prove that ${Y_k,\mathcal F_k}$ is a martingale and that $\mathbb E(Y_k^2)<\infty$.
For the rest of the problem define \(A_n^Y=\sum_{k=1}^n \mathbb E_{\mathcal F_{k-1}}(Y_k-Y_{k-1})^2.\)
b.
(i) Prove that \(A_n^Y \le \int_C^\infty x^{-2}\,dx = C^{-1}, \quad \text{a.s.}\)
(ii) Prove that \(\sup_{n\ge1}\mathbb E(Y_n^2)<\infty \quad\text{and}\quad \lim_{n\to\infty}Y_n \text{ exists and is finite a.s.}\)
c.
Show how to use Kronecker’s Lemma to prove that if $A_n^X\to\infty$ a.s., then \(\frac{X_n}{A_n^X}\to 0 \quad \text{a.s.}\)
Solution
a(i)
Claim
\(\mathbb E_{\mathcal F_{k-1}}(Y_k-Y_{k-1})^2 =\frac{A_k^X-A_{k-1}^X}{(A_k^X)^2} \le \frac{1}{C} \quad\text{a.s.}\)
Proof
Start by explicitly computing the increment: \(Y_k-Y_{k-1} =\sum_{m=1}^k \frac{X_m-X_{m-1}}{A_m^X} -\sum_{m=1}^{k-1} \frac{X_m-X_{m-1}}{A_m^X} =\frac{X_k-X_{k-1}}{A_k^X}.\)
Now substitute into the conditional second moment: \(\mathbb E_{\mathcal F_{k-1}}(Y_k-Y_{k-1})^2 =\mathbb E_{\mathcal F_{k-1}} \frac{(X_k-X_{k-1})^2}{(A_k^X)^2}.\)
Since $A_k^X$ is predictable, $A_k^X\in\mathcal F_{k-1}$, so it can be pulled out: \(=\frac{\mathbb E_{\mathcal F_{k-1}}(X_k-X_{k-1})^2}{(A_k^X)^2}. \tag{1}\)
Now observe directly from the definition of $A_k^X$: \(A_k^X-A_{k-1}^X =\mathbb E_{\mathcal F_{k-1}}(X_k-X_{k-1})^2.\)
Substituting into (1) gives \(\mathbb E_{\mathcal F_{k-1}}(Y_k-Y_{k-1})^2 =\frac{A_k^X-A_{k-1}^X}{(A_k^X)^2}.\)
To bound this, use only monotonicity:
- $A_k^X\ge A_1^X=C$,
- $A_k^X-A_{k-1}^X\le A_k^X$.
Hence \(\frac{A_k^X-A_{k-1}^X}{(A_k^X)^2} \le \frac{1}{A_k^X} \le \frac{1}{C}.\) ∎
Conclusion
Each normalized increment of $Y$ has uniformly bounded conditional variance.
Key Takeaways
- Predictability lets $A_k^X$ exit conditional expectations.
- $A_k^X-A_{k-1}^X$ is exactly the conditional variance increment.
- Bounding uses only monotonicity, no summability yet.
a(ii)
Claim
${Y_k,\mathcal F_k}$ is a martingale and $\mathbb E(Y_k^2)<\infty$ for all $k$.
Proof
Adaptedness.
Each summand $(X_k-X_{k-1})/A_k^X$ is $\mathcal F_k$-measurable, hence $Y_n\in\mathcal F_n$.
Integrability.
From part (a)(i),
\(\mathbb E(\Delta Y_k^2)
=\mathbb E\big[\mathbb E_{\mathcal F_{k-1}}(\Delta Y_k^2)\big]
\le \frac{1}{C}.\)
By Cauchy–Schwarz,
\(\mathbb E|\Delta Y_k|
\le \sqrt{\mathbb E(\Delta Y_k^2)}<\infty.\)
Since $Y_n=\sum_{k=1}^n\Delta Y_k$ is a finite sum,
\(\mathbb E|Y_n|<\infty.\)
Martingale property. \(\mathbb E_{\mathcal F_{n-1}}(Y_n-Y_{n-1}) =\frac{1}{A_n^X} \bigl(\mathbb E_{\mathcal F_{n-1}}X_n-X_{n-1}\bigr) =0.\)
Second moment.
Martingale differences are orthogonal:
\(\mathbb E(\Delta Y_i\Delta Y_j)=0\quad(i<j).\)
Thus
\(\mathbb E(Y_n^2)
=\sum_{k=1}^n \mathbb E(\Delta Y_k^2)<\infty.\)
∎
Conclusion
$Y$ is an $L^2$ martingale.
Key Takeaways
- Martingale integrability is for each fixed $n$, not uniform.
- Orthogonality of increments simplifies $L^2$ calculations.
b(i)
Claim
\(A_n^Y =\sum_{k=1}^n \frac{A_k^X-A_{k-1}^X}{(A_k^X)^2} \le C^{-1}\quad\text{a.s.}\)
Proof
From part (a)(i), \(A_n^Y =\sum_{k=1}^n \frac{A_k^X-A_{k-1}^X}{(A_k^X)^2}.\)
Now pause and reinterpret the sum.
- $A_k^X-A_{k-1}^X$ plays the role of a measure increment.
- $(A_k^X)^{-2}$ is a decreasing function of the current quadratic variation level.
Let $f(x)=x^{-2}$ for $x\ge C$.
Then $f$ is positive and decreasing, and the sum can be read as a right-endpoint Riemann–Stieltjes sum:
\(\sum_{k=1}^n f(A_k^X)\,(A_k^X-A_{k-1}^X).\)
For a decreasing function, right-endpoint rectangles lie below the curve, so \(\sum_{k=1}^n \frac{A_k^X-A_{k-1}^X}{(A_k^X)^2} \le \int_{A_1^X}^\infty x^{-2}\,dx =\int_C^\infty x^{-2}\,dx =C^{-1}.\) ∎
Conclusion
The quadratic variation of $Y$ is uniformly bounded.
Key Takeaways
- Quadratic variation behaves like a Stieltjes measure.
- Right-endpoint + decreasing function ⇒ sum ≤ integral.
- This pattern appears repeatedly in martingale limit theorems.
b(ii)
Claim
\(\sup_n \mathbb E(Y_n^2)<\infty \quad\text{and}\quad Y_n\to Y_\infty \text{ a.s.}\)
Proof
From orthogonality, \(\mathbb E(Y_n^2) =\mathbb E(A_n^Y) \le C^{-1}.\) Hence $\sup_n\mathbb E(Y_n^2)<\infty$. By the martingale convergence theorem, $Y_n$ converges almost surely (and in $L^2$). ∎
Conclusion
$Y_n$ converges a.s. to a finite limit.
Key Takeaways
- $L^2$-bounded martingales converge a.s.
- Bounded quadratic variation ⇒ bounded second moments.
c
Claim
If $A_n^X\to\infty$ a.s., then \(\frac{X_n}{A_n^X}\to 0 \quad\text{a.s.}\)
Proof
Write $D_k=X_k-X_{k-1}$, a martingale difference sequence. From part (b), the series \(\sum_{k=1}^\infty \frac{D_k}{A_k^X}\) converges almost surely. By Kronecker’s Lemma (martingale difference version), \(\frac{1}{A_n^X}\sum_{k=1}^n D_k \to 0 \quad\text{a.s.}\) Since $\sum_{k=1}^n D_k=X_n-X_0$, the claim follows. ∎
Conclusion
Diverging quadratic variation forces sublinear martingale growth.
Key Takeaways
- Kronecker converts convergence of weighted increments into normalized limits.
- This is the bridge from $Y_n$ back to $X_n$.
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