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2018.Q6 — Martingale / Supermartingale Construction and Convergence

2018 Probability Prelim Exam (PDF)

Let ${X_n}, {Y_n}, {Z_n}{n\ge 0}$ be three sequences of random variables,
integrable, nonnegative, and adapted to ${\mathcal F_n}{n\ge 0}$.
Assume
\(\sum_{k=1}^\infty Y_k < \infty \quad \text{a.s.}\) (a) Show that
\(\prod_{k=1}^n (1+Y_k)^2\) is non-decreasing and converges a.s.

(b)(i) Show that
\(\left\{ \frac{X_n}{\prod_{k=1}^{\,n-1}(1+Y_k)^2},\ \mathcal F_n \right\}_{n\ge1}\) is a nonnegative supermartingale.

(b)(ii) Prove that
\(\left\{ \frac{\min(X_n,Z_n)}{\prod_{k=1}^{\,n-1}(1+Y_k)^2},\ \mathcal F_n \right\}_{n\ge1}\) is also a supermartingale.

(c) Conclude that $X_n$ and $\min(X_n,Z_n)$ converge a.s.

Part (a)

Claim

The product
\(P_n := \prod_{k=1}^n (1+Y_k)^2\) is non-decreasing and converges almost surely.

Proof

Since $Y_k \ge 0$, we have $(1+Y_k)^2 \ge 1$. Therefore
\(P_{n+1} = P_n (1+Y_{n+1})^2 \ge P_n,\) so ${P_n}$ is non-decreasing.

To show convergence, consider the logarithm: \(\log P_n = \sum_{k=1}^n 2\log(1+Y_k).\) Because $Y_k \ge 0$, we may use the standard inequality
\(\log(1+Y_k) \le Y_k.\) Thus
\(\log P_n \le 2\sum_{k=1}^n Y_k.\) By the assumption $\sum Y_k < \infty$ a.s., the right-hand side is finite a.s., so
\(\log P_n\) converges a.s. Since $\log$ is continuous and monotone on $(0,\infty)$, this implies
\(P_n \to P_\infty < \infty \quad \text{a.s.}\)

Conclusion

\(\prod_{k=1}^n (1+Y_k)^2\) is non-decreasing and convergent almost surely.

Key Ideas

  • Using logarithms to convert products into sums.
  • Inequality $\log(1+u) \le u$ for $u\ge0$.
  • A non-decreasing positive sequence converges iff it is bounded.
  • Boundedness follows from the assumption $\sum Y_n < \infty$.

Part (b)(i)

Let
\(M_n := \frac{X_n}{\prod_{k=1}^{\,n-1}(1+Y_k)^2}.\)

Claim

${M_n,\mathcal F_n}$ is a nonnegative supermartingale.

Proof

Nonnegativity.
Since $X_n \ge 0$ and each $(1+Y_k)^2 >0$,
\(M_n \ge 0.\)

Adaptedness and integrability are inherited from $X_n$.

Supermartingale property.
Compute: \(E[M_{n+1}\mid \mathcal F_n] = E\left[ \frac{X_{n+1}}{\prod_{k=1}^{\,n}(1+Y_k)^2} \,\middle|\, \mathcal F_n \right] = \frac{1}{\prod_{k=1}^{\,n}(1+Y_k)^2} E[X_{n+1}\mid \mathcal F_n].\)

The problem assumption (from the handwritten text) gives
\(E[X_{n+1}\mid \mathcal F_n] \le X_n (1+Y_n)^2.\)

Substituting: \(E[M_{n+1}\mid \mathcal F_n] \le \frac{X_n (1+Y_n)^2}{\prod_{k=1}^{\,n}(1+Y_k)^2} = \frac{X_n}{\prod_{k=1}^{\,n-1}(1+Y_k)^2} = M_n.\)

Thus ${M_n}$ is a supermartingale.

Conclusion

\(M_n = \frac{X_n}{\prod_{k=1}^{n-1}(1+Y_k)^2}\) is a nonnegative supermartingale.

Key Ideas

  • Constructing a supermartingale by dividing out a known growth factor.
  • Verifying supermartingale condition via
    \(E[X_{n+1}\mid\mathcal F_n] \le X_n (1+Y_n)^2.\)
  • Checking the three components: nonnegativity, integrability, adaptedness.
  • Dividing by the product from part (a), which is monotone and finite.

Part (b)(ii)

Define
\(M_n^* := \frac{\min(X_n, Z_n)}{\prod_{k=1}^{\,n-1}(1+Y_k)^2}.\)

Claim

${M_n^*, \mathcal F_n}$ is a supermartingale.

Proof

Nonnegativity.
Since $X_n\ge0$, $Z_n\ge0$,
\(\min(X_n,Z_n) \ge 0 \quad \Rightarrow\quad M_n^*\ge0.\)

Supermartingale property.
Define \(H_n := \frac{X_n}{\prod_{k=1}^{\,n-1}(1+Y_k)^2}, \qquad J_n := \frac{Z_n}{\prod_{k=1}^{\,n-1}(1+Y_k)^2}.\) Part (b)(i) shows $H_n$ is a supermartingale.
The same argument (replacing $X_n$ with $Z_n$) shows $J_n$ is also a supermartingale.

A standard result (the problem hint) states:

If $H_n$ and $J_n$ are supermartingales, then $\min(H_n, J_n)$
is also a supermartingale.

Therefore, \(M_n^* = \min(H_n, J_n)\) is a supermartingale.

Conclusion

\(\frac{\min(X_n,Z_n)}{\prod_{k=1}^{n-1}(1+Y_k)^2}\) is a supermartingale.

Key Ideas

  • The minimum of two supermartingales is again a supermartingale.
  • Reduce the problem to part (b)(i) applied to $X_n$ and separately to $Z_n$.
  • Nonnegativity + adaptedness carry over through the $\min$ operator.

Part (c)

Claim

Both $X_n$ and $\min(X_n,Z_n)$ converge almost surely.

Proof

From parts (b)(i)–(ii),
\(M_n \quad \text{and} \quad M_n^*\) are nonnegative supermartingales.
By the Martingale Convergence Theorem, each converges a.s.: \(M_n \to M_\infty, \qquad M_n^* \to M_\infty^* \qquad\text{a.s.}\)

From part (a), the product
\(P_{n-1} := \prod_{k=1}^{n-1}(1+Y_k)^2\) converges a.s. to a finite limit $P_\infty >0$.

Now write: \(X_n = M_n \, P_{n-1}.\) Both factors converge a.s., so $X_n$ converges a.s.

Similarly, \(\min(X_n, Z_n) = M_n^* \, P_{n-1}\) converges a.s.

Conclusion

\(X_n \to X_\infty, \qquad \min(X_n,Z_n) \to W_\infty \quad\text{a.s.}\)

Key Ideas

  • Nonnegative supermartingales converge (MCT).
  • Express the original processes as
    \(X_n = M_n \cdot \text{(deterministic convergent factor)}.\)
  • Use part (a)’s product convergence to rebuild the limit of $X_n$.
  • Same reconstruction works for $\min(X_n,Z_n)$.

Summary of Techniques Learned

  1. Product Convergence via Logs
    • When facing $\prod (1+Y_n)^2$, take logs and use $\log(1+u)\le u$.
  2. Building a Supermartingale
    • Divide a process by its known multiplicative growth factor.
    • Check the supermartingale inequality after the division.
  3. Minimum of Supermartingales
    • A very useful structural lemma: $\min(H_n, J_n)$ is supermartingale.
  4. Martingale Convergence Theorem
    • Every nonnegative supermartingale converges almost surely.
  5. Reconstructing the Limit of the Original Sequence
    • If $M_n = X_n / a_n$ and the normalizing factor $a_n$ converges,
      then $X_n$ also converges.
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