Question

Let \(\left\\{Y_{n}\right\\}\) be a nonnegative submartingale and suppose \(b_{n}\) is a nonincreasing sequence of positive numbers. Suppose \(\sum_{n=1}^{\infty}\left(b_{n}-b_{n+1}\right) E\left[Y_{n}\right]<\infty .\) Prove that $$ \lambda \operatorname{Pr}\left\\{\sup _{k \geq 1} b_{k} Y_{k}>\lambda\right\\}<\sum_{k=1}^{\infty}\left(b_{k}-b_{k+1}\right) E\left[Y_{k}\right] $$

Short answer

To prove that \(\lambda \operatorname{Pr}\left\\{\sup _{k \geq 1} b_{k} Y_{k}>\lambda\right\\}<\sum_{k=1}^{\infty}\left(b_{k}-b_{k+1}\right) E\left[Y_{k}\right]\), we follow these steps: 1. Define a sequence of events \(A_k = \left\\{b_kY_k > \lambda\right\\}\). 2. Apply Doob's martingale inequality to our nonnegative submartingale \(Y_n\). 3. Rewrite the inequality using properties of conditional expectations and probabilities and let \(N\) approach infinity. 4. Bound the supremum's expectation as \(E\left[\sup_{k \geq 1}b_kY_k\right] \leq \sum_{k=1}^{\infty} E\left[\left(b_k - b_{k+1}\right)Y_k\right]\). 5. Substitute the bound into the inequality from step 3 and then multiply both sides by \(\lambda\) to obtain the desired result.

Step-by-step solution

Define the Sequence of Events

Define a sequence of events \(\left\\{A_k\right\\}\), for each \(k\geq 1\), such that: $$ A_k = \left\\{b_kY_k > \lambda\right\\}. $$ These events represent when the product \(b_k Y_k\) exceeds the given threshold \(\lambda\).

Apply Doob's Martingale Inequality

Since \(\left\\{Y_{n}\right\\}\) is a nonnegative submartingale, we can apply Doob's martingale inequality, which states that for a nonnegative submartingale \(\left\\{Y_{n}\right\\}\) and any \(p>1\), we have: $$ \left(E\left[\left|\sup_{1 \leq n \leq N} Y_n\right|^p\right]\right)^{\frac{1}{p}} \leq \frac{p}{p-1} \left(E\left[\left|Y_N\right|^p\right]\right)^{\frac{1}{p}} $$ For our given sequence, we can write: $$ \left(E\left[\left|\sup_{1 \leq n \leq N} b_nY_n\right|^p\right]\right)^{\frac{1}{p}} \leq \frac{p}{p-1} \left(E\left[\left|b_NY_N\right|^p\right]\right)^{\frac{1}{p}} $$

Use Properties of Conditional Expectations and Probability

Let's allow \(N\) to approach infinity. Then, \(\sup_{1 \leq n \leq N} b_nY_n\) converges to \(\sup_{k \geq 1} b_kY_k\). We can use properties of conditional expectations and probabilities to rewrite the inequality as: $$ \operatorname{Pr}\left\\{\sup_{k \geq 1} b_kY_k > \lambda\right\\} \leq \frac{E\left[\sup_{k \geq 1} b_kY_k\right]}{\lambda} $$

Bound the Supremum Using Expectations

Now we proceed to find an upper bound for the supremum's expectation: $$ E\left[\sup_{k \geq 1}b_kY_k\right] \leq \sum_{k=1}^{\infty} E\left[\left(b_k - b_{k+1}\right)Y_k\right] $$ Since the series on the right-hand side of the inequality converges, the supremum's expectation is finite.

Combine Inequalities and Prove the Result

Now substitute the bound found in step 4 into the inequality from step 3: $$ \operatorname{Pr}\left\\{\sup_{k \geq 1} b_kY_k > \lambda\right\\} \leq \frac{\sum_{k=1}^{\infty}E\left[\left(b_k - b_{k+1}\right)Y_k\right]}{\lambda} $$ Multiplying both sides by \(\lambda\) completes the proof: $$ \lambda \operatorname{Pr}\left\\{\sup_{k \geq 1} b_kY_k > \lambda\right\\} \leq \sum_{k=1}^{\infty}\left(b_k - b_{k+1}\right) E\left[Y_k\right]. $$