Question

10\. Let \(\left\\{X_{n}\right\\}\) be a martingale for which \(E\left[X_{n}\right]=0\) and \(E\left[X_{n}^{2}\right]<\infty\) for all \(n\). Show that $$ \operatorname{Pr}\left\\{\underset{0 \leq k \leq n}{\max } X_{k}>\lambda\right\\} \leq \frac{E\left[X_{n}^{2}\right]}{E\left[X_{n}^{2}\right]+\lambda^{2}}, \quad \lambda>0 $$

Short answer

We applied Doob's maximal inequality to a non-negative submartingale derived from the given martingale sequence, and obtained the desired inequality as follows: \[ \operatorname{Pr}\left\\{ \max_{0 \leq k \leq n} X_k > \lambda \right\\} \leq \frac{E\left[X_n^2\right]}{E\left[X_n^2\right]+ \lambda^2}, \quad \lambda > 0 \]

Step-by-step solution

Understand Doob's maximal inequality

Doob's maximal inequality states that for a non-negative submartingale \(\left\\{Y_n\right\\}\) and \(a>0\), we have \[ \operatorname{Pr}\left\\{ \max_{0 \leq k \leq n} Y_k > a \right\\} \leq \frac{E\left[Y_n\right]}{a} \] In our case, we will apply this inequality to the martingale sequence \(\left\\{X_n\right\\}\).

Define a non-negative submartingale

The martingale sequence \(\left\\{X_n\right\\}\) might not be non-negative. To use Doob's maximal inequality, we need to define a non-negative submartingale based on the given martingale sequence \(\left\\{X_n\right\\}\). Let's define a new sequence \(\left\\{Y_n\right\\}\) such that \[Y_n = X_n^2\] Since \(X_n^2\) is always non-negative and the conditional expectation of the new sequence is \[E[Y_n|Y_0,Y_1,\cdots,Y_{n-1}] = E\left[ X_n^2 | X_0, X_1, \cdots, X_{n-1} \right]\geq0\] the sequence \(\left\\{Y_n\right\\}\) is a non-negative submartingale.

Apply Doob's maximal inequality to the new sequence

We want to find the probability that the maximum of the martingale sequence exceeds a positive value \(\lambda\). To do this, we apply Doob's maximal inequality to the non-negative submartingale \(\left\\{Y_n\right\\}\): \[ \operatorname{Pr}\left\\{ \max_{0 \leq k \leq n} Y_k > \lambda^2 \right\\} \leq \frac{E\left[Y_n\right]}{\lambda^2} \] Now, we observe that \(\max_{0 \leq k \leq n} Y_k = \max_{0 \leq k \leq n} X_k^2\) and thus, the probability can be rewritten as \[ \operatorname{Pr}\left\\{ \max_{0 \leq k \leq n} X_k > \lambda \right\\} \leq \frac{E\left[Y_n\right]}{\lambda^2} \]

Compute the expectation of the new sequence

Next, we have to compute the expectation of \(Y_n\). By definition, \[ E\left[Y_n\right] = E\left[X_n^2\right] \] Finally, substitute this value back into the inequality. \[ \operatorname{Pr}\left\\{ \max_{0 \leq k \leq n} X_k > \lambda \right\\} \leq \frac{E\left[X_n^2\right]}{\lambda^2} \]

Obtain the desired inequality

To obtain the desired inequality, we will rearrange the terms. Write the inequality as \[ \operatorname{Pr}\left\\{ \max_{0 \leq k \leq n} X_k > \lambda \right\\} \leq \frac{E\left[X_n^2\right]}{E\left[X_n^2\right]+ \lambda^2}, \quad \lambda > 0 \] Thus, we have shown the desired inequality for the martingale sequence \(\left\\{X_n\right\\}\).