I have been trying to prove the following result:
Let $(X_n)_{n \in \mathbb N_0}$ be a submartingale or supermartingale. Use Doob's Inequality and Doob's Decomposition to show that, for all $n \in \mathbb N$ and $\lambda > 0$, $$ \lambda\mathbb P\left[|X|_n^* \geq \lambda\right] \leq 12\mathbb E\left[\left|X_0\right|\right] + 9\mathbb E\left[\left|X_n\right|\right]. $$ where $|X|_n^* = \sup\left\{|X_k| : 0 \leq k \leq n\right\}$.
The version of Doob's inequality we're using is that for any $p \geq 1$, $\lambda > 0$, and martingale or positive submartingale $Y$, $$ \lambda^p \mathbb P\left[|Y|_n^*\geq \lambda\right] \leq \mathbb E\left[\left|Y_n\right|^p\right]. $$ It suffices to prove this result when $X$ is a submartingale. Using Doob's decomposition $X = M+A$, $M$ a martingale and $A$ an increasing predictable process with $A_0 = 0$ (so $A$ is a positive submartingale), one can in fact show a stronger inequality. Indeed, since $A$ is positive and increasing, $|X|_n^* \leq |M|_n^* + A_n$. And since $A_0 = 0$: $$ \mathbb E\left[A_n\right] = \mathbb E\left[X_n\right] - \mathbb E\left[M_n\right] = \mathbb E\left[X_n\right] - \mathbb E\left[M_0\right] = \mathbb E\left[X_n\right] - \mathbb E\left[X_0\right] \leq \mathbb E\left[|X_n|\right] + \mathbb E\left[|X_0|\right] $$ from which it follows that $$ \mathbb E\left[\left|M_n\right|\right] \leq \mathbb E\left[\left|X_n\right|\right] + \mathbb E\left[A_n\right] \leq 2\mathbb E\left[\left|X_n\right|\right] + \mathbb E\left[\left|X_0\right|\right]. $$ Using these inequalities, it follows that \begin{align*} \lambda\mathbb P\left[|X|^*_n\geq \lambda\right] & \leq \lambda\mathbb P\left[|M|_n^*+A_n\geq\lambda\right] \\ &\leq \lambda \mathbb P\left[ |M|^*_n\geq \frac 2 3 \lambda\right] + \lambda\mathbb P\left[A_n\geq\frac 1 3 \lambda\right] \\ &\leq \frac 3 2 \mathbb E\left[\left|M_n\right|\right]+ 3\mathbb E\left[A_n\right] \\ &\leq 6\mathbb E\left[\left|X_n\right|\right]+\frac 9 2 \mathbb E\left[\left|X_0\right|\right] \end{align*} My question is twofold:
- Is there an error in this argument, such as a flaw in my assumptions or an unjustified assumption I'm not noticing? And if not,
- Is there a reason the book I'm using (Klenke's Probability Theory: A Comprehensive Course) uses the coefficients $12$ and $9$ rather than $9/2$ and $6$? Is the stated result somehow more classical or easier to show using more fundamental properties of martingales and the Doob decomposition?
This problem was also discussed here, but this thread doesn't really address the seeming arbitrariness of the coefficients $12$ and $9$. Can anyone provide any insight?
This is only a fragment of an answer because I don't touch upon your proof or the techniques it uses, but it is too long for a comment. My intuition is that the coefficients are arbitrary because they are not optimal. Here is one possible improvement, which I take from the book Brownian Motion, Martingales, and Stochastic Calculus by Jean-François Le Gall (p.263)
Proof (not in the book). Fix $\lambda>0$ and $k\in\mathbb{N}$. Let $A_k=\left\{\omega\in\Omega : \sup_{n\leq k}Y_k(\omega)> \lambda\right\}$. Define the stopping time $T=\inf\left\{n\in\mathbb{N} : Y_n> \lambda\right\}$, and notice that $A_k=\left\{T\leq k\right\}$. Since $(Y_n)_{n\in\mathbb{N}}$ is a supermartingale $$\mathbb{E}Y_0\geq\mathbb{E}Y_{T\land k}\geq \lambda \mathbb{P}(A_k)+\mathbb{E}[Y_k\mathbf{1}_{A_k^c}]$$ Now, let $S=\inf\left\{n\in\mathbb{N} : Y_n<-\lambda\right\}$ and $B_k=\left\{\omega\in\Omega : \inf_{n\leq k} Y_k(\omega)<-\lambda\right\}$. We have $$\mathbb{E}Y_k\leq\mathbb{E}Y_{S\land k}\leq -\lambda \mathbb{P}(B_k)+\mathbb{E}[Y_k\mathbf{1}_{B_k^c}]$$ Rearranging and summing the two inequalities gives $$\lambda\mathbb{P}\left[\sup_{n\leq k}\left|Y_n\right|>\lambda\right]\leq \mathbb{E}Y_0-\mathbb{E}[Y_k\mathbf{1}_{A^c_k}]-\mathbb{E}[Y_k\mathbf{1}_{B_k}]\leq \mathbb{E}|Y_0|+2\mathbb{E}|Y_k|$$ By the way, we also proved that an even better upper bound is $\mathbb{E}Y_0 + 2\mathbb{E}Y_k^-$.