Assumption of non-negativity in Doob's inequality for the maximum of discrete-time submartingales

Question

Was doing some review of discrete time stochastic processes and I have a question about where the non-negativity assumption is used in the proof of the following standard theorem:

Let $(M_n)_{n=0,1,2\ldots}$ be a non-negative submartingale. Define $M^*_n = \max_{m\le n}M_m$ and let $\lambda > 0.$ Then $$P(M_n^*\ge\lambda) \le\frac{1}{\lambda } E(M_n 1(M_n^*\ge \lambda)).$$

The proof, which I've seen in a few places, goes as follows:

Proof: Let $\tau = \min\{m\ge 0\mid M_m\ge\lambda\}$ be the first passage time to $\lambda.$ Observe that $\{M^*_n\ge\lambda\} = \{\tau \le n\}$ since the maximum is greater than $\lambda$ if and only if $M$ has passed $\lambda$ at some time in the past. Then, we can write $$ P(M_n^*\ge\lambda) = P(\tau\le n) = \sum_{k=0}^n P(\tau=k).$$ Since $M_n$ is a submartingale, we have $$ E(M_n\mid\mathcal F_k) \ge M_k$$ for all $0\le k\le n.$ We have $\{\tau=k\}\in\mathcal F_k,$ so the properties of conditional expectation give $$E(M_n1(\tau = k)) \ge E(M_k1(\tau=k)).$$ Finally, since $M_k\ge\lambda$ on $\{\tau=k\},$ we can write $$ E(M_n1(\tau = k))\ge E(M_k1(\tau=k)) \ge \lambda P(\tau=k).$$ Then we plug in this inequality to the first equation, giving $$P(M_n^*\ge\lambda) = \sum_{k=0}^n P(\tau=k) \le \sum_{k=0}^n\frac{1}{\lambda}E(M_n1(\tau=k)) = \frac{1}{\lambda } E(M_n 1(M_n^*\ge \lambda))$$ where in the last step we used the fact that the $\{\tau=k\}$ sets are disjoint for different $k$'s and that $$\bigcup_{k=1}^n\{\tau=k\} = \{\tau\le n\} = \{M^*_n \ge \lambda\}.$$

I cannot see where the assumption that $M_n$ is non-negative is used. I figure I must be missing something (probably something obvious) since all the statements of the theorem I've seen make the assumption.

Answer 1

The inequality holds true for any submartingale (not necessarily non-negative); you can find the statement, for instance, in Measures, Integrals and Martingales by René Schilling, Lemma 19.11.

Answer 2

I figured it out. It was really stupid of course. In both of my sources that used the non-negativity assumption, the actual inequality in the theorem was $$ P(M_n^*\ge\lambda) \le \frac{1}{\lambda}E(M_n1(M^*_n\ge\lambda)) \le\frac{1}{\lambda}E(M_n)$$ so nonnegativity is required for the right-hand inequality, which is trivial under that assumption (thus why I overlooked it). As saz said in their answer, the left-hand inequality is true for all submartingales and the proof I posted shows that.

One might then wonder why they didn't simply use $E(|M_n|)$ instead on the right-hand side and make the theorem apply for all submartingales. I guess 'non-negative' and 'submartingale' just go together cause so many useful submartingales are constructed by squaring martingales, and they only needed the lemma for these kinds of cases.