Let $(\Omega,\mathcal A,\operatorname P)$ be a probability space, $\tau:\Omega\to\Omega$ be a measurable map on $(\Omega,\mathcal A)$ with $\operatorname P\circ\:\tau^{-1}=\operatorname P$, $Y\_n:\Omega\to[-\infty,\infty)$ be $\mathcal E$-measurable for $n\in\mathbb N$ with $\operatorname E\left[Y\_1^+\right]<\infty$ and $$Y\_{m+n}\le Y\_m+Y\_n\circ\tau^m\;\;\;\text{for all }m,n\in\mathbb N\tag1$$ and $$M\_n:=\max(Y\_1,\ldots,Y\_n)\;\;\;\text{for }n\in\mathbb N.$$



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> It's easy to show the following extension of the [maximal ergodic theorem](https://en.wikipedia.org/wiki/Maximal_ergodic_theorem): $$\operatorname E[Y\_1;M\_n\ge0]\ge0\;\;\;\text{for all }n\in\mathbb N.\tag2$$
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The ordinary maximal ergodic theorem is given by the special case, where $$Y\_n=\sum\_{i=0}^{n-1}X\circ\tau^i\;\;\;\text{for all }n\in\mathbb N$$ for some integrable real-valued random variable on $(\Omega,\mathcal A,\operatorname P)$. In that special case, it can be deduced from $(2)$ that $$\operatorname P\left[\sup\_{n\in\mathbb N}\left|\frac{Y\_n}n\right|\ge c\right]\le\frac1c\operatorname E[|Y\_1|]\;\;\;\text{for all }c>0\tag3.$$



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> Can we extend this result to the general case?
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A subadditive maximal ergodic theorem
**The answer is no in general, but yes if the sequence $Y\_n$ is non-negative.**


First, let us focus on the case where $Y\_n$ is non-positive. Then, $\sup \frac{1}{n}|Y\_n|=-\inf \frac{1}{n}Y\_n$. If you assume moreover that all the $Y\_n$ are $L^1$, then by Kingman's subadditive theorem, $\frac{1}{n}Y\_n$ converges to $Y=\inf \frac{1}{n}Y\_n$ almost surely. Notice then that $\mathbb{P}\left (\sup\frac{1}{n} |Y\_n|\geq c\right )=\mathbb{P}\left (Y\leq -c\right )$. Whenever $Y$ has positive probability of taking the value $-\infty$, you cannot bound $\mathbb{P}\left (Y\leq -c\right )$ by something converging to 0 as $c$ goes to infinity.


Here is a concrete counter-example. Let $Y\_n$ be the constant function $Y\_n=-n^2$. Then $Y\_n$ is subadditive and satisifes all your assumptions. You have $\sup \frac{1}{n}|Y\_n|=+\infty$ and so for any $c$, $\mathbb{P}\left (\sup\frac{1}{n} |Y\_n|\geq c\right )=1$, so you don't have $\mathbb{P}\left (\sup\frac{1}{n} |Y\_n|\geq c\right )\leq \frac{1}{c}\mathbb{E}(|Y\_1|)=\frac{1}{c}$.


However, small remark : the answer is yes for non-positive $Y\_n$ if you have the property that $\mathbb{E}(\frac{1}{n}|Y\_n|)\leq \mathbb{E}(|Y\_1|)$. Indeed, using Markov inequality, you get $\mathbb{P}\left (\frac{1}{n} |Y\_n|\geq c\right )\leq \frac{1}{c}\mathbb{E}(\frac{1}{n}|Y\_n|)\leq \frac{1}{c}\mathbb{E}(|Y\_1|)$ and this is true for all $n$, so this is true for the almost sure limit, using dominated convergence.




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About the non-negative case now. Inequality (3) is usually stated without the absolute values in literature :
$$\mathbb{P}\left (\sup\frac{1}{n} Y\_n\geq c\right )\leq \frac{1}{c}\mathbb{E}(|Y\_1|).$$


**This statement is true in general** and so in particular, the answer to your question is yes whenever $Y\_n$ is non-negative.


Indeed, consider a subbaditive sequence $Y\_n$, that is satisfying your condition $Y\_{n+m}\leq Y\_m+Y\_n\circ \tau^m$. Let $Z\_n=\sup\_{k=1,...,n}\frac{1}{k}Y\_k$. Also let $\widetilde{Y}\_n=\sum\_{j=0}^{n-1}Y\_1\circ \tau^j$ and finally, let $\widetilde{Z}\_n=\sup\_{k=1,...,n}\frac{1}{k}\widetilde{Y}\_k$. As you claim, the result is true for the sequence $\widetilde{Y}\_n$.


Note that since $Z\_n$ in non-decreasing, you have $\mathbb{P}\left (\sup\frac{1}{n} Y\_n\geq c\right )=\lim\_n\mathbb{P}(Z\_n\geq c)$ so we just need to prove that $\mathbb{P}(Z\_n\geq c)\leq \frac{1}{c}\mathbb{E}(Y\_1)$.


Now for fixed $n$, for every $x$, there exists $1\leq k(x)\leq n$ such that $Z\_n=\frac{1}{k(x)}Y\_{k(x)}$. Because of subadditivity, you have $Y\_{k(x)}\leq \sum\_{j=0}^{k(x)-1}Y\_1\circ \tau^j(x)=\widetilde{Y}\_{k(x)}(x)$.
So $\frac{1}{k(x)}Y\_{k(x)}\leq \frac{1}{k(x)}\widetilde{Y}\_{k(x)}(x)\leq \widetilde{Z}\_n(x)$. This proves that for any $x$, $Z\_n(x)\leq \widetilde{Z}\_n(x)$ so $\mathbb{P}(Z\_n\geq c)\leq \mathbb{P}(\widetilde{Z}\_n\geq c)$. Using that $\widetilde{Z}\_n$ also is non-decreasing, you get $\mathbb{P}(\widetilde{Z}\_n\geq c)\leq \mathbb{P}\left (\sup\frac{1}{n} \widetilde{Y}\_n\geq c\right )$ and so you can use the result for $\widetilde{Y}\_n$.


