Q. Let $\left(X_{n}\right)_{n}$ be a martingale such that there exists $K$ which satisfies $$ \mathbb{P}\left(X_{n} \leq K\right)=1 $$ Define the process $M_{n}=K-X_{n}$, for $n \in \mathbb{N}$. Prove that $\left(M_{n}\right)_{n}$ converges almost surely.
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Here is a rough idea of what I have in mind so please let me know if I'm on the right track. First I start off by proving that Mn is a Martingale using the two Martingale properties $E[M_n] < \infty$ and $E[M_{n+1}/F_n]=M_n$.
After doing this, I'm quite unsure how to proceed but have a fair idea that I'm supposed to use the M.C.T but again, unsure how to use it so how do I proceed?
Thanks!
"Can we infer that (it $X$ is a submartingale) somehow?"
For simplicity, let the sequence $\{X_n\}$ be such that $\mathbb P[X_n\mid F_{n-1}]$ are i.i.d, and WLOG, let the sequence start at $X_0 = 0$ (as otherwise we can translate the system)
$\mathbb{P}\left(X_{n} \leq K\right)=1$ then $\mathbb{P}\left(X_{n} > K\right)=0$. In other words, given any $\varepsilon >0$, we can bound the probability of $\mathbb{P}\left(X_{n} > K\right)<\varepsilon$. Note that we will cross K in n steps, if at each step we go more than $\frac{K}{n}$ \begin{align*} \mathbb{P}\left(X_{n} > K\right) &\geq \mathbb{P}\left[\left(X_1-X_0 > \frac{K}{n}\right) \wedge\left(X_2-X_1 > \frac{K}{n}\right)\wedge\dots\wedge\left(X_n-X_{n-1} > \frac{K}{n}\right)\right] \end{align*} Now we use the independence property: \begin{align*} \mathbb{P}\left(X_{n} > K\right) &\geq \mathbb{P}\left[X_1-X_0 > \frac{K}{n}\right] \times\mathbb{P}\left[X_2-X_1 > \frac{K}{n} \right]\times\dots\times\mathbb{P}\left[X_n-X_{n-1} > \frac{K}{n}\right] \end{align*} By the identically distributed property: \begin{align*} \left(\mathbb{P}\left[X_{i+1}-X_i > \frac{K}{n}\right]\right)^n &\leq\mathbb{P}\left(X_{n} > K\right) <\varepsilon\\ \mathbb{P}\left[X_{i+1}-X_i > \frac{K}{n}\right] &<\varepsilon^{\frac{1}{n}}\\ \mathbb{P}\left[X_{i+1}> X_i + \frac{K}{n}\right] &<\varepsilon^{\frac{1}{n}}\\ \end{align*} Crucially, this is true for all n. By choosing an appropriate $\varepsilon$, we can set $\mathbb{P}\left[X_{i+1}> X_i \right] =0$ Now we calculate the expectation. Let $Y=X_{i+1}-X_i$: \begin{align*} \mathbb E [X_{i+1} | X_i] &= \int\limits_{y=-\infty}^\infty (X_i + y) f_Y [X_{i+1}-X_i=y ] dy\\ &=X_i + \int\limits_{y=-\infty}^\infty y\ f_Y [X_{i+1}-X_i=y ] dy\\ &=X_i + \int\limits_{y=-\infty}^0 y\ f_Y [X_{i+1}-X_i=y ]dy+\int\limits_{y=0}^\infty y\ f_Y [X_{i+1}-X_i=y ]dy \\ &= X_i + \int\limits_{y=-\infty}^0 y\ f_Y [X_{i+1}-X_i=y ]dy+\int\limits_{y=0}^{K/n} y\ f_Y [X_{i+1}-X_i=y ]dy+\\&\qquad\qquad\int\limits_{y=K/n}^K y\ f_Y [X_{i+1}-X_i=y ]dy \\ &\leq X_i + \int\limits_{y=-\infty}^0 y\ f_Y [X_{i+1}-X_i=y ]dy+\frac{K}{n}+ K \varepsilon^{\frac{1}{n}} \\ &\leq X_i +\frac{K}{n} + K \varepsilon^{\frac{1}{n}} \\ \end{align*} which holds for all $n,\varepsilon$. Therefore, $\mathbb E [X_{i+1} | X_i]\leq X_i$
Now that we have $X_n$ is a submartingale, $M_n = K-X_n$ is a supermartingale, which as per Doob's criterion (and the comment) converges surely