这是一份nottingham诺丁汉大学MATH4061作业代写的成功案例

Let $D_{n}:=E\left(X_{n} \mid \mathcal{B}{n-1}\right)-X{n-1}$ for $n=-1,-2, \ldots$. By assumption, $D_{n} \geq 0$ a.s. Now $\sum_{n \leq-1} E D_{n}=\sum_{n \leq-1} E\left(X_{n}-X_{n-1}\right)=E X_{-1}-K$; note that $E X_{n} \downarrow K$ as $n \rightarrow-\infty$. Since the sum is finite, the series $\sum D_{n}$ converges by monotone convergence to an integrable function.

Let $Z_{n}:=\sum_{k \leq n} D_{k}$ for each $n$. Then $Z_{n}$ is nondecreasing as $n$ increases almost surely, and $Z_{n} \downarrow 0$ a.s. as $n \rightarrow-\infty$. Let $Y_{n}:=X_{n}-Z_{n}$. To show that $Y_{n}$ is a (reversed) martingale, we have
$$
\begin{aligned}
E\left(Y_{n} \mid \mathcal{B}{n-1}\right) &=E\left(X{n}-Z_{n} \mid \mathcal{B}{n-1}\right)=D{n}+X_{n-1}-\sum_{k \leq n} D_{k} \
&=X_{n-1}-\sum_{k<n} D_{k}=X_{n-1}-Z_{n-1}=Y_{n-1}
\end{aligned}
$$
proving the decomposition as desired. Now $Z_{n}$ converges in $\mathcal{L}^{1}$ by monotone or dominated convergence, and $Y_{n}$ converges a.s. and in $\mathcal{L}^{1}$ by Theorem 10.6.1, so $X_{n}$ converges likewise.

MATH4061 COURSE NOTES ：

If $M, P$, and $Q$ are laws on $S, \rho(M, P)<x$ and $\rho(P, Q)<y$, then for any Borel set $A$,
$$
M(A) \leq P\left(A^{x}\right)+x \leq Q\left(\left(A^{x}\right)^{y}\right)+y+x \leq Q\left(A^{x+y}\right)+x+y,
$$
so $\rho(M, Q) \leq x+y$. Letting $x \downarrow \rho(M, P)$ and $y \downarrow \rho(P, Q)$ gives $\rho(M, Q) \leq$ $\rho(M, P)+\rho(P, Q)$.

The metric $\rho$ is called the Prohorov metric, or sometimes the LévyProhorov metric. Now for any laws $P$ and $Q$ on $S$, let $\int f d(P-Q):=$ $\int f d P-\int f d Q$ and