## 统计代写|概率论代写Probability theory代考|Uniform Integrability and Optional Sampling

We extend the optional sampling theorem to unbounded stopping times. We will see that this is possible if the underlying martingale is uniformly integrable (compare Definition 6.16).

Lemma 10.20 Let $\left(X_n\right){n \in \mathbb{N}_0}$ be a uniformly integrable martingale. Then the family $\left(X\tau: \tau\right.$ is a finite stopping time $)$ is uniformly integrable.

Proof By Theorem 6.19, there exists a monotone increasing, convex function $f$ : $[0, \infty) \rightarrow[0, \infty)$ with $\liminf {x \rightarrow \infty} f(x) / x=\infty$ and $L:=\sup {n \in \mathbb{N}0} \mathbf{E}\left[f\left(\left|X_n\right|\right)\right]<$ $\infty$. If $\tau<\infty$ is a finite stopping time, then by the optional sampling theorem for bounded stopping times (Theorem $10.11$ with $\tau=n$ and $\sigma=\tau \wedge n$ ), $\mathbf{E}\left[X_n \mid \mathcal{F}{\tau \wedge n}\right]=$ $X_{\tau \wedge n}$. Since ${\tau \leq n} \in \mathcal{F}{\tau \wedge n}$, Jensen’s inequality yields \begin{aligned} \mathbf{E}\left[f\left(\left|X\tau\right|\right) \mathbb{1}{{\tau \leq n}}\right] &=\mathbf{E}\left[f\left(\left|X{\tau \wedge n}\right|\right) \mathbb{1}{{\tau \leq n}}\right] \ & \leq \mathbf{E}\left[\mathbf{E}\left[f\left(\left|X_n\right|\right) \mid \mathcal{F}{\tau \wedge n}\right] \mathbb{1}{{\tau \leq n}}\right] \ &=\mathbf{E}\left[f\left(\left|X_n\right|\right) \mathbb{1}{{\tau \leq n}}\right] \leq L . \end{aligned}
Hence $\mathbf{E}\left[f\left(\left|X_\tau\right|\right)\right] \leq L$. By Theorem 6.19, the family
$\left{X_\tau, \tau\right.$ is a finite stopping time $}$
is uniformly integrable.
Theorem 10.21 (Optional sampling and uniform integrahility) I.et ( $X_n, n \in$ $\mathbb{N}0$ ) be a uniformly integrable martingale (respectively supermartingale) and let $\sigma \leq \tau$ be finite stopping times. Then $\mathbf{E}\left[\left|X\tau\right|\right]<\infty$ and $X_\sigma=\mathbf{E}\left[X_\tau \mid \mathcal{F}\sigma\right]$ (respectively $X\sigma \geq \mathbf{E}\left[X_\tau \mid \mathcal{F}_\sigma\right]$ ).

Proof First let $X$ be a martingale. We have ${\sigma \leq n} \cap F \in \mathcal{F}{\sigma \wedge n}$ for all $F \in \mathcal{F}\sigma$. Hence, by the optional sampling theorem (Theorem 10.11),
$$\mathbf{E}\left[X_{\tau \wedge n} \mathbb{1}{{\sigma \leq n \backslash \cap F}\right]=\mathbf{E}\left[X{\sigma \wedge n} \mathbb{1}_{{\sigma \leq n} \cap F}\right] .$$

## 统计代写|概率论代写Probability theory代考|Doob’s Inequality

With Kolmogorov’s inequality (Theorem 5.28), we became acquainted with an inequality that bounds the probability of large values of the maximum of a square integrable process with independent centered increments. Here we want to improve this inequality in two directions. On the one hand, we replace the independent increments by the assumption that the process of partial sums is a martingale. On the other hand, we can manage with less than second moments; alternatively, we can get better bounds if we have higher moments.

Let $I \subset \mathbb{N}0$ and let $X=\left(X_n\right){n \in I}$ be a stochastic process. For $n \in \mathbb{N}$, we denote
$$X_n^=\sup \left{X_k: k \leq n\right} \quad \text { and } \quad|X|n^=\sup \left{\left|X_k\right|: k \leq n\right} .$$
Lemma 11.1 If $X$ is a submartingale, then, for all $\lambda>0$,
$$\lambda \mathbf{P}\left[X_n^* \geq \lambda\right] \leq \mathbf{E}\left[X_n \mathbb{1}{\left(X_n^* \geq \lambda\right}}\right] \leq \mathbf{E}\left[\left|X_n\right| \mathbb{1}_{\left{X_n^* \geq \lambda\right]}\right] .$$

Proof The second inequality is trivial. For the first one, let
$$\tau:=\inf \left{k \in I: X_k \geq \lambda\right} \wedge n .$$
By Theorem $10.11$ (optional sampling theorem),
\begin{aligned} \mathbf{E}\left[X_n\right] \geq \mathbf{E}\left[X_\tau\right] &=\mathbf{E}\left[X_\tau \mathbb{1}{\left(X_n^* \geq \lambda\right}}\right]+\mathbf{E}\left[X\tau \mathbb{1}{\left(X_n^<\lambda\right}}\right] \ & \geq \lambda \mathbf{P}\left[X_n^ \geq \lambda\right]+\mathbf{E}\left[X_n \mathbb{1}{\left(X_n^<\lambda\right}}\right] . \end{aligned} (Note that $\tau=n$ if $X_n^<\lambda$.) Now subtract $\mathbf{E}\left[X_n \mathbb{1}_{\left(X_n^<\lambda\right]}\right]$. Theorem 11.2 (Doob’s $L^p$-inequality) Let $X$ be a martingale or a positive submartingale. (i) For any $p \geq 1$ and $\lambda>0$, $$\lambda^p \mathbf{P}\left[|X|_n^ \geq \lambda\right] \leq \mathbf{E}\left[\left|X_n\right|^p\right] .$$ (ii) For any $p>1$,
$$\mathbf{E}\left[\left|X_n\right|^p\right] \leq \mathbf{E}\left[\left(|X|_n^*\right)^p\right] \leq\left(\frac{p}{p-1}\right)^p \mathbf{E}\left[\left|X_n\right|^p\right] .$$
Proof We follow the proof in [145]. As all the statements in (i) and (ii) are trivially true if $\mathbf{E}\left[|X|_n^p\right]=\infty$, we may and will assume that $\mathbf{E}\left[\left|X_n\right|^p\right]<\infty$.

# 概率论代考

## 统计代写|概率论代写概率论代考|均匀可积性和可选抽样

.

，因此$\mathbf{E}\left[f\left(\left|X_\tau\right|\right)\right] \leq L$。根据定理6.19，族
$\left{X_\tau, \tau\right.$是有限停止时间$}$

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