数学代写|随机分析作业代写stochastic analysis代考|Definition of Martingale

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数学代写|随机分析作业代写stochastic analysis代考|Definition of Martingale

数学代写|随机分析作业代写stochastic analysis代考|probability space

Let $\mathbf{T}$ be an ordered set. We only consider the case that $\mathbf{T}$ is a subset of $[-\infty, \infty]$ in this book. In almost all cases, $\mathbf{T}$ is ${0,1,2, \ldots, m}, m \geqq 1$, or $\mathbf{Z}{\geq 0}={0,1,2, \ldots}$, or $[0, T], T>0$, or $[0, \infty)$, although we consider the case that $\mathbf{T}$ is $\mathbf{Z}{\leqq 0}={n \in$ $\mathbf{Z} ; n \leqq 0}$.
Let $(\Omega, \mathcal{F}, P)$ be a probability space.
Definition 2.1.1 We say that $\left{\mathcal{F}{t}\right}{t \in \mathbf{T}}$ is a filtration, if the following two conditions are satisfied.
(1) $\mathcal{F}{t}, t \in \mathbf{T}$ are sub- $\sigma$-algebras. (2) $\mathcal{F}{s} \subset \mathcal{F}{t}$ for any $s, t \in \mathbf{T}$ with $s{t}\right}_{t \in \mathbf{T}}$ be a filtration. We say that $X=\left{X_{t}\right}_{t \in \mathbf{T}}$ is an $\left{\mathcal{F}{t}\right}{t \in \mathbf{T}^{-}}$ martingale, if the following two conditions are satisfied.
(1) $X_{t}$ is an $\mathcal{F}{t}$-measurable integrable random variable for each $t \in \mathbf{T}$. (2) If $s, t \in \mathbf{T}, s{t} \mid \mathcal{F}{s}\right]=X{s}
$$
Definition 2.1.3 Let $\left{\mathcal{F}{t}\right}{t \in \mathbf{T}}$ be a filtration. We say that $X=\left{X_{t}\right}_{t \in \mathbf{T}}$ is an $\left{\mathcal{F}{t}\right}{t \in \mathbf{T}^{-}}$ submartingale (supermartingale), if the following two conditions are satisfied.
(1) $X_{t}$ is an $\mathcal{F}{t}$-measurable integrable random variable for each $t \in \mathbf{T}$. (2) If $s, t \in \mathbf{T}, s{t} \mid \mathcal{F}{s}\right] \geqq(\leqq) X{s}
$$
In the case that it is obvious what filtration $\left{\mathcal{F}{t}\right}{t \in \mathbf{T}}$ we handle, we call $\left{\mathcal{F}{t}\right}{t \in \mathbf{T}^{-}}$ martingales (sub-, supermartingales) martingales (sub-, supermartingales) for simplicity.
The following is obvious.

数学代写|随机分析作业代写stochastic analysis代考|martingale

Proposition 2.1.1 (1) If $X=\left{X_{t}\right}_{t \in \mathbf{T}}$ and $Y=\left{Y_{t}\right}_{t \in \mathbf{T}}$ are martingales, and if $a, b \in \mathbf{R}$, then $a X+b Y=\left{a X_{t}+b Y_{t}\right}_{t \in \mathbf{T}}$ is a martingale.

(2) If $X=\left{X_{t}\right}_{t \in \mathbf{T}}$ and $Y=\left{Y_{t}\right}_{t \in \mathbf{T}}$ are submartingales (supermartingales), and if $a, b \geqq 0$, then $a X+b Y=\left{a X_{t}+b Y_{t}\right}_{t \in \mathbf{T}}$ is also a submartingale (supermartingale).
Also, we have the following.
Proposition 2.1.2 (1) Let $X=\left{X_{t}\right}_{t \in \mathbf{T}}$ be a martingale, and $\varphi: \mathbf{R} \rightarrow \mathbf{R}$ be a convex function. Suppose moreover that $E\left[\left|\varphi\left(X_{t}\right)\right|\right]<\infty$ for all $t \in \mathbf{T}$. Then $\left{\varphi\left(X_{t}\right)\right}_{t \in \mathbf{T}}$ is a submartingale.
(2) Let $X=\left{X_{t}\right}_{t \in \mathbf{T}}$ be a submartingale, and $\varphi: \mathbf{R} \rightarrow \mathbf{R}$ be a non-decreasing convex function. Suppose moreover that $E\left[\left|\varphi\left(X_{t}\right)\right|\right]<\infty$ for all $t \in \mathbf{T}$. Then $\left{\varphi\left(X_{t}\right)\right}_{t \in \mathbf{T}}$ is a submartingale.

Proof In both cases, by Jensen’s inequality we see that for any $t, s \in \mathbf{T}$ with $s<t$
$$
E\left[\varphi\left(X_{t}\right) \mid \mathcal{F}{s}\right] \geqq \varphi\left(E\left[X{t} \mid \mathcal{F}{s}\right]\right) \geqq \varphi\left(X{s}\right)
$$
Therefore we have our assertions.

数学代写|随机分析作业代写stochastic analysis代考|Definition of Martingale

随机分析代写

数学代写|随机分析作业代写STOCHASTIC ANALYSIS代考|PROBABILITY SPACE

让吨是一个有序集。我们只考虑这样的情况吨是的一个子集[−∞,∞]在这本书中。在几乎所有情况下,$\mathbf{T}$ is a subset of $[-\infty, \infty]$ in this book. In almost all cases, $\mathbf{T}$ is ${0,1,2, \ldots, m}, m \geqq 1$, or $\mathbf{Z}{\geq 0}={0,1,2, \ldots}$, or $[0, T], T>0$, or $[0, \infty)$, although we consider the case that $\mathbf{T}$ is $\mathbf{Z}{\leqq 0}={n \in$ $\mathbf{Z} ; n \leqq 0}$.
Let $(\Omega, \mathcal{F}, P)$ be a probability space.
Definition 2.1.1 We say that $\left{\mathcal{F}{t}\right}{t \in \mathbf{T}}$ is a filtration, if the following two conditions are satisfied.
(1) $\mathcal{F}{t}, t \in \mathbf{T}$ are sub- $\sigma$-algebras. (2) $\mathcal{F}{s} \subset \mathcal{F}{t}$ for any $s, t \in \mathbf{T}$ with $s{t}\right}_{t \in \mathbf{T}}$ be a filtration. We say that $X=\left{X_{t}\right}_{t \in \mathbf{T}}$ is an $\left{\mathcal{F}{t}\right}{t \in \mathbf{T}^{-}}$ martingale, if the following two conditions are satisfied.
(1) $X_{t}$ is an $\mathcal{F}{t}$-measurable integrable random variable for each $t \in \mathbf{T}$. (2) If $s, t \in \mathbf{T}, s{t} \mid \mathcal{F}{s}\right]=X{s}
$$
Definition 2.1.3 Let $\left{\mathcal{F}{t}\right}{t \in \mathbf{T}}$ be a filtration. We say that $X=\left{X_{t}\right}_{t \in \mathbf{T}}$ is an $\left{\mathcal{F}{t}\right}{t \in \mathbf{T}^{-}}$ submartingale (supermartingale), if the following two conditions are satisfied.
(1) $X_{t}$ is an $\mathcal{F}{t}$-measurable integrable random variable for each $t \in \mathbf{T}$. (2) If $s, t \in \mathbf{T}, s{t} \mid \mathcal{F}{s}\right] \geqq(\leqq) X{s}
$$
In the case that it is obvious what filtration $\left{\mathcal{F}{t}\right}{t \in \mathbf{T}}$ we handle, we call $\left{\mathcal{F}{t}\right}{t \in \mathbf{T}^{-}}$ martingales (sub-, supermartingales) martingales (sub-, supermartingales) 为简单起见。
以下是显而易见的。

数学代写|随机分析作业代写STOCHASTIC ANALYSIS代考|MARTINGALE

命题 2.1.11如果$X=\left{X_{t}\right}_{t \in \mathbf{T}}$ and $Y=\left{Y_{t}\right}_{t \in \mathbf{T}}$ are martingales, and if $a, b \in \mathbf{R}$, then $a X+b Y=\left{a X_{t}+b Y_{t}\right}_{t \in \mathbf{T}}$ is a martingale.。

2如果$X=\left{X_{t}\right}_{t \in \mathbf{T}}$ and $Y=\left{Y_{t}\right}_{t \in \mathbf{T}}$ are submartingales (supermartingales), and if $a, b \geqq 0$, then $a X+b Y=\left{a X_{t}+b Y_{t}\right}_{t \in \mathbf{T}}$ is also a submartingale (supermartingale).
Also, we have the following.
此外,我们还有以下内容。
命题 2.1.21(1) Let $X=\left{X_{t}\right}_{t \in \mathbf{T}}$ be a martingale, and $\varphi: \mathbf{R} \rightarrow \mathbf{R}$ be a convex function. Suppose moreover that $E\left[\left|\varphi\left(X_{t}\right)\right|\right]<\infty$ for all $t \in \mathbf{T}$. Then $\left{\varphi\left(X_{t}\right)\right}_{t \in \mathbf{T}}$ is a submartingale.
(2) Let $X=\left{X_{t}\right}_{t \in \mathbf{T}}$ be a submartingale, and $\varphi: \mathbf{R} \rightarrow \mathbf{R}$ be a non-decreasing convex function. Suppose moreover that $E\left[\left|\varphi\left(X_{t}\right)\right|\right]<\infty$ for all $t \in \mathbf{T}$. Then $\left{\varphi\left(X_{t}\right)\right}_{t \in \mathbf{T}}$ is a submartingale.

证明 在这两种情况下,根据 Jensen 不等式,我们看到对于任何$t, s \in \mathbf{T}$ with $s<t$
$$
E\left[\varphi\left(X_{t}\right) \mid \mathcal{F}{s}\right] \geqq \varphi\left(E\left[X{t} \mid \mathcal{F}{s}\right]\right) \geqq \varphi\left(X{s}\right)
$$
因此我们有我们的断言。

数学代写|随机分析作业代写stochastic analysis代考

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