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数学代写|统计计算作业代写Statistical Computing代考|Brownian motion

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数学代写|统计计算作业代写Statistical Computing代考|Properties

In this section we briefly state some of the most important properties of Brownian motion. Using definition $6.1$, we can immediately derive the following results:

• We have $B_{t}=B_{t}-B_{0} \sim \mathcal{N}\left(0, t I_{d}\right)$ and in particular the expectation of $B_{t}$ is $\mathbb{E}\left(B_{t}\right)=0 .$
• For the one-dimensional case we find $B_{t} \sim \mathcal{N}(0, t)$ and thus the standard deviation of $B_{t}$ is $\sqrt{t}$.
• Similarly, for any dimension, we find that $B_{t}$ has the same distribution as $\sqrt{t} B_{1}$. Thus, the magnitude of $\left|B_{t}\right|$ only grows like $\sqrt{t}$ as $t$ increases.

Another basic result, relating the one-dimensional case to the $d$-dimensional case, is given in the following lemma.

数学代写|统计计算作业代写STATISTICAL COMPUTING代考|Direct simulation

As we have seen at the start of this chapter, we cannot hope to simulate all of the infinitely many values $\left(B_{t}\right){t \geq 0}$ on a computer simultaneously. Instead, we restrict ourselves to simulating the values of $B$ for times $0=t{0}<t_{1}<\cdots<t_{n}$. For a Brownian motion, this can be easily done by using the first two conditions from definition $6.1$ : we have $B_{0}=0$ and $B_{t_{i}}$ can be computed from $B_{t_{i-1}}$ by adding an $\mathcal{N}\left(0, t_{i}-t_{i-1}\right)$ distributed random value, independent of all values computed so far. This method is described in the following algorithm.

input:
sample times $0=t_{0}<t_{1}<\cdots<t_{n}$
randomness used:
an i.i.d. sequence $\left(\varepsilon_{i}\right){i=1,2, \ldots, n}$ with distribution $\mathcal{N}(0,1)$ output: a sample of $B{t_{0}}, \ldots, B_{t_{n}}$, that is a discretised path of a Brownian motion
1: $B_{0} \leftarrow 0$
2: for $i=1,2, \ldots, n$ do
3: $\quad$ generate $\varepsilon_{i} \sim \mathcal{N}(0,1)$
4: $\quad \Delta B_{i} \leftarrow \sqrt{t_{i}-t_{i-1}} \varepsilon_{i}$
5: $\quad B_{t_{i}} \leftarrow B_{t_{i-1}}+\Delta B_{i}$
$6:$ end for
7: return $\left(B_{t_{i}}\right)_{i=0,1, \ldots, n}$

数学代写|统计计算作业代写STATISTICAL COMPUTING代考|Interpolation and Brownian bridges

If we have already simulated values of a Brownian motion $B$ for a set of times, it is possible to refine the simulated path afterwards by simulating values of $B$ for additional times. This method is called interpolation of the Brownian path. Since these additional simulations need to be compatible with the already sampled values, some care is needed when implementing this method.

Assume that we know the values of $B$ at times $0=t_{0}<t_{1}<\cdots<t_{n}$ and that we want to simulate an additional value for $B$ at time $s$ with $t_{i-1}<s<t_{i}$ for some $i \in{2,3, \ldots, n}$. As an abbreviation we write $r=t_{i-1}$ and $t=t_{i}$. By the Markov property of Brownian motion (lemma 6.5, part (b)), the increment $B_{s}-B_{r}$ is independent of $\left(B_{u}\right){0 \leq u \leq r}$. Similarly, $\left(B{u}-B_{t}\right){u \geq t}$ is independent of $\left(B{u}\right){0 \leq u \leq t}$ and thus of $B{s}-B_{r}$. Consequently, we only need to take the value $B_{t}=B_{t_{i}}$ into account when sampling the increment $B_{s}-B_{r}$; by independence the remaining $B_{t_{j}}$ with $j \neq i$ do not affect the distribution of $B_{s}-B_{r}$.

数学代写|统计计算作业代写STATISTICAL COMPUTING代考|PROPERTIES

• 我们有乙吨=乙吨−乙0∼ñ(0,吨一世d)尤其是期望乙吨是和(乙吨)=0.
• 对于一维情况，我们发现乙吨∼ñ(0,吨)因此标准差乙吨是吨.
• 同样，对于任何维度，我们发现乙吨具有相同的分布吨乙1. 因此，幅度|乙吨|只会长得像吨作为吨增加。

数学代写|统计计算作业代写STATISTICAL COMPUTING代考|DIRECT SIMULATION

input:
sample times $0=t_{0}<t_{1}<\cdots<t_{n}$ randomness used:
an i.i.d. sequence $\left(\varepsilon_{i}\right){i=1,2, \ldots, n}$ with distribution $\mathcal{N}(0,1)$ output: a sample of $B{t_{0}}, \ldots, B_{t_{n}}$, that is a discretised path of a Brownian motion
1: $B_{0} \leftarrow 0$
2: for $i=1,2, \ldots, n$ do
3: generate $\varepsilon_{i} \sim \mathcal{N}(0,1)$
$\begin{array}{ll}\text { 4: } & \Delta B_{i} \leftarrow \sqrt{t_{i}-t_{i-1}} \varepsilon_{i} \ 5: & B_{t_{i}} \leftarrow B_{t_{i-1}}+\Delta B_{i}\end{array}$
6: end for
7: return $\left(B_{t_{i}}\right)_{i=0,1, \ldots, n}$

数学代写|统计计算作业代写STATISTICAL COMPUTING代考|INTERPOLATION AND BROWNIAN BRIDGES

Assume that $B$ is one-dimensional with $B_{r}=a$. Since $\left(B_{u+r}-B_{r}\right){u \geq 0}$ is a Brownian motion independent of $B{r}$, we have $B_{s} \sim \mathcal{N}(a, s-r)$ and $B_{t}-B_{s} \sim$ $\mathcal{N}(0, t-s)$, independently of each other. Thus, the joint density of $\left(B_{s}, B_{t}\right)$ is