## MATH4320 Stochastic Process课程简介

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## PREREQUISITES

Course Content: Dynamical processes throughout science and economics are often influenced by random fluctuations. Mathematically, a dynamical model that explicitly includes random fluctuation is a stochastic process. Math 4320 will introduce you to both the theory and the applications of stochastic processes. We will first review probability theory before examining new material. In particular, we will discuss background in probability theory with emphasis on conditional expectations and conditional distributions. Then we will cover more advanced topics such as discrete-time Markov chains, Poisson process, continuous-time Markov chains. Grading \& Make-up Policy/Assignment \& Exam Details: Please consult your instructor’s syllabus regarding any and all grading/assignment guidelines.

## MATH4320 Stochastic Process HELP（EXAM HELP， ONLINE TUTOR）

Problem 1. Calculate each of the following probabilities. (Give your answers as rational numbers or decimals to at least 3 places. You might find a table of the standard normal cumulative distribution useful for this.)
(A) $P{W(t)=0$ for some $2 \leq t \leq 3}$.
(B) $P{W(2)>W(1)>W(3)}$.

Okay, here are the calculations for Problem 1:

(A) $P{W(t)=0$ for some $2 \leq t \leq 3}$.

Let $W(t)$ be a standard Brownian motion. The probability that $W(t) = 0$ for some $2 \leq t \leq 3$ is the same as the probability that $W(2) = 0$ or $W(3) = 0$, since $W(t)$ is continuous and has independent increments.

Using the standard normal cumulative distribution table or a calculator, we can find that $P{W(2) = 0} = P{W(3) = 0} = 0.5$. Therefore, the probability that $W(t) = 0$ for some $2 \leq t \leq 3$ is:

$$P{W(t)=0 \text{ for some }2 \leq t \leq 3} = P{W(2) = 0 \text{ or } W(3) = 0} = 0.5 + 0.5 – 0.5 \times 0.5 = 0.75$$

(B) $P{W(2)>W(1)>W(3)}$.

Since $W(t)$ is a continuous process, the event $W(2) > W(1) > W(3)$ is the same as the event $W(2) – W(3) > W(1) – W(3) > 0$. Since $W(2) – W(3)$ and $W(1) – W(3)$ are independent standard normal random variables, we can use the standard normal cumulative distribution table to calculate the probabilities.

Let $Z_1 = W(2) – W(3)$ and $Z_2 = W(1) – W(3)$. Then we want to calculate $P{Z_1 > Z_2 > 0}$.

From the properties of the normal distribution, we know that $Z_1$ and $Z_2$ are also standard normal random variables. We can use the fact that the joint distribution of two independent standard normal random variables $(Z_1, Z_2)$ is bivariate normal with mean vector $\boldsymbol{\mu} = (0, 0)$ and covariance matrix $\boldsymbol{\Sigma} = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix}$.

Therefore, we can convert the event $Z_1 > Z_2 > 0$ to a region in the $(Z_1, Z_2)$ plane and use the bivariate normal distribution to calculate the probability. The region is a triangle with vertices $(0, 0)$, $(+\infty, 0)$, and $(+\infty, +\infty)$. The probability of this region can be calculated using a bivariate normal table or a calculator.

Using a standard normal table or calculator, we find that $P{Z_1 > Z_2 > 0} \approx 0.125$. Therefore, the probability that $W(2) > W(1) > W(3)$ is:

$$P{W(2)>W(1)>W(3)} \approx P{Z_1 > Z_2 > 0} \approx 0.125$$

So the answer is approximately 0.125.

Problem 2. Show that for any $t>0$ the random variable $M(t)-W(t)$ has the same distribution as the random variable $|W(t)|$. HINT: Begin by calculating
$$P{M(t) \geq a \text { and } W(t) \leq a-b}$$
for $a, b>0$. You may find the reflection principle helpful.

Let $M(t)$ and $W(t)$ be a standard Brownian motion and its maximum process, respectively. We want to show that for any $t > 0$, the random variable $M(t) – W(t)$ has the same distribution as $|W(t)|$.

To begin, let $a, b > 0$. Then we have:

\begin{align*} P{M(t) \geq a \text{ and } W(t) \leq a-b} &= P{M(t) – W(t) \geq a – (a-b)} \ &= P{M(t) – W(t) \geq b} \ &= P{W(t) \leq -b} + P{W(t) \geq b} \ &= 2P{W(t) \geq b} \quad (\text{by the reflection principle}) \ &= 2[1 – P{W(t) \leq b}] \ &= 2[1 – \Phi(b)] \quad (\text{where } \Phi \text{ is the standard normal CDF}) \ \end{align*}

Next, we want to show that $M(t) – W(t)$ has the same distribution as $|W(t)|$. To do this, we will show that the CDFs of these two random variables are equal.

Let $F_{M-W}(x)$ and $F_{|W|}(x)$ denote the CDFs of $M(t) – W(t)$ and $|W(t)|$, respectively. Then we have:

\begin{align*} F_{M-W}(x) &= P{M(t) – W(t) \leq x} \ &= P{M(t) \leq W(t) + x} \ &= P{M(t) \geq -x \text{ and } W(t) \leq M(t) – x} \ &= \int_{0}^{\infty} P{M(t) \geq -x \text{ and } W(t) \leq a – x} f_{W(t)}(a) da \ &= \int_{0}^{\infty} P{M(t) \geq -x \text{ and } W(t) \leq a – x} \frac{1}{\sqrt{2 \pi t}} e^{-a^2/(2t)} da \ &= \int_{0}^{x} P{M(t) \geq -x \text{ and } W(t) \leq a – x} \frac{1}{\sqrt{2 \pi t}} e^{-a^2/(2t)} da \ &\quad + \int_{x}^{\infty} P{M(t) \geq -x \text{ and } W(t) \leq a – x} \frac{1}{\sqrt{2 \pi t}} e^{-a^2/(2t)} da \ &= \int_{0}^{x} 2[1 – \Phi(x-a)] \frac{1}{\sqrt{2 \pi t}} e^{-a^2/(2t)} da \ &\quad + \int_{x}^{\infty} 2\Phi(a-x) \frac{1}{\sqrt{2 \pi t}} e^{-a^2/(2t)} da \ &= 2\int

## Textbooks

• An Introduction to Stochastic Modeling, Fourth Edition by Pinsky and Karlin (freely
available through the university library here)
• Essentials of Stochastic Processes, Third Edition by Durrett (freely available through
the university library here)
To reiterate, the textbooks are freely available through the university library. Note that
you must be connected to the university Wi-Fi or VPN to access the ebooks from the library
links. Furthermore, the library links take some time to populate, so do not be alarmed if
the webpage looks bare for a few seconds.

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