MATH4320 Stochastic Process - 统计代写答疑辅导

分类： MATH4320 Stochastic Process

数学代写|MATH4320 Stochastic Process

Posted on 2023年4月3日2023年4月3日 by statistics-lab

Statistics-lab™可以为您提供uh.edu MATH4320 Stochastic Process随机过程课程的代写代考和辅导服务！

MATH4320 Stochastic Process课程简介

As an AI language model, I do not have access to specific course information or instructor contact details. However, I can provide you with some general advice on how to obtain the updated information for your course.

Firstly, check your school or university’s website for the latest course syllabus and textbook information. This information is usually available on the course page or through the online learning management system (LMS) used by your institution.

If the information is not available online, you can try contacting your course instructor or academic advisor directly via email or through the online messaging system provided by your institution. They should be able to provide you with the most up-to-date information regarding the course content, required textbooks, and any other relevant details.

Additionally, attending the first class meeting and orientation can also provide you with important updates on the course syllabus, textbooks, and other course-related information.

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）

问题 1.

Exercise 1. Imagine that you are playing poker (or any other game where only the player knows their own strength of position). Let’s say that you have a very strong hand that is almost, but not quite, unbeatable. In fact, you can compute that the probability that your opponent can beat you (i.e. the opponent has a “stronger hand”) is 4\%. Let’s say that you know the following: if your opponent has a stronger hand, they will raise your bet with probability $90 \%$; if they have a weak hand, they will raise your bet with probability $10 \%$. If your opponent raises your bet, what is the probability that they can beat you?

Solution: Let us write the events of strong hand by $S$, weak hand by $W$, and an opponent raising by $R$. We know from the problem that
$$
\mathbb{P}(S)=0.04, \quad \mathbb{P}(R \mid S)=0.9, \quad \mathbb{P}(R \mid W)=0.1,
$$
and we are asked to compute $\mathbb{P}(S \mid R)$. But this is
$$
\mathbb{P}(S \mid R)=\frac{\mathbb{P}(S \cap R)}{\mathbb{P}(R)} .
$$
We know $\mathbb{P}(S \cap R)=\mathbb{P}(R \cap S)=\mathbb{P}(R \mid S) \mathbb{P}(S)=0.9 \cdot 0.04=0.036$.
We have $\mathbb{P}(R)=\mathbb{P}(R \mid S) \mathbb{P}(S)+\mathbb{P}(R \mid W) \mathbb{P}(W)=0.9 \cdot 0.04+0.1 \cdot 0.96=0.132$.
Therefore $\mathbb{P}(S \mid R)=0.036 / 0.132 \approx 0.2727$.

问题 2.

Exercise 2. Imagine that we have a bag of 50 coins, 49 of which are normal, but one one is weird and has two sides that are both “heads”. We draw a coin at random from the bag and flip it six times. What is the probability that we picked the weird coin, given that we flipped six heads?

Solution: We denote $W$ as the event that we have picked the weind coin, and $H$ as the event of flipping six heads in a row. The question is asking us to compute $P(W \mid H)=\mathbb{P}(W \cap H) / \mathbb{P}(H)$. First we can see that $W \cap H=W$, since the weird coin is guaranteed to turn up six heads. Thus $\mathbb{P}(W \cap H)=1 / 50$.
Next, we see that
$$
\mathbb{P}(H)=\mathbb{P}(H \mid W) \mathbb{P}(W)+\mathbb{P}\left(H \mid W^c\right) \mathbb{P}\left(W^c\right)=1 \cdot \frac{1}{50}+\frac{1}{64} \cdot \frac{49}{50}=\frac{113}{3200} \approx 0.0353 .
$$
Therefore
$$
\mathbb{P}(W \mid H)=\frac{1 / 50}{113 / 3200}=\frac{64}{113} \approx 56.6 \%
$$
This is probably higher than we might have expected intuitively. Given the observation of six heads, we’re more likely than not to have gotten a biased coin, even though the biased coin is so rare.

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.

此图像的alt属性为空；文件名为%E7%B2%89%E7%AC%94%E5%AD%97%E6%B5%B7%E6%8A%A5-1024x575-10.png — MATH4320 Stochastic Process

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数学代写|MATH4320 Stochastic Process

Posted on 2023年3月31日2023年3月31日 by statistics-lab

Statistics-lab™可以为您提供uh.edu MATH4320 Stochastic Process随机过程课程的代写代考和辅导服务！

MATH4320 Stochastic Process课程简介

Additionally, attending the first class meeting and orientation can also provide you with important updates on the course syllabus, textbooks, and other course-related information.

PREREQUISITES

MATH4320 Stochastic Process HELP（EXAM HELP， ONLINE TUTOR）

问题 1.

Let $\mathcal{F}$ be a $\sigma$-field on some set $\Omega$.
(1) Show that if $A_1, A_2, \ldots$ are in $\mathcal{F}$, then so is $\cap_{k=1}^{\infty} A_k$.
(2) Show that if $A_1, A_2$ are in $\mathcal{F}$, then so is their symmetric difference $A_1 \triangle A_2:=$ $A_1 \cup A_2-A_1 \cap A_2$.

(1) We prove this by induction. First, note that $A_1\in \mathcal{F}$ by assumption. Now suppose that $\bigcap_{k=1}^n A_k\in\mathcal{F}$ for some $n\geq 1$. Then $A_{n+1}\in\mathcal{F}$ by assumption, so we have $\bigcap_{k=1}^{n+1} A_k = \left(\bigcap_{k=1}^{n} A_k\right) \cap A_{n+1}\in\mathcal{F}$, since $\mathcal{F}$ is a $\sigma$-field and therefore closed under countable intersections.

(2) We have $A_1, A_2\in\mathcal{F}$ by assumption. Since $\mathcal{F}$ is a $\sigma$-field, it is closed under set complements, unions, and intersections. Therefore, we have $A_1^c, A_2^c\in\mathcal{F}$, and hence \begin{align*} A_1\triangle A_2 &= (A_1\cup A_2) \setminus (A_1\cap A_2)\ &= (A_1\cup A_2) \cap (A_1^c \cup A_2^c)\ &= (A_1\cap A_1^c) \cup (A_1\cap A_2^c) \cup (A_2\cap A_1^c) \cup (A_2\cap A_2^c)\ &= (A_1\setminus A_2) \cup (A_2\setminus A_1)\ &= (A_1\cap A_2^c)\cup (A_1^c\cap A_2). \end{align*} Since $\mathcal{F}$ is closed under intersections and unions, we have $A_1\cap A_2^c\in\mathcal{F}$ and $A_1^c\cap A_2\in\mathcal{F}$, and hence $A_1\triangle A_2\in\mathcal{F}$ as desired.

问题 2.

Let $f: U \rightarrow E$ be a function, where $U$ and $E$ are arbitrary sets. For any subset $A \subseteq E$, define
$$
f^{-1}(A)={u \in U ; f(u) \in A}
$$
(i) Show that for all $u \in U$,
$$
1_A(f(u))=1_{f^{-1}(A)}(u)
$$
(ii) Prove that if $\mathcal{E}$ is a $\sigma$-field on $E$, then the collection of subsets of $U$
$$
f^{-1}(\mathcal{E}):=\left{f^{-1}(A) ; A \in \mathcal{E}\right}
$$
is a $\sigma$-field on $U$.

(i) We have $1_A(f(u)) = 1$ if $f(u) \in A$ and $0$ otherwise. On the other hand, $1_{f^{-1}(A)}(u) = 1$ if $u \in f^{-1}(A)$ and $0$ otherwise. But $u \in f^{-1}(A)$ if and only if $f(u) \in A$. Therefore, we have $1_A(f(u)) = 1_{f^{-1}(A)}(u)$ for all $u \in U$.

(ii) We need to show that $f^{-1}(\mathcal{E})$ is a $\sigma$-field on $U$, i.e., it satisfies the following three conditions:

(a) $U \in f^{-1}(\mathcal{E})$; (b) If $A \in f^{-1}(\mathcal{E})$, then $A^c \in f^{-1}(\mathcal{E})$; (c) If $A_1, A_2, \ldots$ are in $f^{-1}(\mathcal{E})$, then $\bigcup_{k=1}^{\infty} A_k \in f^{-1}(\mathcal{E})$.

(a) Since $\mathcal{E}$ is a $\sigma$-field on $E$, we have $E \in \mathcal{E}$, and hence $U = f^{-1}(E) \in f^{-1}(\mathcal{E})$.

(b) Suppose $A \in f^{-1}(\mathcal{E})$. Then $A = f^{-1}(B)$ for some $B \in \mathcal{E}$. Since $\mathcal{E}$ is a $\sigma$-field, we have $B^c \in \mathcal{E}$. Therefore,

A^c = f^{-1}(B^c) = \left(f^{-1}(B)\right)^c \in f^{-1}(\mathcal{E}).Ac=f−1(Bc)=(f−1(B))c∈f−1(E).

(c) Suppose $A_1, A_2, \ldots$ are in $f^{-1}(\mathcal{E})$. Then $A_k = f^{-1}(B_k)$ for some $B_k \in \mathcal{E}$ for each $k$. Since $\mathcal{E}$ is a $\sigma$-field, we have $\bigcup_{k=1}^{\infty} B_k \in \mathcal{E}$. Therefore,

\bigcup_{k=1}^{\infty} A_k = f^{-1}\left(\bigcup_{k=1}^{\infty} B_k\right) \in f^{-1}(\mathcal{E}).k=1⋃∞Ak=f−1(k=1⋃∞Bk)∈f−1(E).

This completes the proof that $f^{-1}(\mathcal{E})$ is a $\sigma$-field on $U$.

Textbooks

数学代写|MATH4320 Stochastic Process

Posted on 2023年3月6日2023年3月6日 by statistics-lab

Statistics-lab™可以为您提供uh.edu MATH4320 Stochastic Process随机过程课程的代写代考和辅导服务！

MATH4320 Stochastic Process课程简介

Additionally, attending the first class meeting and orientation can also provide you with important updates on the course syllabus, textbooks, and other course-related information.

PREREQUISITES

MATH4320 Stochastic Process HELP（EXAM HELP， ONLINE TUTOR）

问题 1.

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:

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

So the answer is 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.

问题 2.

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