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
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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）
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 $\mathbb{P}(S \mid R)=0.036 / 0.132 \approx 0.2727$.
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.
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