**MY-ASSIGNMENTEXPERT™**可以为您提供lse.ac.uk **ST302 Stochastic processes**随机过程课程的代写代考和辅导服务！

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#### 这是伦敦政经学校随机过程课程的代写成功案例。

**ST302**课程简介

Availability

This course is compulsory on the BSc in Actuarial Science. This course is available on the BSc in Business Mathematics and Statistics, BSc in Data Science, BSc in Financial Mathematics and Statistics, BSc in Mathematics with Economics and BSc in Mathematics, Statistics and Business. This course is available as an outside option to students on other programmes where regulations permit and to General Course students.

Pre-requisites

Students must have completed either Probability, Distribution Theory and Inference (ST202) or Probability and Distribution Theory (ST206).

## Prerequisites

Teaching

This course will be delivered through a combination of classes, lectures, and Q&A sessions totalling a minimum of 29 hours across Michaelmas Term.

The course includes a reading week in Week 6 of Michaelmas Term.

Formative coursework

Compulsory written answers to two sets of problems.

Indicative reading

Lecture notes will be provided. Relevant books include R Durrett, Essentials of Stochastic Processes; T Mikosch, Elementary Stochastic Calculus with Finance in View; Institute of Actuaries core reading notes.

**ST302 Stochastic processes** HELP（EXAM HELP， ONLINE TUTOR）

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

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 weird coin, and $H$ as the event of fipping 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.

This in fact turns the standard argument about probability on its head. We all learn at some point that it’s wrong to think that a coin that just fipped six heads is more likely to turn up tails by “the law of averages”, and usually in that context we learn that the next coin fip is equally likely to be a head or a tail. But in this case, we see that the next coin flip is actually more likely to be a head – in fact in this scenario, if we flip six heads, the probability of flipping a seventh head is $64 / 113 * 1+49 / 113 * 1 / 2 \approx 78.3 \%$ !

A box has slips of paper numbered from 1 to 10. A paper is drawn at random and let $X_1$ be the number drawn. This paper is replaced, and the papers are remixed. A second paper with number $X_2$ is drawn. Find the distributions of $\left(X_1, X_2\right)$. Are $X_1$ and $X_2$ independent?

Answer the same questions if the first number is not replaced before the second is drawn.

Solution: In the first scenario, every pair of $(k, l) \in[10]^2$ is equally likely, so

$$

\mathbb{P}\left(X_1=k, X_2=l\right)=\frac{1}{100},

$$

and it is not hard to see that $\mathbb{P}\left(X_1=k, X_2=l\right)=\mathbb{P}\left(X_1=k\right) \mathbb{P}\left(X_2=l\right)$, so $X_1, X_2$ are independent. In the second case, we are not allowed to have $X_1=X_2$, so we have for any pair $(k, l)$ with $k \neq l$ is equally likely, but pairs of the form $(k, k)$ have probability zero. Therefore we have

$$

\mathbb{P}\left(X_1=k, X_2=l\right)= \begin{cases}\frac{1}{90}, & k \neq l, \ 0, & k=l .\end{cases}

$$

But $X_1, X_2$ are clearly not independent, since $\mathbb{P}\left(X_1=X_2=k\right)=0$, but $\mathbb{P}\left(X_1=k\right) \mathbb{P}\left(X_2=k\right)=$ $(1 / 10)^2=1 / 100$.

**MY-ASSIGNMENTEXPERT™**可以为您提供LSE.AC.UK **ST302 STOCHASTIC PROCESSES**随机过程课程的代写代考和辅导服务！