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# 数学代写|蒙特卡罗模拟代考Monte Carlo Method代考|MATH483 RANDOM EXPERIMENTS

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## 数学代写|蒙特卡罗模拟代考Monte Carlo Method代考|RANDOM EXPERIMENTS

The basic notion in probability theory is that of a random experiment: an experiment whose outcome cannot be determined in advance. The most fundamental example is the experiment where a fair coin is tossed a number of times. For simplicity suppose that the coin is tossed three times. The sample space, denoted $\Omega$, is the set of all possible outcomes of the experiment. In this case $\Omega$ has eight possible outcomes:
$$\Omega={H H H, H H T, H T H, H T T, T H H, T H T, T T H, T T T},$$
where, for example, HTH means that the first toss is heads, the second tails, and the third heads.

Subsets of the sample space are called events. For example, the event $A$ that the third toss is heads is
$$A={H H H, H T H, T H H, T T H} .$$
We say that event A occurs if the outcome of the experiment is one of the elements in $A$. Since events are sets, we can apply the usual set operations to them. For example, the event $A \cup B$, called the union of $A$ and $B$, is the event that $A$ or $B$ or both occur, and the event $A \cap B$, called the intersection of $A$ and $B$, is the event that $A$ and $B$ both occur. Similar notation holds for unions and intersections of more than two events. The event $A^c$, called the complement of $A$, is the event that $A$ does not occur. Two events $A$ and $B$ that have no outcomes in common, that is, their intersection is empty, are called disjoint events. The main step is to specify the probability of each event.

## 数学代写|蒙特卡罗模拟代考Monte Carlo Method代考|CONDITIONAL PROBABILITY AND INDEPENDENCE

How do probabilities change when we know that some event $B \subset \Omega$ has occurred? Given that the outcome lies in $B$, the event $A$ will occur if and only if $A \cap B$ occurs, and the relative chance of $A$ occurring is therefore $\mathbb{P}(A \cap B) / \mathbb{P}(B)$. This leads to the definition of the conditional probability of $A$ given $B$ :
$$\mathbb{P}(A \mid B)=\frac{\mathbb{P}(A \cap B)}{\mathbb{P}(B)} .$$
For example, suppose that we toss a fair coin three times. Let $B$ be the event that the total number of heads is two. The conditional probability of the event $A$ that the first toss is heads, given that $B$ occurs, is $(2 / 8) /(3 / 8)=2 / 3$.

Rewriting (1.3) and interchanging the role of $A$ and $B$ gives the relation $\mathbb{P}(A \cap$ $B)=\mathbb{P}(A) \mathbb{P}(B \mid A)$. This can be generalized easily to the product rule of probability, which states that for any sequence of events $A_1, A_2, \ldots, A_n$,
$$\mathbb{P}\left(A_1 \cdots A_n\right)=\mathbb{P}\left(A_1\right) \mathbb{P}\left(A_2 \mid A_1\right) \mathbb{P}\left(A_3 \mid A_1 A_2\right) \cdots \mathbb{P}\left(A_n \mid A_1 \cdots A_{n-1}\right),$$
using the abbreviation $A_1 A_2 \cdots A_k \equiv A_1 \cap A_2 \cap \cdots \cap A_k$.
Suppose that $B_1, B_2, \ldots, B_n$ is a partition of $\Omega$. That is, $B_1, B_2, \ldots, B_n$ are disjoint and their union is $\Omega$. Then, by the sum rule, $\mathbb{P}(A)=\sum_{i=1}^n \mathbb{P}\left(A \cap B_i\right)$ and hence, by the definition of conditional probability, we have the law of total probability:
$$\mathbb{P}(A)=\sum_{i=1}^n \mathbb{P}\left(A \mid B_i\right) \mathbb{P}\left(B_i\right)$$

## 数学代写|蒙特卡罗模拟代考Monte Carlo Method代考|RANDOM EXPERIMENTS

$\Omega=H H H, H H T, H T H, H T T, T H H, T H T, T T H, T T T$

$$A=H H H, H T H, T H H, T T H .$$

## 数学代写|蒙特卡罗模拟代考Monte Carlo Method代考|CONDITIONAL PROBABILITY AND INDEPENDENCE

$$\mathbb{P}(A \mid B)=\frac{\mathbb{P}(A \cap B)}{\mathbb{P}(B)} .$$

$$\mathbb{P}\left(A_1 \cdots A_n\right)=\mathbb{P}\left(A_1\right) \mathbb{P}\left(A_2 \mid A_1\right) \mathbb{P}\left(A_3 \mid A_1 A_2\right) \cdots \mathbb{P}\left(A_n \mid A_1 \cdots A_{n-1}\right),$$

$$\mathbb{P}(A)=\sum_{i=1}^n \mathbb{P}\left(A \mid B_i\right) \mathbb{P}\left(B_i\right)$$

## MATLAB代写

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