Posted on Categories:Probability theory, 数学代写, 概率论

# 数学代写|概率论代考Probability Theory代写|Math461 Properties of Probability Measures

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## 数学代写|概率论代考Probability Theory代写|Properties of Probability Measures

In this section we collect and prove a number of useful properties of probability measures. Throughout the section, the sample space is denoted by $\Omega$ and $A, B, \ldots$ are events in $\Omega$.
Lemma 1.3.1. (a) For events $A_1, A_2, \ldots$ which are pairwise disjoint, we have
$$P\left(\bigcup_{i=1}^{\infty} A_i\right)=\sum_{i=1}^{\infty} P\left(A_i\right) .$$
(b) $P\left(A^c\right)=1-P(A)$.
(c) If $A \subseteq B$, then $P(A) \leq P(B)$.
More precisely, we have that $P(B)=P(A)+P(B \backslash A)$.
(d) $P(A \cup B)=P(A)+P(B)-P(A \cap B)$.

Proof. (a) We have
\begin{aligned} P\left(\bigcup_{i=1}^{\infty} A_i\right)=\sum_{\omega \in \cup_i A_i} P(\omega) & =\sum_{\omega \in A_1} P(\omega)+\sum_{\omega \in A_2} P(\omega)+\cdots \ & =\sum_{i=1}^{\infty} P\left(A_i\right) . \end{aligned}
(b) Take $A_1=A, A_2=A^c$ and $A_j=\emptyset$, for all $j \geq 3$. It follows from (a) that $1=P(\Omega)=P\left(A \cup A^c\right)=P(A)+P\left(A^c\right)$, proving (b).
(c) We can write $B=A \cup(B \backslash A)$. This is a union of disjoint events, and the result now follows from (a).
(d) We can write $A \cup B=A \cup(B \backslash A)$, which is a disjoint union. Hence we find that
\begin{aligned} P(A \cup B) & =P(A)+P(B \backslash A)=P(A)+P(B \backslash(A \cap B)) \ & =P(A)+P(B)-P(A \cap B) \end{aligned}
where the last equality follows from (c).

## 数学代写|概率论代考Probability Theory代写|Conditional Probabilities

When we talk and think about probability, the concept of independence plays a crucial role. For instance, when we flip a coin twice, we are inclined to say that the outcome of the first flip ‘says nothing’ about the outcome of the second. Somehow, we believe that information about the first flip gives us no information about the outcome of the second. We believe that the two outcomes are independent of each other.

On the other hand, when we throw a die, and consider the event $E_3$ that the outcome is equal to 3 , and the event $E_{\leq 4}$ that the outcome is at most 4 , then information about $E_{\leq 4}$ does, in fact, change the probability of $E_3$. Indeed, if I tell you that $E_{\leq 4}$ does not occur, then we know for sure that $E_3$ cannot occur either, and hence the new probability of $E_3$ had better be 0 . If I tell you that $E_{\leq 4}$ does occur, then there are four possibilities left. The new probability that $E_3$ occurs should therefore be $\frac{1}{4}$, see Example $1.4 .2$ below.

The last argument can be carried out in much more general terms, as follows. Suppose I tell you that in a certain sample space $\Omega$, we have two events $A$ and $B$, with probabilities $P(A)$ and $P(B)$ respectively. This means that a fraction $P(A)$ of all probability mass is concentrated in the event $A$, and similarly for $B$. Now suppose that I know that the event $B$ occurs. Does this new information change the probability of the event $A$ ? Well, we now know that only outcomes in $B$ matter, and we can disregard the rest of the sample space. Hence we only need to look at the probabilities of elements in $B$. The new probability that $A$ occurs should now be the fraction of probability mass in $B$ that is also in $A$. That is, it should be the sum of the probabilities of all outcomes in $B \cap A$, divided by the probability of $B$.

# 概率论代写

## 数学代写|概率论代考Probability Theory代写|Properties of Probability Measures

$$P\left(\bigcup_{i=1}^{\infty} A_i\right)=\sum_{i=1}^{\infty} P\left(A_i\right) .$$
(二) $P\left(A^c\right)=1-P(A)$.
(c) 如果 $A \subseteq B$ ，然后 $P(A) \leq P(B)$.

(四) $P(A \cup B)=P(A)+P(B)-P(A \cap B)$.

$$P\left(\bigcup_{i=1}^{\infty} A_i\right)=\sum_{\omega \in \cup_i A_i} P(\omega)=\sum_{\omega \in A_1} P(\omega)+\sum_{\omega \in A_2} P(\omega)+\cdots \quad=\sum_{i=1}^{\infty} P\left(A_i\right) .$$
(b) 采取 $A_1=A, A_2=A^c$ 和 $A_j=\emptyset ，$ 对全部 $j \geq 3$. 从 (a) 可以看出 $1=P(\Omega)=P\left(A \cup A^c\right)=P(A)+P\left(A^c\right)$ ，证明 (b) 。
(c) 我们可以写 $B=A \cup(B \backslash A)$. 这是不相交事件的并集，结果现在来自 (a)。
(d) 我们可以写 $A \cup B=A \cup(B \backslash A)$ ，这是一个不相交的联盟。因此我们发现
$$P(A \cup B)=P(A)+P(B \backslash A)=P(A)+P(B \backslash(A \cap B)) \quad=P(A)+P(B)-P(A \cap B)$$

## 数学代写|概率论代考Probability Theory代写|Conditional Probabilities

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## MATLAB代写

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