## 统计代写|概率论代写Probability theory代考|MATHS2103

2022年10月17日

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## 统计代写|概率论代写Probability theory代考|Elementary Conditional Probabilities

Example 8.1 We throw a die and consider the events
$A:={$ the face shows an odd number $}$,
$B:={$ the face shows three or smaller $}$.
Clearly, $\mathbf{P}[A]=\frac{1}{2}$ and $\mathbf{P}[B]=\frac{1}{2}$. However, what is the probability that $A$ occurs if we already know that $B$ occurs?

We model the experiment on the probability space $(\Omega, \mathcal{A}, \mathbf{P})$, where $\Omega=$ ${1, \ldots, 6}, \mathcal{A}=2^{\Omega}$ and $\mathbf{P}$ is the uniform distribution on $\Omega$. Then
$$A={1,3,5} \quad \text { and } B={1,2,3} .$$
If we know that $B$ has occurred, it is plausible to assume the uniform distribution on the remaining possible outcomes; that is, on ${1,2,3}$. Thus we define a new probability measure $\mathbf{P}_B$ on $\left(B, 2^B\right)$ by
$$\mathbf{P}_B[C]=\frac{# C}{# B} \quad \text { for } C \subset B \text {. }$$

By assigning the points in $\Omega \backslash B$ probability zero (since they are impossible if $B$ has occurred), we can extend $\mathbf{P}_B$ to a measure on $\Omega$ :
$$\mathbf{P}[C \mid B]:=\mathbf{P}_B[C \cap B]=\frac{#(C \cap B)}{# B} \text { for } C \subset \Omega \text {. }$$
In this way, we get $\mathbf{P}[A \mid B]=\frac{#{1,3}}{#{1,2,3}}=\frac{2}{3} . \diamond$
Motivated by this example, we make the following definition.
Definition $8.2$ (Conditional probability) Let $(\Omega, \mathcal{A}, \mathbf{P})$ be a probability space and $B \in \mathcal{A}$. We define the conditional probability given $B$ for any $A \in \mathcal{A}$ by
\mathbf{P}[A \mid B]=\left{\begin{aligned} \frac{\mathbf{P}[A \cap B]}{\mathbf{P}[B]}, & \text { if } \mathbf{P}[B]>0, \ 0, & \text { otherwise. } \end{aligned}\right.
Remark $8.3$ The specification in (8.1) for the case $\mathbf{P}[B]=0$ is arbitrary and is of no importance. $\diamond$

## 统计代写|概率论代写Probability theory代考|Conditional Expectations

Let $X$ be a random variable that is uniformly distributed on $[0,1]$. Assume that if we know the value $X=x$, the random variables $Y_1, \ldots, Y_n$ are independent and $\operatorname{Ber}_x$-distributed. So far, with our machinery we can only deal with conditional probabilities of the type $\mathbf{P}[\cdot \mid X \in[a, b]], a<b$ (since $X \in[a, b]$ has positive probability). How about $\mathbf{P}\left[Y_1=\ldots=Y_n=1 \mid X=x\right]$ ? Intuitively, this should be $x^n$. We thus need a notion of conditional probabilities that allows us to deal with conditioning on events with probability zero and that is consistent with our intuition. In the next section, we will see that in the current example this can be done using transition kernels. First, however, we have to consider a more general situation.
In the following, $\mathcal{F} \subset \mathcal{A}$ will be a sub- $\sigma$-algebra and $X \in \mathcal{L}^1(\Omega, \mathcal{A}, \mathbf{P})$. In analogy with Lemma 8.10, we make the following definition.

Definition 8.11 (Conditional expectation) A random variable $Y$ is called a conditional expectation of $X$ given $\mathcal{F}$, symbolically $\mathbf{E}[X \mid \mathcal{F}]:=Y$, if:
(i) $Y$ is $\mathcal{F}$-measurable.
(ii) For any $A \in \mathcal{F}$, we have $\mathbf{E}\left[X \mathbb{1}_A\right]=\mathbf{E}\left[Y \mathbb{1}_A\right]$.
For $B \in \mathcal{A}, \mathbf{P}[B \mid \mathcal{F}]:=\mathbf{E}\left[\mathbb{1}_B \mid \mathcal{F}\right]$ is called a conditional probability of $B$ given the $\sigma$-algebra $\mathcal{F}$.
Theorem 8.12 $\mathbf{E}[X \mid \mathcal{F}]$ exists and is unique (up to equality almost surely).
Since conditional expectations are defined only up to equality a.s., all equalities with conditional expectations are understood as equalities a.s., even if we do not say so explicitly.

# 概率论代考

## 统计代写|概率论代写概率论代考|基本条件概率

$A:={$脸上显示一个奇数$}$，
$B:={$脸上显示三个或更小的$}$ .

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## MATLAB代写

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