• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 商科代写|商业数学代写business mathematics代考|Probability and Expected Value

First, let us discuss games of chance. Games of chance include, but are not limited to, games involving flipping coin, rolling dice, drawing cards from a deck, spinning a wheel, and such. Suppose you are flipping a coin with a friend. If you both flip the same, both heads or both tails, you win $\$ 1$. If you flip a head and a tail each, you lose$\$1$. If you play this game with this bet 100 times over the course of the evening, how much do you expect to win? Or lose? To answer such questions, we need two concepts: the probability of an event and the expected value of a random variable.

In situations like this, it makes sense to count the number of times something occurs. An efficient way to do this is to use the frequency definition of the probability of an event. The probability of the event two heads or two tails is the number of ways we can achieve these results divided by the total number of possible outcomes. That is, we define both coins being flipped that land on both heads or both tails as a favorable event $A$. We can define event $A$ as the set of outcomes that include ${\mathrm{HH}, \mathrm{TT}}$.
Favorable outcomes are those that consist of ${\mathrm{HH}, \mathrm{TT}}$.
$$\begin{array}{r} \text { Probability of an event }=\frac{\text { Favorable outcomes }}{\text { Total outcomes }} \ \text { Probability of an event }{A}=\frac{\text { Number of outcomes of }{A}}{\text { Total outcomes }} \end{array}$$

Of course, the probability of an event (flip of a fair coin) must be equal or greater than zero, and equal to or less than 1. And the sum of the probabilities of all possible events must equal 1 . That is,
\begin{aligned} &0 \leq p_{i} \leq 1 \ &\sum_{i=1}^{n} p_{i}=1, i=1,2, \ldots, n \end{aligned}
We need to compute all the possible outcomes of flipping two coins, and then determine how many result in the same results defined by event $A$. A tree is useful for visualizing the outcomes. These outcomes constitute the sample space. On the first flip, the possible results are $\mathrm{H}$ or $\mathrm{T}$. And on the second flip, the same outcomes are still available. We assume that these events are equally likely to occur based on flipping and obtaining either a head or tail of each flip.

First, we define a random variable as a rule that assigns a number to every outcome of a sample.

We use $E[X]$, which is stated as the expected value of $X$. We define $E[X]$ as follows:
Expected value, $E[X]$, is the mean or average value.
Further, we provide the following two formulas: in the discrete case, $E[X]=\sum_{i=1}^{n} x_{i} p\left(x_{i}\right)$ and in the continuous case, $E[X]=\int_{-\infty}^{+\infty} x * f(x) d x$.

There are numerous ways to calculate the average value. We present a few common methods that you could use in decision theory.

If you had 2 quiz grades, an 82 and a 98 , almost intuitively you would add the two numbers and divide by 2 , giving an average of 90 .

Average scores: Two scores that were earned were 82 and 98 . Compute the average.
$$E[X]=\frac{(82+98)}{2}=90$$
If after 5 quizzes, you had three 82 s and two 98 s, you would add them and divide by $5 .$
$$\text { Average }=\frac{3(82)+2(98)}{5}=88.4$$
Rearranging the terms, we obtain
$$\text { Average }=\frac{3}{5}(82)+\frac{2}{5}(98)$$
In this form, we have two payoffs, 82 and 98 , each multiplied by the weights, $3 / 5$ and $2 / 5$. This is analogous to the definition of expected value.

Suppose a game has outcomes $a_{1}, a_{2}, \ldots, a_{n}$, each with a payoff $w_{1}, w_{2}, \ldots, w_{n}$ and a corresponding probability $p_{1}, p_{2}, \ldots, p_{n}$ where $p_{1}+p_{2}+\ldots+p_{n}=1$ and $0 \leq p_{i} \leq 1$, then the quantity
$$E=w_{1} p_{1}+w_{1} p_{2}+\ldots+w_{1} p_{n}$$
is the expected value of the game. Note that expected value is analogous to weighted average, but the weights must be probabilities $\left(0 \leq p_{i} \leq 1\right)$ and the weights must sum to 1 .

## 商科代写|商业数学代写business mathematics代考|Probability and Expected Value

Probability of an event $=\frac{\text { Favorable outcomes }}{\text { Total outcomes }}$ Probability of an event $A=\frac{\text { Number of outcomes of } A}{\text { Total outcomes }}$

$$0 \leq p_{i} \leq 1 \quad \sum_{i=1}^{n} p_{i}=1, i=1,2, \ldots, n$$

$$E[X]=\frac{(82+98)}{2}=90$$

$$\text { Average }=\frac{3(82)+2(98)}{5}=88.4$$

$$\text { Average }=\frac{3}{5}(82)+\frac{2}{5}(98)$$

$$E=w_{1} p_{1}+w_{1} p_{2}+\ldots+w_{1} p_{n}$$

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## MATLAB代写

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