# 电子工程代写|计算数学基础代写Mathematical Foundations of Computing代考|ECE3020

## 电子工程代写|计算数学基础代写Mathematical Foundations of Computing代考|The Sample Mean

The sample mean $\bar{x}$ of a sample with $n$ elements is defined as
$$\bar{x}=\frac{1}{n} \sum_{i=1}^{n} x_{i}=\frac{1}{n} \sum_{i} x_{i}$$
Alternatively, given a sample in tabular form, we can sum over the different possible values of $x$ :
$$\bar{x}=\frac{1}{n} \sum_{x} n(x) x$$
Adopting a frequentist approach to probability, where $P(x)$, the probability of a value $x$, is defined as the limiting value of $n(x) / n$, we see that
$$\lim {n \rightarrow \infty} \bar{x}=\lim {n \rightarrow \infty} \frac{1}{n} \sum_{x} n(x) x=\sum_{x} x P(x)=\mu$$
This shows that as the sample size becomes very large, its mean is the expected value of a data item, which is the strong law of large numbers described in Section 1.7.4.

The sample mean $\bar{x}$ is a random variable whose value depends on the values taken on by the elements in the sample. It turns out that the expected value of this random variable, that is, $E(\bar{x})$, is also $\mu$. (This is not the same as the limiting value of the sample mean.) To see this, we start with the definition of the sample mean: $\bar{x}=\frac{1}{n}\left(x_{1}+x_{2}+\ldots+x_{n}\right)$, so that $n \bar{x}=\left(x_{1}+x_{2}+\ldots+x_{n}\right)$. Taking expectations of both sides gives us
$$E(n \bar{x})=E\left(x_{1}+x_{2}+\ldots+x_{n}\right)$$
From the sum rule of expectations, we can rewrite this as
$$E(n \bar{x})=E\left(x_{1}\right)+E\left(x_{2}\right)+\ldots+E\left(x_{n}\right)=n \mu$$
Therefore,
$$E(\bar{x})=E\left(\frac{n \bar{x}}{n}\right)=\frac{E(n \bar{x})}{n}=\mu$$
as stated.

## 电子工程代写|计算数学基础代写Mathematical Foundations of Computing代考|The Sample Median

The median value of a sample is the value such that $50 \%$ of the elements are larger than this value. For a sample with an odd number of elements, it is the middle element after sorting. For a sample with an even number of elements, it is the mean of the two middle elements after sorting.

For samples that contain outliers, the median is a better representative than the mean because it is relatively insensitive to outliers. The median is also an unbiased estimator of the population mean. However, it can be shown that if the underlying distribution is normal, the asymptotic variance of the median of a sample is $1.57$ times larger than the asymptotic variance of the sample mean. Hence, if the underlying distribution is normal, the same accuracy in estimating the population parameter can be obtained by collecting 100 observations and computing their mean or by collecting 157 samples and computing their median. However, if the underlying distribution is unimodal and sharper-peaked than normal (also called leptokurtic), the median is a more efficient estimator than the mean because, in such situations, the variance of the mean is higher than the variance of the median due to the presence of outliers.

Unlike the mean or the median, which seek to represent the central tendency of a sample, we now consider ways of representing the degree to which the elements in a sample differ from each other. These are also called measures of variability.

The simplest measure of variability is the range, which is the difference between the largest and smallest elements. The range is susceptible to outliers and therefore not reliable. A better measure is to sort the data values and then determine the data values that lie at $q \%$ and $1-q \%$. The difference between the two values is the range of values in which the central $1-2 q \%$ of the sample lies. This conveys nearly the same information as the range but with less sensitivity to outliers. A typical value of $q$ is 25 , in which case this measure is also called the interquartile range. In the context of delay bounds and service-level agreements, a typical value of $q$ is 5 (so that the span is $5 \%-95 \%$ ).

## 电子工程代写|计算数学基础代写Mathematical Foundations of Computing代考|The Sample Mean

$$\bar{x}=\frac{1}{n} \sum_{i=1}^{n} x_{i}=\frac{1}{n} \sum_{i} x_{i}$$

$$\bar{x}=\frac{1}{n} \sum_{x} n(x) x$$

$$\lim n \rightarrow \infty \bar{x}=\lim n \rightarrow \infty \frac{1}{n} \sum_{x} n(x) x=\sum_{x} x P(x)=\mu$$

$$E(n \bar{x})=E\left(x_{1}+x_{2}+\ldots+x_{n}\right)$$

$$E(n \bar{x})=E\left(x_{1}\right)+E\left(x_{2}\right)+\ldots+E\left(x_{n}\right)=n \mu$$

$$E(\bar{x})=E\left(\frac{n \bar{x}}{n}\right)=\frac{E(n \bar{x})}{n}=\mu$$

## 电子工程代写|计算数学基础代写Mathematical Foundations of Computing代考|The Sample Median

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