# Expectation Operator Rules, Covariance and Standard Error of the Mean

Yao Yao on September 10, 2014

## 1. Expectation Operator Rules

\begin{align} \operatorname{E}[aX] &= a \operatorname{E}[X] \newline \operatorname{E}[X + c] &= \operatorname{E}[X] + c \newline \operatorname{E}[X + Y] &= \operatorname{E}[X] + \operatorname{E}[Y], \text{X and Y may be independent or not} \newline \operatorname{E}[XY] &= \operatorname{E}[X] * \operatorname{E}[X], \text{only if X and Y are independent} \end{align}

Here proves the last rule:

Suppose the joint pdf of $X$ and $Y$ is $j(x,y)$, then

$\operatorname{E}[XY] = \iint xy \, j(x,y)\,\mathrm{d}x\,\mathrm{d}y$

If $X$ and $Y$ are independent, then by definition $j(x,y) = f(x)g(y)$ where $f$ and $g$ are the marginal PDFs for $X$ and $Y$. Then

\begin{align} \operatorname{E}[XY] &= \iint xy \,j(x,y)\,\mathrm{d}x\,\mathrm{d}y = \iint x y f(x) g(y)\,\mathrm{d}y\,\mathrm{d}x \newline &= \left[\int x f(x)\,\mathrm{d}x\right]\left[\int y g(y)\,\mathrm{d}y\right] = \operatorname{E}[X]\operatorname{E}[Y] \end{align}

## 2. Covariance Again

\begin{align} Cov(X,Y) &= \operatorname{E}\left[\left(X - \operatorname{E}\left[X\right]\right) \left(Y - \operatorname{E}\left[Y\right]\right)\right] \newline &= \operatorname{E}\left[X Y - X \operatorname{E}\left[Y\right] - \operatorname{E}\left[X\right] Y + \operatorname{E}\left[X\right] \operatorname{E}\left[Y\right]\right] \newline &= \operatorname{E}\left[X Y\right] - \operatorname{E}\left[X\right] \operatorname{E}\left[Y\right] - \operatorname{E}\left[X\right] \operatorname{E}\left[Y\right] + \operatorname{E}\left[X\right] \operatorname{E}\left[Y\right] \newline &= \operatorname{E}\left[X Y\right] - \operatorname{E}\left[X\right] \operatorname{E}\left[Y\right] \newline &= \frac{\sum_{i=1}^{n}(x_i - \bar{x})(y_i - \bar{y})}{n-1} \end{align}

If $X$ and $Y$ are independent, $Cov(X, Y) = 0$

## 3. Standard Error of the Mean

If $X_1, X_2 , \ldots, X_n$ are n independent observations from a population that has a mean $\mu$ and standard deviation $\sigma$, $\bar{X} = \frac{1}{n} \sum_n {x_i}$ is itself a random variable, and satisfy

• $E[\bar{X}] = \mu$
• $Var(\bar{X}) = \frac{\sigma^2}{n}$

The standard error of the mean (SEM) is the standard deviation of the sample-meanâ€™s estimate of a population mean, i.e.

$\text{SE}_\bar{x}\ = \sqrt{Var(\bar{X})} = \frac{\sigma}{\sqrt{n}}$

## 4. Proof of $Var(\bar{X}) = \frac{\sigma^2}{n}$

### 4.1 Proof I

Suppose $T = (X_1 + X_2 + \cdots + X_n)$, then

\begin{align} Var(T) &= E[T^2] - E[T]^2 = Var(X_1 + X_2 + \cdots + X_n) = n Var(X) = n\sigma^2 \newline Var(\bar{X}) &= Var(\frac{T}{n}) = E[(\frac{T}{n})^2] - E[\frac{T}{n}]^2 \newline &= \frac{1}{n^2}(E[T^2] - E[T]^2) = \frac{1}{n^2} n\sigma^2 = \frac{\sigma^2}{n} \end{align}

### 4.2 Proof II

Suppose $T = (X_1 + X_2 + \cdots + X_n)$, then

\begin{align} E[T^2] &= E \left [ \sum_{i=1}^{n}(X_i^2) + \sum_{i,j=1,...,n}^{i \neq j}(X_i X_j) \right ] \newline &= nE[X^2] + (n^2-n)E[X]^2, \text{ for } X_i, X_j \, \text{ are independent }, E[X_i X_j] = E[X_i] E[X_j] \newline &= nE[X^2] + (n^2-n)\mu^2 \newline \because E[X^2] &= \sigma^2 + E[X]^2 \newline &= \sigma^2 + \mu^2 \newline \therefore E[T^2] &= n\sigma^2 + n\mu^2 + (n^2-n)\mu^2 \newline &= n\sigma^2 + n^2\mu^2 \newline \therefore Var(\bar{X}) &= \frac{1}{n^2}(E[T^2] - E[T]^2) \newline &= \frac{1}{n^2}(n\sigma^2 + n^2\mu^2 - n^2\mu^2) \newline &= \frac{\sigma^2}{n} \end{align}