Mean of a Random Variable

• Suppose that a probability distribution of a random variable $X$ is specified.
• For a measure of central tendency of the random variable $X$ we use the terms expectation, expected value, and average value for the same concept.
• Intuitively, the expected value of $X$ is the average value that the random variable takes on.
• However, some of the values of the random variable $X$ could be more (or less) probable than the other in the distribution unless the random variable is distributed uniformly.
• Hence, in order to consider an average value of $X$ we need to take its probability into account.

• If I repeat the experiment many times, what would be the average number of an outcome of a random variable?
• Definition 4.1:
  
• The expected value is used as a measure of centering or location of the distribution of a random variable $X$.
• By the uniform distribution assumption, i.e. all values of $X$ are equally likely to occur in population with size $N$, $f(x) = \frac{1}{N}$ for all $x$,
  $$E(X)=\sum_x xf(x)=\sum_x x(\frac{1}{N})=(\frac{1}{N})\sum_i x_i=\mu=\bar{x}$$

• Example: If two coins are tossed 16 times and $X$ is the number of heads that occur per toss, then the value of $X$ can be 0, 1, 2.
• The experiment yields no heads, one head, and two heads a total of 4, 7, and 5 times, respectively.
• The average number of heads per toss is then

$$0*\frac{4}{16}+1*\frac{7}{16}+2*\frac{5}{16}$$

where $\frac{4}{16},\frac{7}{16},\frac{5}{16}$ are relative frequencies

$x$	0	1	2
$f(x)$	4/16	7/16	5/16

$$0*\frac{4}{16}+1*\frac{7}{16}+2*\frac{5}{16}=\frac{17}{16}=1.0625$$

• Example 4.1: A lot contain 4 good components and 3 defective components.
  • A sample of 3 is taken by a quality inspector.
  • Find the expected value of the number of good components in this sample.
• Solution: $X$ represents the number of good components
  $$f(x)=\frac{\left( \begin{array}{c} 4\\ x\\ \end{array}\... ...( \begin{array}{c} 7\\ 3\\ \end{array}\right) },x=0,1,2,3$$
  $$\mu=E(X)=0*f(0)+1*f(1)+2*f(2)+3*f(3)=\frac{12}{7}$$

• Example 4.3: Let $X$ be the random variable that denotes the life in hours of a certain electronic device. The probability density function is as the following.
  $$f(x)=\left\lbrace \begin{array}{lc} \frac{20000}{x^3}, & x>100 \\ 0, & elsewhere \\ \end{array}\right\rbrace$$
  Find the expected life of this type of device.
• Solution:
  $$\mu=E(X)=\int_{100}^\infty x \frac{20000}{x^3}dx=-\frac{20000}{x}\textbar_{100}^\infty=200$$

• Mean of $g(X)$ (any real-valued function): If $X$ is a discrete random variable with $f(x)$, for $x = -1,0,1,2$, and $g(X) = X^2$ then
  $$\begin{array}{l} P[g(X) = 0] = P(X = 0) = f(0), \\ P[g(X) =... ...f(-1)+ f(1), \\ P[g(X) = 4] = P(X = 2) = f(2), \\ \end{array}$$

• The probability distribution of $g(X)$ can be written

  $g(x)$	0	1	4
  $P[g(X)=4]$	f(0)	f(-1)+f(1)	f(2)

• $$\begin{array}{rl} E(g(X))& = 0*f(0)+1*[f(-1)+f(1)]+4*f(2) \\... ...(0)+(1)^2*f(1)+(2)^2*f(2) \\ & = \sum_x g(x)*f(x) \end{array}$$

• Theorem 4.1::
  
• Example 4.5: Let $X$ be a random variable with density function
  $$f(x)=\left\lbrace \begin{array}{lc} \frac{x^2}{3}, & -1<x<2 \\ 0, & elsewhere \\ \end{array}\right\rbrace$$

• Find the expected value of $g(X) = 4X + 3$.
• Solution:
  $$E[g(X)]=E(4X+3)=\frac{1}{3}\int_{-1}^2 (4x^3+3x^2)dx=8$$

• Theorem 4.2::
  
• Example 4.7: Find $E(Y/X)$ for the density function
  $$f(x,y)=\left\lbrace \begin{array}{lc} \frac{x(1+3y^2)}{4},... ...<x<2,~,0<y<1 \\ 0, & elsewhere \\ \end{array}\right\rbrace$$

• Solution:
  $$E(\frac{Y}{X})=\int_{0}^1 \int_{0}^2 \frac{y}{x} \frac{x(1+3y^2)}{4}dxdy=\frac{5}{8}$$

• If $g(X, Y) = X$ is
  $$E(X)=\left\lbrace \begin{array}{l} \sum_x \sum_y xf(x,y)=\... ...dy =\int_{-\infty}^\infty xg(x)dx\\ \end{array}\right\rbrace$$
  where $g(x)$ is the marginal distribution of $X$
• If $g(X, Y) = Y$ is
  $$E(Y)=\left\lbrace \begin{array}{l} \sum_x \sum_y yf(x,y)=\... ...dy =\int_{-\infty}^\infty yh(y)dy\\ \end{array}\right\rbrace$$
  where $h(y)$ is the marginal distribution of $Y$