Weighted Mean Revisited

If a man is randomly selected, what is the probability that his daily income is 400 BDT?

• \(AM = \frac{61500}{125}=492\)
• In expectation, \(rf\) is the weight (Total weight = 1)

Expectation from Mean

\(\displaystyle \bar X=\frac{\Sigma x_if_i}{\Sigma f_i}\)

• \(= \frac{x_1f_1+x_2f_2+\cdots+x_3f_3}{N}\) (letting \(\Sigma f_i=N\))
• \(= x_1\frac{f_1}{N}+x_2\frac{f_2}{N}+\cdots +x_2\frac{f_n}{N}\)
• \(= x_1p_1+x_2p_2+\cdots+x_np_n\)
• \(= \sum x_i\cdot p(x_i)\)

Expectation

Probabilities are weights

\[\begin{eqnarray} E(X) &=& x_1 \times P(x_1) + x_2 \times P(x_2)+ \cdots +x_n \times P(x_n) \nonumber \\ &=& \displaystyle \sum_{i=1}^n x_i \times P(x_i) \nonumber \\ \end{eqnarray}\]

• \(E(X^2) = \sum x^2 \times p(x)\)

Expectation Example

Find

1. \(E(X)\)
2. \(E(X^2)\)

Properties of Expectation

Aslo refer to the properties of AM (\(E(X)=\bar X\))

• Expectation of a constant, \(E(c)=c\)
• \(E(aX)=aE(X)\)
• \(E(X-a) = E(X) - E(a)\)
• \(E(aX+b) = aE(X)+b\)
• \(E(X+Y) = E(X)+E(Y)\)
• \(E(XY) = E(X) E(Y)\) (if independent, relate with probability)
• \(E(X^2)\ge \{E(X)\}^2\)
• \(E\left(\frac 1 X \right) \ge \frac 1 {E(X)}\)

Variance

Recall this (click)

We knew, \(\sigma^2 = \bar {X^2} - (\bar X)^2=\) Mean of square - square of mean

• \(V(X)=E(X^2)-\{E(X)\}^2\)
• Original Formula: \(V(X) = E\{x-E(X)\}^2\) (Match)
• Expand and prove these are equal

Variance Properties

• \(V(c)=0\) ponder, why?
• \(V(X+a)=V(X)\)
• \(V(aX+c) = a^2V(X)\) (depends only on scale, recall)
• \(V(X+Y) = V(X)+V(Y)\)

Covariance

\(Cov(X,Y) = \frac{\Sigma (x-\bar x)(y-\bar x)}{n}\)

Write in terms of \(E(X)\)

• \(Cov(X,Y) = E [\{x-E(X)\}\{y-E(Y)\}]\) (Expand)
• \(E(XY)-E(X)E(Y)\)

Prove E(X) Properties

Go back to see the properties

E(a)

\(E(a) = \sum a \cdot p(x_i)\)

• \(=a \cdot \sum p(x_i)\)
• \(=a \times 1 = a\)
• Others can be proven similarly

Double Summation Revisited

• If sum over \(x_i\), we get only 1 column at a time
• If sum over \(y_i\), we get only 1 row at a time
• SO to get grand total, we must use \(\sum \sum (x_i+y_j)\)

E(X+Y) & E(XY)

E(X²) ≥ {E(X)}²

\(V(X)\ge 0\)

• \(\therefore E(X^2) - \{E(X)\}^2 \ge 0\)

AM & HM

\(E(\frac 1 X) \ge \frac 1 {E(X)}\)

• \(AM \ge HM\)
• \(HM =\) opposite of mean of opposite \(\to \frac{1}{\frac{\sum \frac 1 {x_i}}{n}} = \frac 1 {E(\frac 1 x)}\)
• \(E(X) \ge \frac{1}{E(\frac 1 x)}\)
• If \(0.5 \gt \frac 1 3 \to 3 \gt \frac 1 {0.5}\)

Variance of constant (a)

\(V(X) = E\{X-E(X)\}^2\)

• \(V(a) = E\{a-E(a)\}^2\)
• \(E(a-a)^2\)
• \(E(0)=0\)

V(aX+b)

\(V(X) = E\{X-E(X)\}^2\)

\(V(aX+b)=\)

• \(E\{(aX+b)-E(aX+b)\}^2\)
• \(E\{aX+b-aE(X)-b\}^2\)
• \(E\{aX-aE(X)\}^2\)
• \(a^2E\{X-E(X)\}^2\)
• \(a^2V(X)\)
• Does variance depend on origin and scale?
• \(V(X+Y)=?\)

Cov(X,Y) Properties

For independent X, Y

• \(\to E(XY) = E(X)E(Y)\)
• \(Cov(X,Y) = E(XY) - E(X)E(Y) = 0\)
• Correlation, \(r = \frac{Cov(X,Y)}{\sigma_x \sigma_y}=0\)

Example 01: E(X) & V(X)

1. Find k
2. Find \(E(X)\) and \(V(X)\)

Example 02: \(P(x) = \frac 1 n\)

\(P(x) = \frac 1 n; x = 1,2,3, \cdots,n\) Find \(E(X)\) and \(V(X)\)

Example 03

\(P(x)=\frac {3-|4-x|} k; x = 2,3,4,5,6\)

Find

1. k
2. V(2X-1)

Example 04: \(P(x) = \frac{3-|4-x|}{k}\)

\(P(x) = \frac{3-|4-x|}{k}; x = 2, 3, 4, 5, 6\)

Find

1. k
2. V(2x-1)

Example 05: y = 3x+5. Find V(3y-2)

• V(3y-2) = \(3^2V(y)\)
• \(9V(3x+5)\)
• \(9 \cdot 3^2 \cdot V(x)\)

Example 06: \(f(x) = \frac x 8\)

\(f(x) = \frac x 8; 0 <x<4\)

Find

• E(X),
• E(2x^2+3)
• V(X)