• $F_X:\mathbb{R}\to \mathbb{R}$
• $F_X(x)=P(X\le x)$ for $x\in \mathbb{R}$

• taking jump of $p(x_i)$ upon reaching it
• for $x_1<x_2<\cdots <x_n$
• then $F$ constant in $[x_i,x_{i+1})$
• to know cdf, knowing pdf of following range is enough
  • $(-\infty ,x_1),[x_1,x_2),[x_2,x_3),\dots [x_{n-1},x_n),[x_n,\infty )$

• $F_X$: monolithic / non-decreasing
• $0\le F_X\le 1$
• always starts at / near 0 for small $x$
  • and end at or near $1$ for large $x$
• discrete r.v.'s cdf: a step function
• $F(x)$: right-continuous
  • i.e. can approach the limit only from the right

if $X$ w/ pmf by

• $p(1)=\frac{1}{4};p(2)=\frac{1}{2};p(3)=\frac{1}{8};;p(4)=\frac{1}{8}$
• then cdf:
  • $(-\infty ,1);0$
  • $[1,2)=\frac{1}{4}$
    • jump of $p_X(1)$
  • $[2,3)=\frac{3}{4};$
    • jump of $p_X(2)$
  • $[3,4)=\frac{7}{8};$
    • jump of $p_X(3)$
  • $[4,\infty )=1;$
    • jump of $p_X(4)$

no. of mortgage approved per week: given below

• prob: fewer than 4 mortgage approved
  • $p(0)+p(1)+p(2)+p(3)=0.7$
• prob: more than 2, but no more that 5 approved
  • $p(3)+p(4)+p(5)=0.55$
  • or: $P(X\le 5)-P(X\le 2)=F(5)-F(2)$
• cdf chart: trivial

given cdf, show pdf $$F(X)=\{\begin{array}{llc}0,& & \text{if }x<0;\\ 0.1,& & \text{if }0\le x<1;\\ 0.2,& & \text{if }1\le x<2;\\ 0.4,& & \text{if }2\le x<3;\\ 0.7,& & \text{if }3\le x<4;\\ 0.85,& & \text{if }4\le x<5;\\ 0.95,& & \text{if }5\le x<6;\\ 1,& & \text{if }6\le x;\end{array}$$

• simply by $p(x)=F(x)-F(x-1)$
• and $p(x_1)=F(x_1)$

• $E(X)={\sum }_{x}x{p}_X(x)$
• aka $E[X],{\mu }_X$
• $X$: a random var
• but $E(x)$: a deterministic number

• and some of weight: 1

• and $w,x,y,z$ for the values

1. weighted average of possible values of $X$
   • weight: $P(X=x)$
2. measure of central location of r.v. $X$
3. expectation of average val of r.v. over large no. of experiments
   • further connected w/ ch. 8: law of large numbers

• and $P(X=0)=1-p$ and $P(X=1)=p$
• random variable: called Bernoulli r.v. of parameter $p$
• denoted by $X\sim Be(p)$
• $E(X)=p$

• i.e. ${\sum }_{i=1}^6\frac{i}{6}=\frac{7}{2}=3.5$

• if: boy-girl chance are independent & equally likely
• then how many children should this couple expect?
• assume: get sex 1 for first try
• and let $E(X^{\prime })$: no. of tries until getting sex $2$
• $E(X)=1+E(X^{\prime })$
• $E(X^{\prime })=\frac{1}{2}\cdot 1+\frac{1}{2}(1+E(X^{\prime }))=1+\frac{1}{2}E(X^{\prime })$
• $\frac{1}{2}E(X^{\prime })=1$
• $E(X^{\prime })=2$
• $E(X)=1+E(X^{\prime })=3$
• prof's way
  1. $p(i)$: prob. that first $i-1$ children are girls, and then boy
  2. compute $E(X)$ $$\begin{array}{rl}E(X)& =\sum _{i=2}^{\infty }i{P}_X(i)\\ & =\sum _{i=2}^{\infty }i(\frac{1}{2}{)}^{i-1}\\ & =\sum _{i=1}^{\infty }i(\frac{1}{2}{)}^{i-1}-1\end{array}$$
  • and, from calculus:
    • $1+x+x^2+x^3+\cdots ={\sum }_{i=1}^{\infty }{x}^i=\frac{1}{1-x}$ if $|x|<1$
    • $0+1+2x+3x^2+\cdots ={\sum }_{i=1}^{\infty }i{x}^i=\frac{1}{(1-x{)}^2}$
      • derivation
    • as $x=1/2$, $E(X)=\frac{1}{(1-x{)}^2}-1=4-1=3$

if $P(X=-1)=0.2,P(X=0)=0.5,P(X=1)=0.3$

• $Y=X^2$
• then $P(Y=1)=P(X=-1)+P(X=1)=0.5$

given $X$ and function $g(X)$, how can we compute $E[g(X)]$?

• standard: compute pmf of $g(X)$, and compute $E[g(X)]$ by definition
  • $Y\in \{0,1\}$
  • $P(Y=0)=0.5$
  • $P(Y=1)=0.5$
  • $E(Y)=0.5$
• alternatively: you can break it down $$\begin{array}{rl}E(Y)=E(X^2)& =0^2\times 0.5+1^2\times 0.5\\ & =0^2\times 0.5+1^2\times (0.2+0.3)\\ & =0^2\times p_X(0)+(-1{)}^2\times p_X(-1)+1^2\times p_X(1)\\ & =0\times p_X(0)+1\times p_X(-1)+1\times p_X(1)\end{array}$$
• in general: $E(g(x))=\sum g(x)p_X(x)$
• 👨‍🏫 no need to compute pmf for $g(x)$ separately

theorem: r.v. $X$ w/ value $x_i$ for $i\ge 1$ for $p_X(x_i)$

• then for any real value function $g$

$$\begin{array}{rl}E[g(X)]& =\sum _{i}g(x_i)p_X(x_i)\\ & =\sum _{x}g(x_i)p_X(x)\end{array}$$

proof:

• idea: group together all terms in ${\sum }_{i}g(x_i)p(x_i)$ w/ same value for $g(x_i)$, then represent it as sum per each distinct value of $y_j=g(x_i)$

$$\begin{array}{rl}\sum _{i}g(x_i)p(x_i)& =\sum _j\sum _{i:g(x_i)=y_j}g(x_i)p(x_i)=\sum _j\sum _{i:g(x_i)=y_j}{y}_{j}p(x_i)\\ & =\sum _j{y}_j\sum _{i:g(x_i)=y_j}p(x_i)=\sum _j{y}_{j}P(g(X)=y_j)\\ & =E[g(X)]\end{array}$$

$E(X^2)={\sum }_x{x}^2{p}_X(x)$

• is called a second moment of $X$
• $E(X^k)$ for $k\ge 1$: called k-th moment of $X$

let $\mu =E(X),g(x)=(x-\mu {)}^k$ then

• $E(X-\mu {)}^k=E[(X-\mu {)}^k]$ (notation)
• is called $k$-th central moment

remarks

• $E(X)$: aka first moment / mean of $X$
• first central moment: $0$
• second central moment: $E[(X-\mu {)}^2]$
  • aka variance of $X$

let $a,b$: constants, then

• $E[aX+b]=aE(X)+b$
• 👨‍🏫 expectation: a linear operator

proof:

• $g(x)=ax+b$ $$\begin{array}{rl}E[aX+b]& =\sum _x[ax+b]p(x)\\ & =\sum _x[axp(x)+bp(x)]\\ & =\sum _{x}axp(x)+\sum _{x}bp(x)\\ & =a\sum _{x}xp(x)+b\sum _{x}p(x)\\ & =aE(X)+b\end{array}$$