• $E(X)={\sum }_{k=1}^{\infty }P(X\ge k)={\sum }_{k=0}^{\infty }P(X>k)$
• tail sum formula fo expectation

• $\text{var}(X)=E(X-\mu {)}^2$
• i.e. measure of scattering / spread of the values of $X$

• $\sigma )X=\sqrt{\text{var}(X)}$

• w/ $X$ discrete r.v., $E(X)=\mu$ $$\begin{array}{rl}\text{var}(X)& =E(x-\mu {)}^2=\sum _x(x-\mu {)}^{2}p(x)\\ & =\sum _x(x^2-2x\mu +{\mu }^2)p(x)\\ & =\sum _x{x}^{2}p(x)-2x\mu p(x)+{\mu }^{2}p(x)\\ & =\sum _x{x}^{2}p(x)-\sum _{x}2x\mu p(x)+\sum _x{\mu }^{2}p(x)\\ & =\sum _x{x}^{2}p(x)-2\mu \sum _{x}xp(x)+{\mu }^2\sum _{x}p(x)\\ & =\sum _x{x}^{2}p(x)-2\mu E(x)+{\mu }^2\\ & =\sum _x{x}^{2}p(x)-2E(x{)}^2+E(x{)}^2\\ & =\sum _x{x}^{2}p(x)-E(x{)}^2\\ & =E(x^2)-E(x{)}^2\end{array}$$

• $\text{var}(X)\ge 0$ as it's squared
• $\text{var}(X)=0$ iff $X$ is a degenerate r.v.
  • i.e. taking only one value: mean
  • special case of some inequality
• following tha formula: $E(X^2)\ge [E(X){]}^2\ge 0$

• similarly, just sqrt at each side

calculate $\text{var}(X)$ is $X$: outcome when a fair die is rolled

consider the prob. distribution on stock $A,B$

• find expected return:

• $E(A)=0.2\cdot 0.01+0.3\cdot 0.02+0.3\cdot 0.03+0.2\cdot 0.04=2.5\%$
• $E(B)=0.2\cdot 0.10+0.3\cdot 0.06+0.3\cdot 0.02-0.2\cdot 0.02=4\%$
• find variance & SD of return
  • $\text{var}(X)=E(X^2)-({\mu }_x{)}^2$

1. each trial results in an event, or not

• occurrence of event: success
• non-occurrence of event: failure
• $p:=P(\text{success})$
• $q:=1-p=P(\text{failure})$

2. each trial w/ success probability $p$, failure with $q=1-p$
3. repeating trials independently

• such trial: Bernoulli (p) trials

• performing experiment only once: $$X=\{\begin{array}{llc}1& & \text{success}\\ 0& & \text{failure}\end{array}$$

  • $P(X=1)=p$, $P(X=0)=1-p$
  • $E(X)=p$; $\text{var}(X)=p(1-p)$
    • as $E(X^2)-(E(X){)}^2=p-p^2$

• denote by $Be(p)$

• aka $X\sim Be(p)$

• aka $X\sim Ber(p)$

• aka $X\sim Bernoulli(p)$

• distribution of Bernoulli r.v.: Bernoulli distribution

• experiment $n$ times and define
  • $X=$ no. of success in $n$ $Be(p)$ trials
  • $P(X=k)=\left(\genfrac{}{}{0ex}{}{n}{k}\right)p^k{q}^{n-k}$
• denoted by $Bin(n,p)$
  • binomial r.v., Binomial distribution
  • $X\sim Bin(n,p)$
• is it a valid pmf?
  • i.e. show ${\sum }_{k=0}^{n}p(X=k)=1$ $$\begin{array}{rl}\sum _{k=0}^{n}p(X=k)& =\sum _{k=0}^n\left(\genfrac{}{}{0ex}{}{n}{k}\right)p^k{q}^{n-k}\\ & =(p+q{)}^n\\ & =(1{)}^n\\ & =1\end{array}$$

• let $X$: no. of $Be(p)$ trials to obtain the first success
• $P(X=k)=p{q}^{k-1}$
• and $E(X)=\frac{1}{p}$
• and $\text{var}(X)=\frac{1-p}{p^2}$
• denoted by $Geom(p)$

• ${\sum }_{i=0}^{\infty }{x}^i=\frac{1}{1-x}$ for $|x|<1$
• ${\sum }_{i=1}^{\infty }i{x}^{i-1}=\frac{1}{(1-x{)}^2}$ for $|x|<1$
• ${\sum }_{i=2}^{\infty }i(i-1)x^{i-2}=\frac{2}{(1-x{)}^3}$ for $|x|<1$

• $X^{\prime }$: no. of failures in $Be(p)$ trials
  • in order to obtain the first success
• and $X=X^{\prime }+1$
• $P(X^{\prime }=k)=p{q}^k$ for $k=0,1,\dots$
• $E(X^{\prime })=\frac{1-p}{p}$
• $\text{Var}(X^{\prime })=\frac{1-p}{p^2}$
• conventionally: geometric seq. means $X$ (at least in this course)

• $X$: no. of $Be(p)$ to get $r$ success
• $X=r,r+1,\dots$
• $P(X=k)=\left(\genfrac{}{}{0ex}{}{k-1}{r-1}\right)p^r{q}^{k-r}$
  • $k\ge r$
  • 👨‍🎓 better: $\left(\genfrac{}{}{0ex}{}{k-1}{r-1}\right)p^{r-1}{q}^{(k-1)-(r-1)}p$
• $E(X)=\frac{r}{p}$
• $\text{var}(X)=\frac{r(1-p)}{p^2}$
• denoted by: $NB(r,p)$
• $Geom(p)=NB(1,p)$
• 👨‍🏫 it's called negative binomial, because it is extended from negative binomial theorem
  • $(x+y{)}^{-m}$
  • or: $\left(\genfrac{}{}{0ex}{}{k-1}{r-1}\right)=(-1{)}^{r-1}(\genfrac{}{}{0ex}{}{-(k-r+1)}{r-1})$