STT 861 Theory of Prob and STT I Lecture Note  3
20170920
Random variable, independent random variable and their examples; Bernoulli distribution, Binomial distribution.
Portal to all the other notes
 Lecture 01  2017.09.06
 Lecture 02  2017.09.13
 Lecture 03  2017.09.20 > This post
 Lecture 04  2017.09.27
 Lecture 05  2017.10.04
 Lecture 06  2017.10.11
 Lecture 07  2017.10.18
 Lecture 08  2017.10.25
 Lecture 09  2017.11.01
 Lecture 10  2017.11.08
 Lecture 11  2017.11.15
 Lecture 12  2017.11.20
 Lecture 13  2017.11.29
 Lecture 14  2017.12.06
Lecture 03  Sept 20 2017
Example 1.3.8
Chevalier de Méré.
Game #1: throw a sixsided die 4 times. Win if get a 6 at least once.
Game #2: throw two sixsided dice 24 times. Win if get double 6 at least one.
Find their probabilities.
Answer:
\[\begin{align*} P(\textrm{win at game 1})&=1P(\textrm{lose at game 1}) \\ &= 1(\frac{5}{6})^4 \\ &=0.5177 \end{align*}\] \[\begin{align*} P(\textrm{win at game 2})&=1P(\textrm{lose at game 2}) \\ &= 1(\frac{35}{36})^{24} \\ &=0.4914 \end{align*}\]The numbers are close but different.
Blaise Pascal – First calculator
Example 1.3.9
Try it at home
Random Variables (RV or r.v.)
Definition: A r.v. $X$ is a function from a prob space $\Omega$ to a set of numbers $N$. If $N$ is a subset of integer $\mathbb{N}$ then we say that X is discrete; If $N$ is a subset of the real number $\mathbb{R}$, then we saythat $X$ is continuous.
Example 1
What is the chance that it will take $k$ tries until one success (e.g. one head in coin tosses). Assume $P(1 success)=p$.
Answer:
The sequence of trials leading to this event is:
00…0001, $k1$ fails, 1 success.
Because of the independence. The prob of that event
\[P(that~event)=(1p)^{k1}p\]This shows that a great choice for our probability space is $\Omega={\omega_1, \omega_2, … ,\omega_k,…}$, where $\omega_k$ is the elementary outcome “it takes $k$ trials until the first success”.
Next question: find the prob. $P(\omega_k)$ for every $k$.
Answer:
\[P(\omega_k)=p(1p)^{k1}\]Let’s check the axioms of prob. for our prob measure $P$:
\[P(A\cup B)=P(A)+P(B)\]automatically satisfied.
\[P(A)\geq0\]automatically satisfied, since $P(\omega_k)\geq0$.
\[P(\Omega)=1?\] \[\begin{align*} \sum_{k=1}^{\infty}P[\omega_k]&=\sum_{k=1}^{\infty}p(1p)^{k1}\\ &= p\sum_{k=0}^{\infty}(1p)^{k'}\\ &= p\frac{1}{1(1p)}=p\frac{1}{p}=1 \end{align*}\](because it is the sum of the whole geometry series with ratio $1p$) thus the last axiom is satisfied.
Note: looking forward to Chapter 1.4, let $X$ be the # of trails needed until the first success. Let $\Omega$ be the space of all sequence of successes and failures.
$X$ is defined on this $\Omega$ via the formula:
$X=k$ if $\omega=\omega_k$, $X(\omega_k)=k$.
\[P[X=k]=P(\omega_k)=p(1p)^{k1}\]This defines the prob distribution of the r.v. $X$ be sum of all the values $P[X=k]$ is 1. We say that $p_k=P(X=k)$ is the prob mass function of $X$.
Example from Chapter 1.3

Disease: $P(D)=0.006$, fairly rare

Test for this disease. $V$= a person is pos for this test.
Given $P(VD)=0.98$ sensitivity
Given $P(VD^C)=0.01$ false positive. This is e.q. to $(1P(V^CD^C)=0.99)$ specificity.
Q1: find $P(V)$.
A1:
\[\begin{align*} P(V) &= P(V\cap D)+P(V\cap D^C) \\ &=P(VD)P(D)+P(VD^C)P(D^C) \\ &=0.98\times 0.006 + 0.01\times(10.006) \\ &=0.00588+0.00994=0.01582 \end{align*}\]Q2: find $P(DV)$
A2: Use the definition of conditional probability
\[P(DV)=\frac{P(V\cap D)}{P(V)}=\frac{0.00588}{0.01582}=0.3717\]Example 2
2 dice, so makes same for $\Omega$ to have 36 elementary outcomes. Assume dice are fair & the two tosses are independence. Then let $X$ be sum of the outcome of the 2 dice.
Possible outcome for $X$ and their prob: See Homework 1.
The notation $X=x$ is really an event & we add up the prob of the individual $\omega$’s inside that event to find $p_x$.
We also say that $p_x$ define the prob distribution of $X$.
Independent RV
Definition: $X$ & $Y$ are indenpendent r.v. if
\[P(\{X=x\}\cap \{Y=y\})=P(\{X=x\})\times P( \{Y=y\})=p_x\times p_y\]Note: the pair $(X,Y)$ is known as a bivariate r.v. we use the notation $p_{x,y}$ for $P(X=x \& Y=y)$. The $p_{x,y}$ is the prob mass function of
Note: if $X$ & $Y$ are not independent, then $p_{x,y}$ not always equal to $p_xp_y$.
Example 3
Let $X$ be result of one die toss, let $Y$ be the result of another die.
If the two tosses do not influence each other, then the two r.v.s can be legitimately be modeled as independent.
Special discrete distributions
Bernoulli Distribution
Definition: A r.v. $X$ is called a Bernoulli r.v. with parameter $p$ if $X=0$ with prob $1p$, and $X=1$ with prob $p$.
$p$ is also called success probability. Indeed 0 & 1 can model the failure & success outcomes of a trail. The PMF of $X$ is $p_0=1p$ and $p_1=p$.
Binomial Distribution
Definition: A r.v. $X$ is a Binomial r.v. with parameters $n$ & $p$ if it is the number of successes in a sequence of $n$ independent Bernoulli trails.
“Exactly $x$ heads in 10 coin tosses.”
Here we see that if we denote by $x_1, x_2, x_3, …, x_n$ the corresponding sequence of Bernoulli r.v.
\[X=\sum_{i=1}^{n}x_i\](This is definition)
Indeed, all the failures count as 0 in the above sum and all the successes count as 1.
Big question: It’s clear that $X$ takes on the values $x=0,1,2,…,n$ and none others. But what is the PMF of $X$ for all those possible $x$’s.
Need to think about the event $\{X=k\}$. We can determine each elementary outcomes in the event $\{X=k\}$ by choosing the position of the $k$ “ones” in a sequence of length $n$. We choose a subset of size $k$ in a set of size $n$. That’s one of the outcomes in $\{X=k\}$. There are $C^k_n$ ways of doing so. (experiments are independence and the fact that multiplication is commutative.)
\[P(X=k)=C^k_np^k(1p)^{nk}\](This is theorem)
Note:
\[\sum_{k=0}^{n}C_n^kp^k(1p)^{nk}=1\]Formula:
\[(a+b)^n=\sum_{k=0}^{n}a^nb^{nk}\]Notation: $X\sim Bernoulli(p)$, $X\sim Binom(n,p)$.
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