diff --git a/source/docinfo.ptx b/source/docinfo.ptx index fb0b305..95d059c 100644 --- a/source/docinfo.ptx +++ b/source/docinfo.ptx @@ -13,6 +13,18 @@ \newcommand{\Z}{\mathbb Z} \newcommand{\Q}{\mathbb Q} \newcommand{\R}{\mathbb R} + + + \DeclareMathOperator{\Bin}{Bin} + \DeclareMathOperator{\Geom}{Geom} + \DeclareMathOperator{\Poiss}{Poiss} + \DeclareMathOperator{\Exp}{Exp} + + + \DeclareMathOperator{\E}{E} + \DeclareMathOperator{\Var}{Var} + \DeclareMathOperator{\Cov}{Cov} + diff --git a/source/main.ptx b/source/main.ptx index 815c739..d4e6d06 100644 --- a/source/main.ptx +++ b/source/main.ptx @@ -24,6 +24,7 @@ + diff --git a/source/notes/1-15.ptx b/source/notes/1-15.ptx index e79e0cf..3acd9f6 100644 --- a/source/notes/1-15.ptx +++ b/source/notes/1-15.ptx @@ -1,7 +1,7 @@
- Thursday 1/15 + Thursday, Jan 15

diff --git a/source/notes/1-20.ptx b/source/notes/1-20.ptx index aec8516..ba7c803 100644 --- a/source/notes/1-20.ptx +++ b/source/notes/1-20.ptx @@ -1,7 +1,7 @@

- Tuesday 1/20 + Tuesday, Jan 20

@@ -25,7 +25,7 @@

- The idea is that X is a variable representing a real number value which depends on the outcome of an experiment. + The idea is that X is a variable representing a real number value which depends on the outcome of an experiment.

@@ -33,7 +33,144 @@

An experiment consists of planting 50 seeds in a garden, then growing them for 3 months. Let H_i be the height of plant i. - + Let D be the number of seeds that didn't sprout. + Let + + A \amp = \text{avg height of all 50 plants} + \amp = \frac{H_1 + H_2 + \dotsb + H_{50}}{50} + +

+ + + +

+ A random variable has its own probability distribution. +

+ + + +

+ Roll a fair D6 twice. + Let S be the sum of the rolls. + Then \Omega = \{(1, 1), (1, 2), \dotsc, (6, 6)\} has 36 elements. + Since the die is fair, the distribution on \Omega is uniform, i.e., \Pr(\omega) = \frac{1}{36} for any \omega \in \Omega. +

+ +

+ S takes on the values 2, 3, 4, \dotsc, 12, with probabilities: +

+ + + Distribution for <m>S</m> + + + + x + \Pr(S = x) + + + + 2 + 1/36 + + + + 3 + 2/36 + + + + 4 + 3/36 + + + + \vdots + \vdots + + + + 7 + 6/36 + + + + 8 + 5/36 + + + + \vdots + + + + + 12 + 1/36 + + +
+ +

+ Note that the distribution on \Omega is uniform, but the distribution on S is not. +

+ + + + + +

+ Let \Omega be a sample space and A \subset \Omega an event. + Let + + X = \begin{cases} 1 \amp x \in A \\ 0 \amp x \notin A \end{cases} + + X is called an indicator random variable, and we say "X indicates A". +

+ +

+ The distribution on X is: + + \Pr(X = 1) \amp = \Pr(\{ x \mid x \in A\}) = \Pr(A) + \Pr(X = 0) \amp = 1 - \Pr(A) + +

+ + + + + +

+ Suppose we flip a coin n times. + Let H_i indicate heads on flip i. + Let S be the total number of heads in all flips. + Then: + + S = H_1 + H_2 + \dotsb + H_{n} + + If p is the probability of the coin coming up heads on a flip, then S has the binomial distribution with parameters n, p. + We'll use the notation S \sim \Bin(n, p) and: + + b(k) = b(k; n, p) = \Pr(S = k). + +

+ +

+ Suppose the coin has p = 0.3 and we flip it n = 4 times. + Find b(2) = b(2; 4, 0.3). +

+ +

+ The relevant flip sequences are: + + THHT, HHTT, TTHH, THTH, HTHT, HTTH + + Each individual sequence has a probability of (0.3)(0.3)(0.7)(0.7) = 0.0441. + So the total probability is: + + b(2; 4, 0.3) = (6)(0.0441) = 0.2646. + + That is, (number of flip sequences)(probability of each sequence).