TOP Notes until first midterm.

Author: localhost433

notes/courses/MATH-UA-333/01-combinatorics.md

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+---
+title: Basic Combinatorics
+date: 2025-09-04
+---
+
+## Counting principles
+
+Let $n_1,\dots,n_k$ be the numbers of options in $k$ independent choices (e.g. rice, protein, beans, vegetables). The multiplication principle says that the total number of possible outcomes is
+$$
+n_1 \times n_2 \times \cdots \times n_k.
+$$
+
+When different cases must be considered separately, the addition principle says that if there are $m_1,\dots,m_\ell$ possibilities in each case, then the total number is $m_1 + \cdots + m_\ell$.  
+
+## Permutations
+
+A **permutation** of $n$ distinct objects is an ordering. There are $n! = n\times(n-1)\times\cdots\times 1$ ways to arrange $n$ objects.
+> Example: five people solving five exam problems can assign problems to people in $5!=120$ ways.
+
+## Combinations and binomial coefficients
+
+To choose $k$ objects from $n$ without regard to order, count the ordered selections and then divide by $k!$. The number of subsets of size $k$ is
+$$
+{n \choose k} = \frac{n!}{k!\,(n-k)!},
+$$
+
+called a **binomial coefficient**.
+> Example: from 30 students, the number of ways to choose 3 for a presentation is ${30 \choose 3}$.  
+
+The binomial theorem expands $(a+b)^n$ as
+$$
+(a+b)^n=\sum_{k=0}^{n} {n\choose k} a^{n-k} b^{k}.
+$$
+
+Useful identities include ${n \choose k}={n \choose n-k}$ and the recursive sum $\sum_{i=0}^{k-1} {n-i\choose k-i-1}={n\choose k}$.
+
+## Multinomial coefficients
+
+To divide $n$ items into $r$ groups of sizes $n_1,\dots,n_r$ with $n_1+\cdots+n_r=n$, the number of ways is 
+$$
+\frac{n!}{n_1!n_2!\cdots n_r!}.
+$$
+
+For instance, arranging 30 students into ten presentation groups of three has ${30 \choose 3,3,\dots,3} = \tfrac{30!}{(3!)^{10}}$ possibilities.  
+
+## Multiset permutations
+
+If a word has repeated letters, e.g.\ P,E,P,P,E,R, the number of distinct rearrangements is $\tfrac{6!}{3!\,2!}$; count all $6!$ permutations and divide by the $3!$ ways to reorder the three P’s and $2!$ ways to reorder the two E’s.
+
+## Stars and bars: integer partitions
+
+To distribute $n$ identical candies among $r$ kids so each gets at least one, find the number of positive integer solutions to $x_1+\cdots+x_r=n$. Imagine placing $r-1$ dividers into $n-1$ slots to separate $n$ candies; there are ${n-1 \choose r-1}$ solutions. If zero candies are allowed, give each child one candy first and then distribute the remaining $n+r-1$ candies; the answer becomes ${n+r-1\choose r-1}$.
+

notes/courses/MATH-UA-333/02-axioms.md

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+---
+title: Axioms of Probability
+date: 2025-09-06
+---
+
+## Sample space, events and set operations
+
+A **sample space** $S$ is the set of all possible outcomes, and an **event** is a subset of $S$.  Sets can be combined using intersection ($A\cap B$), union ($A\cup B$), complement ($A^c$) and difference ($B\setminus A$); De Morgan’s laws $(A\cap B)^c=A^c\cup B^c$ and $(A\cup B)^c=A^c\cap B^c$ help relate complements.
+
+## Probability measure axioms
+
+A probability measure $P$ assigns numbers to events such that $0\le P(E)\le 1$, $P(\emptyset)=0$, $P(S)=1$, and $P(E_1\cup E_2\cup\cdots)=P(E_1)+P(E_2)+\cdots$ for disjoint events.  Probabilities can be interpreted as areas in a Venn diagram.
+
+### Identities and the inclusion–exclusion principle
+
+From additivity one derives useful identities:
+
+- Complement: $P(A^c)=1-P(A)$.
+- Decomposition: $P(B)=P(A\cap B)+P(B\setminus A)$.
+- Union: $P(A\cup B)=P(A)+P(B)-P(A\cap B)$, and more generally for $n$ events the inclusion–exclusion formula sums intersections of increasing size with alternating signs.  The union bound states $P(E_1\cup\cdots\cup E_n)\le P(E_1)+\cdots+P(E_n)$.
+
+## Uniform probability on finite spaces
+
+If $S$ has finitely many equally likely outcomes, then $P(E)=\frac{\#E}{\#S}$.
+> Example: two dice have $36$ equally likely outcomes; the event “sum = 7” contains six outcomes, so the probability is $6/36=1/6$.  Drawing 7 cards from a 52‑card deck, the probability of getting a four‑of‑a‑kind is 
+$$
+P(E)=\frac{13\cdot {48\choose 3}}{{52\choose 7}}.
+$$
+
+In the birthday problem with 23 people, the probability that at least two share a birthday is $1-\frac{365\cdot364\cdots(365-22)}{365^{23}}$, which exceeds $\tfrac{1}{2}$.
+
+## Examples and applications
+
+- **Coin‑flip experiment:** sample space $\{HH,HT,TH,TT\}$; event “flips differ” has two outcomes, so $P=2/4=1/2$.
+- **Presentation order:** the six permutations of three presenters form the sample space; the event “Alice goes first” has two outcomes, giving probability $1/3$.
+- **Derangements:** permuting $n$ students’ seats, let $A_k$ be the event that student $k$ stays put.  By inclusion–exclusion, the probability at least one student remains fixed is $\sum_{k=1}^n(-1)^{k-1}{n\choose k}\frac{(n-k)!}{n!}$.
+

notes/courses/MATH-UA-333/03-conditional-independence.md

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+---
+title: Conditional Probability and Independence
+date: 2025-09-11
+---
+
+## Conditional probability
+
+For events $E$ and $F$ with $P(F)>0$, the **conditional probability** of $E$ given $F$ is
+$$
+P(E\mid F)=\frac{P(E\cap F)}{P(F)}.
+$$
+
+Thinking of $F$ as a new sample space helps interpret $P(\cdot\mid F)$.  The product rule follows immediately: $P(E\cap F)=P(E\mid F)P(F)$, and more generally for events $E_1,\dots,E_n$ one can write
+$$
+P(E_1\cap\cdots\cap E_n)=P(E_1\mid E_2\cap\cdots\cap E_n)\,P(E_2\mid E_3\cap\cdots\cap E_n)\cdots P(E_n).
+$$
+
+The **law of total probability** states that if $F_1,\dots,F_m$ form a partition of $S$, then
+$$
+P(E)=\sum_{i=1}^m P(E\mid F_i)\,P(F_i).
+$$
+
+**Bayes’ formula** expresses the reversed conditional:
+$$
+P(F\mid E)=\frac{P(E\mid F)\,P(F)}{P(E)}=\frac{P(E\mid F)\,P(F)}{P(E\mid F)\,P(F)+P(E\mid F^c)\,P(F^c)}.
+$$
+
+> Examples
+> - Alice flips a fair coin to decide between two classes; the probability she gets an A overall is $0.9\cdot 0.5 + 0.8\cdot 0.5=0.85$.
+> - In a Covid test with 50% prevalence, sensitivity 96% and false‑positive rate 2%, if you test positive then $P(\text{Covid}\mid\text{positive}) = \frac{0.96\cdot 0.5}{0.96\cdot 0.5+0.02\cdot 0.5} = \frac{48}{49}$.
+> - From an urn containing two unusual dice, drawing and rolling a 3 gives probability $P(\text{dice A}\mid\text{roll 3})=2/3$.
+
+## Independence
+
+Events $E$ and $F$ are **independent** if 
+$$
+P(E\cap F)=P(E)\,P(F).
+$$
+This means that knowing $F$ does not change the probability of $E$.  Independence is distinct from disjointness (mutually exclusive events): independent events can both occur.  If $E$ and $F$ are disjoint and independent, then one must have probability zero.  Independence extends to families of events: events $A_1,\dots,A_n$ are *pairwise independent* if every pair is independent, and *mutually independent* if $P(A_{i_1}\cap\cdots\cap A_{i_k})=P(A_{i_1})\cdots P(A_{i_k})$ for every $k\ge 2$.  Pairwise independence does not imply mutual independence; for three coins, the events “first flip is heads,” “second flip is tails” and “all flips are the same” are pairwise independent but not mutually independent.
+
+## Applications
+
+- **Product law**: rolling two dice, the event “sum=8” and the event “first roll is 2” are not independent since $P(E\cap F)=\tfrac{1}{36}\ne \tfrac{5}{36}\cdot \tfrac{1}{6}$; but “sum=7” and “first roll is 2” are independent.
+- **Law of total probability and Bayes’ rule** appear in computing disease tests or decision trees as above.
+- **Repeated trials**: if you repeatedly roll a die, the probability that the first 6 appears on the $n$‑th roll is $(5/6)^{\,n-1}\,(1/6)$ and the probability of exactly $k$ successes in $n$ rolls is given by the binomial distribution${n \choose k}(1/6)^k(5/6)^{\,n-k}$.
+

notes/courses/MATH-UA-333/04-05-drv.md

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+---
+title: Discrete Random Variables
+date: 2025-09-16/18/23/25
+---
+
+## Expectation
+For a discrete random variable $X$ with p.m.f. $p_X(t)=P[X=t]$,
+$$
+\E[X] = \sum_{t} t p_X(t), \quad \E[f(X)] = \sum_t f(t) p_X(t),
+$$
+
+**Interpretations**
+- Weighted average over the values of $X$: $\sum_t t P[X=t]$.
+- Weighted average over outcomes $\omega$ in the sample space $S$: $\sum_{\omega\in S} X(\omega) P(\{\omega\})$.
+
+> Examples:
+> - Fair die $X\in\{1,\dots,6\}$: $\E[X]=3.5$.
+> - Heads in three fair flips $Y\in\{0,1,2,3\}$: $\E[Y]=1.5$.
+> - Indicator $\mathbf{1}_A$: $\E[\mathbf{1}_A]=P(A)$.
+
+## Linearity of expectation
+For random variables $X,Y$ and scalar $a\in\mathbb{R}$,
+$$
+\E[aX+Y] = a \E[X] + \E[Y].
+$$
+No independence required.
+
+**Counting-by-indicators trick**
+If $X=\sum_{i=1}^n \mathbf{1}_{A_i}$ counts how many events $A_i$ occur, then
+$$
+\E[X] = \sum_{i=1}^n P(A_i).
+$$
+> Example: number of fixed points in a random permutation - $\E[X]=1$.
+
+## Variance
+Variance measures spread around the mean:
+$$
+\operatorname{Var}(X) = \E[(X-\E X)^2] = \E[X^2] - (\E X)^2.
+$$
+Useful identities:
+- For constants $a,b$: $\operatorname{Var}(aX+b)=a^2 \operatorname{Var}(X)$.
+- For an indicator: $\operatorname{Var}(\mathbf{1}_A)=P(A)\big(1-P(A)\big)$.
+- If $X=\sum_{i=1}^n \mathbf{1}_{A_i}$ with independent indicators, then
+  $$
+  \operatorname{Var}(X)=\sum_{i=1}^n P(A_i)\big(1-P(A_i)\big).
+  $$
+
+## Inclusion–Exclusion via indicators
+Using $\mathbf{1}_{A^c} = 1 - \mathbf{1}_A$ and expanding
+$\mathbf{1}_{\cup_i E_i} = 1 - \prod_{i=1}^n (1-\mathbf{1}_{E_i})$, then taking expectations yields
+$$
+P\Big( \bigcup_{i=1}^n\ E_i \Big)
+= \sum_{k=1}^n (-1)^{k-1}
+\sum_{1 \leq i_1 < \cdots < i_k \leq n}
+P(\E_{i_1} \cap \cdots \cap \E_{i_k}).
+$$
+
+### Bernoulli
+
+If $X\in\{0,1\}$ takes value $1$ with probability $p$, we write $X\sim\operatorname{Ber}(p)$.  Its pmf is $p_X(1)=p$, $p_X(0)=1-p$; the expectation is $p$, and the variance is $p(1-p)$.
+
+### Binomial
+
+Let $X$ be the total number of successes in $n$ independent Bernoulli$(p)$ trials.  Then $X\sim \operatorname{Bin}(n,p)$ and can be written $X=X_1+\cdots+X_n$ with $X_i\sim\operatorname{Ber}(p)$.  The pmf is
+$$
+P(X=k) = {n\choose k} p^k (1-p)^{ n-k}.
+$$
+
+Using linearity of expectation, $\E[X]=\sum_{i=1}^n \E[X_i]=np$, and the variance is $\operatorname{Var}(X)=np(1-p)$.
+> Examples: the number of heads in 3 fair flips follows $\operatorname{Bin}(3,\tfrac12)$; when rolling 10 dice, the number of sixes follows $\operatorname{Bin}(10,\tfrac16)$.
+
+### Poisson
+
+A **Poisson** random variable $X\sim \operatorname{Poi}(\lambda)$ has pmf
+$$
+P(X=k) = \frac{\lambda^k e^{-\lambda}}{k!}, \quad k=0,1,2,\dots.
+$$
+
+From the series expansion of $e^\lambda$, these probabilities sum to one. The expectation and variance are both $\lambda$. A Poisson variable arises as an approximation to $\operatorname{Bin}(n,p)$ when $n$ is large and $p$ is small with $np=\lambda$. 
+
+> For example, the number of 5‑star characters obtained in 500 draws with a 0.6% rate is approximately $\operatorname{Poi}(3)$.
+
+### Geometric
+
+If you repeatedly flip a $p$‑biased coin until the first head appears, the total number of flips $X$ (including the successful flip) follows a **geometric** distribution with parameter $p$, denoted $X\sim \operatorname{Geom}(p)$.  The pmf is
+$$
+P(X=k) = (1-p)^{k-1} p,\quad k=1,2,\dots,
+$$
+
+since the first $k-1$ flips must be tails and the $k$‑th flip a head. Summation identities yield
+$$
+\E[X] = \frac{1}{p},\qquad \operatorname{Var}(X) = \frac{1-p}{p^2}.
+$$

notes/courses/MATH-UA-333/index.json

@@ -0,0 +1,25 @@
+[
+    {
+        "slug": "01-intro",
+        "title": "1 - Basic combinatorics",
+        "date": "2025-09-04"
+    }
+    ,
+    {
+        "slug": "02-axioms",
+        "title": "2 - Axioms of probability",
+        "date": "2025-09-06"
+    }
+    ,
+    {
+        "slug": "03-conditional-independence",
+        "title": "3 - Conditional Probability and Independence",
+        "date": "2025-09-11"
+    }
+    ,
+    {
+        "slug": "04-05-drv",
+        "title": "4/5 - Discrete random variables",
+        "date": "2025-09-16/18/23/25"
+    }
+]

notes/metadata/courses.json

@@ -7,10 +7,10 @@
   }
   ,
   {
-    "slug": "J-N5",
-    "title": "Japanese N5 Level",
-    "semester": "Nan",
-    "instructor": "Mr. Yano"
+    "slug": "MATH-UA-333",
+    "title": "Theory of Probability (MATH-UA 333)",
+    "semester": "Fall 2025",
+    "instructor": "Prof. Yu"
   }
   ,
   {
@@ -19,11 +19,4 @@
     "semester": "Fall 2025",
     "instructor": "Prof. Shatah"
   }
-  ,
-  {
-    "slug": "Convex",
-    "title": "Convex Optimization",
-    "semester": "NAN",
-    "instructor": "Self-study"
-  }
 ]