auto-update(nvim): 2025-02-10 03:00:55

documents/by-course/pstat-120a/course-notes/main.typ

@@ -12,10 +12,7 @@
 
 = Introduction
 
-PSTAT 120A is an introductory course on probability and statistics. However, it
-is a theoretical course rather an applied statistics course. You will not learn
-how to read or conduct real-world statistical studies. Leave your $p$-values at
-home, this ain't your momma's AP Stats.
+PSTAT 120A is an introductory course on probability and statistics with an emphasis on theory.
 
 = Lecture #datetime(day: 6, month: 1, year: 2025).display()
 
@@ -772,8 +769,6 @@ us generalize to more than two colors.
   Both approaches given the same answer.
 ]
 
-= Discussion section #datetime(day: 22, month: 1, year: 2025).display()
-
 = Lecture #datetime(day: 23, month: 1, year: 2025).display()
 
 == Independence
@@ -1144,6 +1139,244 @@ exactly one sequence that gives us success.
   $
 ]
 
+= Notes on textbook chapter 3
+
+Recall that a random variable $X$ is a function $X : Omega -> RR$ that gives
+the probability of an event $omega in Omega$. The _probability distribution_ of
+$X$ gives its important probabilistic information. The probability distribution
+is a description of the probabilities $P(X in B)$ for subsets $B in RR$. We
+describe the probability density function and the cumulative distribution
+function.
+
+A random variable $X$ is discrete if there is countable $A$ such that $P(X in
+A) = 1$. $k$ is a possible value if $P(X = k) > 0$.
+
+A discrete random variable has probability distribution entirely determined by
+p.m.f $p(k) = P(X = k)$. The p.m.f. is a function from the set of possible
+values of $X$ into $[0,1]$. Labeling the p.m.f. with the random variable is
+done by $p_X (k)$.
+
+By the axioms of probability,
+
+$
+  sum_k p_X (k) = sum_k P(X=k) = 1
+$
+
+For a subset $B subset RR$,
+
+$
+  P(X in B) = sum_(k in B) p_X (k)
+$
+
+Now we introduce another major class of random variables.
+
+#definition[
+  Let $X$ be a random variable. If $f$ satisfies
+
+  $
+    P(X <= b) = integral^b_(-infinity) f(x) dif x
+  $
+
+  for all $b in RR$, then $f$ is the *probability density function* of $X$.
+]
+
+The probability that $X in (-infinity, b]$ is equal to the area under the graph
+of $f$ from $-infinity$ to $b$.
+
+A corollary is the following.
+
+#fact[
+  $ P(X in B) = integral_B f(x) dif x $
+]
+
+for any $B subset RR$ where integration makes sense.
+
+The set can be bounded or unbounded, or any collection of intervals.
+
+#fact[
+  $ P(a <= X <= b) = integral_a^b f(x) dif x $
+  $ P(X > a) = integral_a^infinity f(x) dif x $
+]
+
+#fact[
+  If a random variable $X$ has density function $f$ then individual point
+  values have probability zero:
+
+  $ P(X = c) = integral_c^c f(x) dif x = 0, forall c in RR $
+]
+
+#remark[
+  It follows a random variable with a density function is not discrete. Also
+  the probabilities of intervals are not changed by including or excluding
+  endpoints.
+]
+
+How to determine which functions are p.d.f.s? Since $P(-infinity < X <
+infinity) = 1$, a p.d.f. $f$ must satisfy
+
+$
+  f(x) >= 0 forall x in RR \
+  integral^infinity_(-infinity) f(x) dif x = 1
+$
+
+#fact[
+  Random variables with density functions are called _continuous_ random
+  variables. This does not imply that the random variable is a continuous
+  function on $Omega$ but it is standard terminology.
+]
+
+#definition[
+  Let $[a,b]$ be a bounded interval on the real line. A random variable $X$ has
+  the *uniform distribution* on $[a,b]$ if $X$ has density function
+
+  $
+    f(x) = cases(
+      1/(b-a)", if" x in [a,b],
+      0", if" x in.not [a,b]
+    )
+  $
+
+  Abbreviate this by $X ~ "Unif"[a,b]$.
+]
+
+= Notes on week 3 lecture slides
+
+== Negative binomial
+
+Consider a sequence of Bernoulli trials with the following characteristics:
+
+- Each trial success or failure
+- Prob. of success $p$ is same on each trial
+- Trials are independent (notice they are not fixed to specific number)
+- Experiment continues until $r$ successes are observed, where $r$ is a given parameter
+
+Then if $X$ is the number of trials necessary until $r$ successes are observed,
+we say $X$ is a *negative binomial* random variable.
+
+#definition[
+  Let $k in ZZ^+$ and $0 < p <= 1$. A random variable $X$ has the negative
+  binomial distribution with parameters ${k,p}$ if the possible values of $X$
+  are the integers ${k,k+1, k+2, ...}$ and the p.m.f. is
+
+  $
+    P(X = n) = vec(n-1, k-1) p^k (1-p)^(n-k) "for" n >= k
+  $
+
+  Abbreviate this by $X ~ "Negbin"(k,p)$.
+]
+
+#example[
+  Steph Curry has a three point percentage of approx. $43%$. What is the
+  probability that Steph makes his third three-point basket on his $5^"th"$
+  attempt?
+
+  Let $X$ be number of attempts required to observe the 3rd success. Then,
+
+  $
+    X ~ "Negbin"(k = 3, p = 0.43)
+  $
+
+  So,
+  $
+    P(X = 5) &= vec(5-1,3-1)(0.43)^3 (1 - 0.43)^(5-3) \
+    &= vec(4,2) (0.43)^3 (0.57)^2 \
+    &approx 0.155
+  $
+]
+
+== Poisson distribution
+
+This p.m.f. follows from the Taylor expansion
+
+$
+  e^lambda = sum_(k=0)^infinity lambda^k / k!
+$
+
+which implies that
+
+$
+  sum_(k=0)^infinity e^(-lambda) lambda^k / k! = e^(-lambda) e^lambda = 1
+$
+
+#definition[
+  For an integer valued random variable $X$, we say $X ~ "Poisson"(lambda)$ if it has p.m.f.
+
+  $ P(X = k) = e^(lambda) lambda^k / k! $
+
+  for $k in {0,1,2,...}$ for $lambda > 0$ and
+
+  $
+    sum_(k = 0)^infinity P(X=k) = 1
+  $
+]
+
+The Poisson arises from the Binomial. It applies in the binomial context when
+$n$ is very large ($n >= 100$) and $p$ is very small $p <= 0.05$, such that $n
+p$ is a moderate number ($n p < 10$).
+
+Then $X$ follows a Poisson distribution with $lambda = n p$.
+
+$
+  P("Bin"(n,p) = k) approx P("Poisson"(lambda = n p) = k)
+$
+
+for $k = 0,1,...,n$.
+
+#example[
+  The number of typing errors in the page of a textbook.
+
+  Let
+
+  - $n$ be the number of letters of symbols per page (large)
+  - $p$ be the probability of error, small enough such that
+  - $lim_(n -> infinity) lim_(p -> 0) n p = lambda = 0.1$
+
+  What is the probability of exactly 1 error?
+
+  We can approximate the distribution of $X$ with a $"Poisson"(lambda = 0.1)$
+  distribution
+
+  $
+    P(X = 1) = (e^(-0.1) (0.1)^1) / 1! = 0.09048
+  $
+]
+
+#example[
+  The number of reported auto accidents in a big city on any given day
+
+  Let
+
+  - $n$ be the number of autos on the road
+  - $p$ be the probability of an accident for any individual is small such that
+    $lim_(n->infinity) lim_(p->0) n p = lambda = 2$
+
+  What is the probability of no accidents today?
+
+  We can approximate $X$ by $"Poisson"(lambda = 2)$
+
+  $
+    P(X = 0) = (e^(-2) (2)^0) / 0! = 0.1353
+  $
+]
+
+A discrete example:
+
+#example[
+  Suppose we have an election with candidates $B$ and $W$. A total of 10,000
+  ballots were cast such that
+
+  $
+    10,000 "votes" cases(5005 space B, 4995 space W)
+  $
+
+  But 15 ballots had irregularities and were disqualified. What is the
+  probability that the election results will change?
+
+  There are three combinations of disqualified ballots that would result in a
+  different election outcome: 13 $B$ and 2 $W$, 14 $B$ and 1 $W$, and 15 $B$
+  and 0 $W$. What is the probability of these?
+]
+
 = Lecture #datetime(day: 3, month: 2, year: 2025).display()
 
 == CDFs, PMFs, PDFs