2021-06-22 – Notes

12:47

mapping cone as a pushout and mapping fiber as pullback :

Link to Diagram

\(M_f\to X\to Y\) yields a LES in homotopy \(\pi_* M_f \to \pi_* X \to \pi_* Y\).
\(X\to Y\to C_f\) yields \(\pi_*(Y, X) = \pi_{*-1} C_f\)?
- This should be an easy consequence of the LES in homotopy.

Link to Diagram

Binomial and Poisson distributions tend to normal distributions?
Normal distribution rule: 68, 95, 99.7.
\begin{align*} {\mathbb{E}}[r(x)] \coloneqq\int r(x) f(x) \,dx && \text{or } \sum_k r(k) P(x=k) .\end{align*}
\({\mathsf{Var}}(x) = E[x^2] - E^2[x]\)
Marginal densities: if \(f(x, y)\) is a distribution,
\begin{align*} f_x(x) \coloneqq\int f(x, y) \,dy && P(x\times y \in A) = \iint_A f(x, y) \,dx\,dy .\end{align*}
- Independent if \(f(x, y) = f_x(x) f_y(y)\).
If \(x\) has density \(f_x\) and \(y \coloneqq r(x)\) for a continuous increasing function, setting \(s \coloneqq r^{-1}\), \(y\) has density \begin{align*} g_y(y) = f_x(s(y)) s'(y) .\end{align*}
Given a failure rate \(\lambda\) in failures over time and a time span \(t\), the probability of failing \(k\) times in \(t\) units of time is \(\operatorname{Poisson}(\lambda t)\).
- The time between failures is \(\operatorname{Exponential}(\lambda)\).

Prop: the number of integer compositions of \(n\) into exactly \(k\) parts is \({n-1 \choose k-1}\), so \(\sum a_i = n\) with \(a_i\geq 1\) for all \(i\).
- Lay out \(n\) copies of 1 with \(n-1\) blanks between them. Choose to put a plus sign or a comma in each slot: choose \(k-1\) commas to produce \(k\) blocks.
Prop: the number of compositions of \(n\) into any number of parts is \(2^{n-1}\).
- Sum over compositions with exactly \(k\) parts and use the identity \(\sum_{k=1}^{n-1} {n-1 \choose k-1} = 2^{n-1}\).
Prop: the number of weak compositions of \(n\) into exactly \(k\) parts is \({n+k-1 \choose k-1}\).
- Add one to each piece to get a (strong) composition of \(n+k\) into \(k\) parts.
Quotient set of compositions by permutations to get integer partitions.
Stirling numbers : the number of ways to partition an \(n\) element set into exactly \(k\) nonempty unlabeled disjoint blocks whose disjoint union is the original set.
- Prop: recurrence for partitions :
  \begin{align*}s(n+1, k) = s(n, k-1) + ks(n, k).\end{align*}
  - Proof: check if a given partition of \(n+1\) into \(k\) parts contains the singleton \(\left\{{n+1}\right\}\).
    
    If so, delete and count partitions of \(n\) into \(k-1\) parts.
    
    If not, add \(\left\{{n+1}\right\}\) to any of the \(k\) parts.