--- date: 2021-06-22 tags: [ web/quick-notes ] --- # 2021-06-22 ## 12:47 Tags: #higher-algebra/category-theory #homotopy [mapping cone](mapping%20cone.md) as a [pushout](pushout.md) and [mapping fiber](mapping%20fiber) as [pullback](pullback) : \begin{tikzcd} && {} \\ \\ {X \fiberprod{Y} Y^I} && \textcolor{rgb,255:red,92;green,92;blue,214}{X} && CX \\ \\ {Y^I} && \textcolor{rgb,255:red,92;green,92;blue,214}{Y} && {CX \disjoint_X Y} && {} \arrow[from=5-3, to=5-5] \arrow[from=3-5, to=5-5] \arrow["\iota", hook, from=3-3, to=3-5] \arrow["\lrcorner"{anchor=center, pos=0.125, rotate=180}, draw=none, from=5-5, to=3-3] \arrow[two heads, from=5-1, to=5-3] \arrow[from=3-1, to=3-3] \arrow[from=3-1, to=5-1] \arrow["\lrcorner"{anchor=center, pos=0.125}, draw=none, from=3-1, to=5-3] \arrow["f", color={rgb,255:red,92;green,92;blue,214}, from=3-3, to=5-3] \arrow[shorten <=21pt, dashed, two heads, from=3-1, to=5-3] \arrow[shorten >=20pt, dashed, hook, from=3-3, to=5-5] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=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) - $M_f\to X\to Y$ 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. ## Probability Review Tags: #analysis/probability #projects/review #undergraduate \begin{tikzcd}[ampersand replacement=\&] \& {} \& \mu \& {\sigma^2} \& {f(x)} \\ {} \&\&\&\&\& {} \\ {\operatorname{Uniform}(n)} \&\& {{1\over 2}(n+1)} \& {{1\over 2}(n^2-1)} \& {{1\over n}} \\ {\operatorname{Uniform}(a, b)} \&\& {{1\over 2}(a+b)} \& {{1\over 12}(b-a)^2} \& {{1\over b-a}} \\ {\operatorname{Bernoulli}(p)} \&\& p \& {p(1-p)} \& {\fourcase{p}{x=1}{1-p}{x=0}} \\ {\operatorname{Binomial}(n, p)} \& {} \& np \& {np(1-p)} \& {{n\choose x}p^x (1-p)^{nx}} \\ {\operatorname{Geometric}(p)} \&\& {{1\over p}} \& {{1-p \over p^2}} \& {(1-p)^{x-1}p} \\ {\operatorname{Poisson}(\lambda t)} \&\& {\lambda t} \& {\lambda t} \& {(\lambda t)^x e^{-\lambda t} \over x!} \\ {\operatorname{Exponential}(\lambda)} \&\& {{1\over \lambda}} \& {{1\over \lambda^2}} \& {\lambda e^{-\lambda x}} \\ \& {} \arrow[no head, from=2-1, to=2-6] \arrow[no head, from=10-2, to=1-2] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=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) - Binomial and Poisson distributions tend to normal distributions? - Normal distribution rule: 68, 95, 99.7. - \[ \EE[r(x)] \da \int r(x) f(x) \dx && \text{or } \sum_k r(k) P(x=k) .\] - $\Var(x) = E[x^2] - E^2[x]$ - Marginal densities: if $f(x, y)$ is a distribution, \[ f_x(x) \da \int f(x, y) \dy && P(x\times y \in A) = \iint_A f(x, y) \dx \dy .\] - Independent if $f(x, y) = f_x(x) f_y(y)$. - If $x$ has density $f_x$ and $y \da r(x)$ for a continuous increasing function, setting $s \da r\inv$, $y$ has density \[ g_y(y) = f_x(s(y)) s'(y) .\] - 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 $\operatorname{Exponential}(\lambda)$. ## Combinatorics Review Tags: #projects/review #undergraduate - **Prop**: the number of [integer compositions](integer%20compositions) 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](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](integer%20partitions). - [Stirling numbers](Stirling%20numbers) : 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](partitions) : \[s(n+1, k) = s(n, k-1) + ks(n, k).\] - Proof: check if a given partition of $n+1$ into $k$ parts contains the singleton $\ts{n+1}$. - If so, delete and count partitions of $n$ into $k-1$ parts. - If not, add $\ts{n+1}$ to any of the $k$ parts.