Researcher in computer science, math, machine learning, (differential) privacy, AI, etc. at Anthropic. Kiwi🇳🇿 in California🇺🇸 http://stein.ke/
![[2] Marcus Hardt and Jonathan Ullman. Preventing false discovery in interactive data analysis is
hard. In Foundations of Computer Science (FOCS), 2014 IEEE 55th Annual Symposium on, pages
454–463. IEEE, 2014.
From: https://arxiv.org/abs/1510.03349](https://cdn.bsky.app/img/feed_thumbnail/plain/did:plc:thfnaqgktdwqbm42radc322i/bafkreiby7m2jg4i3jadjakltkpaqnttb23q22fyddpi6y5mt6sp4f7deye)
![Screenshot of bsky post with author's identity not shown. [I'm not trying to pile on.]
Sure, LLMs are useful for:
1. Fraud
2. Plagiarism
3. Cognitive off-loading
Which of those use-cases are you promoting?
3:59pm March 10 2026
51 reposts, 14 quotes, 257 likes, 6 saves](https://cdn.bsky.app/img/feed_thumbnail/plain/did:plc:thfnaqgktdwqbm42radc322i/bafkreih5vi5hu53uukdtj7l6bqxr4cgnayxyr22hxe6a7ydrpo6foy45zm)
![[Screenshot of an app]
Code Verification
Call customer service at 1-866-4220306 (outside the U.S. call 1-210-677-0065) to retrieve your One-time identification Code .](https://cdn.bsky.app/img/feed_thumbnail/plain/did:plc:thfnaqgktdwqbm42radc322i/bafkreiffo5kk6fobzmxoz3dnfwonqj36xejsutsj6wh6kuv4eqwzijh3gi)
![def modinv(a, p):
"""Computes inverse of a modulo p i.e. returns b such that (a*b)%p==1"""
u, v = a % p, p
x, y, w, z = 1, 0, 0, 1 # maintain ax+py==u and aw+pz==v
while u != 0: # extended Euclidean algorithm
q, r = v // u, v % u
x, y, w, z = w-qx, z-qy, x, y
u, v = r, u
if v != 1: return None # not invertible
return w % p
def modmatrixinverse(A, p):
"""Computes matrix inverse of A modulo p i.e. returns B s.t. A@B%p==I"""
A = numpy.array(A, dtype=int)
m, n = A.shape
assert m == n, "matrix must be square"
B = numpy.eye(n, dtype=int) # = nxn identity matrix
for i in range(n): # Gaussian elimination
b, j = None, i-1
while b is None:
j += 1
if j >= n: return None # not invertible
b = modinv(A[j,i], p)
if j > i: # swap rows i and j
A[j,:], A[i,:] = A[i,:], A[j,:]
B[j,:], B[i,:] = B[i,:], B[j,:]
A[i,:] = (A[i,:] * b) % p # divide A[i,:] by A[i,i]
B[i,:] = (B[i,:] * b) % p
for j in range(n): # zero out the rest of column i
if j != i:
B[j,:] = (B[j,:] - B[i,:] * A[j,i]) % p
A[j,:] = (A[j,:] - A[i,:] * A[j,i]) % p
return B](https://cdn.bsky.app/img/feed_thumbnail/plain/did:plc:thfnaqgktdwqbm42radc322i/bafkreifpv2rkjmgnwfgoaumjmxgv3gkd3rwufyinjvthca4dzcc2nbntwu)
![\textbf{H\"older's inequality from Jensen's inequality:}
\noindent
Let $x,y\in\mathbb{R}^d$ be arbitrary. Let $p,q \in (1,\infty)$ satisfy $\frac1p+\frac1q=1$. Without loss of generality, assume $x_i \ne 0$ for all $i \in [d]$.
Note that $v \mapsto |v|^p$ is convex.
For $i \in [d]$, define $u_i = |x_i|^q/\|x\|_q^q$ and $v_i = x_iy_i/|x_i|^q$. By Jensen's inequality,
\[\left|\frac{\sum_i x_i y_i}{\|x\|_q^q}\right|^p = \left|\sum_i u_i v_i\right|^p \le \sum_i u_i |v_i|^p = \frac{\sum_i |x_i|^{q+p-pq}|y_i|^p}{\|x\|_q^q} = \frac{\|y\|_p^p}{\|x\|_q^q}, \]
which rearranges to \[\left|\sum_i x_i y_i\right| \le \|x\|_q \cdot \|y\|_p.\]](https://cdn.bsky.app/img/feed_thumbnail/plain/did:plc:thfnaqgktdwqbm42radc322i/bafkreicxf5yshu7vcv6j625wl3ulltwhdsgxugxejil247ily2obs7n37i)
![Full LaTeX source: https://pastebin.com/mA6KjUJs
\begin{proposition}[Triangle-like inequality for KL divergence]\label{prop:kl-triangle}
Let $P$, $R$, and $Q$ be probability distributions with $P$ being absolutely continuous with respect to $R$ and $R$ being absolutely conotinuous with respect to $Q$.
Let $\kappa \in (1,\infty)$.
Then
\[
\dr{\text{KL}}{P}{Q} \le \frac{\kappa}{\kappa-1} \dr{\text{KL}}{P}{R} + \dr{\kappa}{R}{Q},
\]
where $\dr{\text{KL}}{P}{Q} := \ex{X \gets P}{\log(P(X)/Q(X)}$ denotes the KL divergence and\\$\dr{\kappa}{R}{Q} = \frac{1}{\kappa-1} \log \ex{X \gets R}{(R(X)/Q(X))^{\kappa-1}}$ denotes the R\'enyi divergence of order $\kappa$.
\end{proposition}](https://cdn.bsky.app/img/feed_thumbnail/plain/did:plc:thfnaqgktdwqbm42radc322i/bafkreibz3gfiqvyvbld7jgfnb6kzac2j6qrxirofpwrxiv5mpz6j7r6md4)
![LaTeX source: https://pastebin.com/qtw8DPh8
\begin{lemma}[Quotient Rule for Higher Derivatives] ~\\
Let $h(x) := \frac{f(x)}{g(x)}$ with $g(x) \ne 0$. Then, %$h'(x) = \frac{f'(x)g(x)-f(x)g'(x)}{g(x)^2}$. More generally,
for all $n \in \mathbb{N}$,
\[
\left(\begin{array}{c}
h(x) \\
h'(x) \\
h''(x) \\
h'''(x) \\
\vdots \\
h^{(k)}(x) \\
\vdots \\
h^{(n-1)}(x)
\end{array}\right)
=
\left(\begin{array}{ccccc}
g(x) & 0 & 0 & \cdots & 0 \\
g'(x) & g(x) & 0 & \cdots & 0 \\
g''(x) & 2 g'(x) & g(x) & \cdots & 0 \\
g'''(x) & 3 g''(x) & 3 g'(x) & \cdots & 0 \\
\vdots & \vdots & \vdots & \ddots & 0 \\
g^{(k)}(x) & {k \choose 1} g^{(k-1)}(x) & {k \choose 2} g^{(k-2)}(x) & \cdots & 0 \\
\vdots & \vdots & \vdots & \ddots & 0 \\
g^{(n-1)}(x) & {n-1 \choose 1} g^{(n-2)}(x) & {n-1 \choose 2} g^{(n-3)}(x) & \cdots & g(x)
\end{array}\right)^{-1}
\left(\begin{array}{c}
f(x) \\
f'(x) \\
f''(x) \\
f'''(x) \\
\vdots \\
f^{(k)}(x) \\
\vdots \\
f^{(n-1)}(x)
\end{array}\right) .
\]
\end{lemma}](https://cdn.bsky.app/img/feed_thumbnail/plain/did:plc:thfnaqgktdwqbm42radc322i/bafkreidykasrfyvrd4w77drvh2wesxkxkqmswvvzom5fzm27p7wb7da7fe)
![creenshot from Google Scholar with the followng text. But no link to the paper.
[CITATION] The value of the definite integral∫∞−∞ eat2+ bt ect+ d dt
LJ Mordell - Quarterly J. of Math, 1920
Save Cite Cited by 35 Related articles](https://cdn.bsky.app/img/feed_thumbnail/plain/did:plc:thfnaqgktdwqbm42radc322i/bafkreif65lvthgv7ln6q6xxgdtf3wrl5sollmu5q6e7i7f2hd7qvyhyreu)
![\begin{proposition}[Mean of Logit-Normal Distribution]
For all $\mu\in\mathbb{R}$ and all $\sigma>0$,
\begin{align*}
&\ex{X \gets \mathcal{N}(\mu,\sigma^2)}{\frac{1}{1+e^{-X}}} = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty \frac{1}{1+e^{-(\mu+\sigma x)}} e^{-x^2/2} \mathrm{d}x \\
&= \frac12 + \frac{\sum_{n=1}^\infty e^{-\sigma^2n^2/2} \sinh(n\mu)\tanh(n\sigma^2/2) + \frac{2\pi}{\sigma^2} e^{-(2n-1)^2\pi^2/2\sigma^2} \frac{\sin((2n-1)\pi\mu/\sigma^2)}{\sinh((2n-1)\pi^2/\sigma^2)}}{1 + 2 \sum_{m=1}^\infty e^{-\sigma^2m^2/2} \cosh(m\mu)}
\end{align*}
\end{proposition}](https://cdn.bsky.app/img/feed_thumbnail/plain/did:plc:thfnaqgktdwqbm42radc322i/bafkreiah5ittkezmk46nyanv7imc2ha6c6klzuqvh5nj7fx33w5rsl4gxi)
![\begin{theorem}
For all $c>1$, \[\frac{1}{2\pi} \int_{-\pi}^{\pi} \frac{1}{c+\sin(\theta)} \mathrm{d} \theta = \frac{1}{\sqrt{c^2-1}}.\]
\end{theorem}
\begin{proof}
Clearly
\[\frac{\mathrm{d}}{\mathrm{d}\theta} \left[ \frac{2}{\sqrt{c^2-1}} \tan^{-1}\left(\frac{1 + c \tan (\theta/2)}{\sqrt{c^2-1}}\right) \right] = \frac{1}{c+\sin(\theta)}.\]
The result now follows from the fundamental theorem of calculus.
\end{proof}](https://cdn.bsky.app/img/feed_thumbnail/plain/did:plc:thfnaqgktdwqbm42radc322i/bafkreih422niyuwsosyjc5yndqxohza3mo5snurp4vofn4g54igvqi3icu)
![For all $a,x \in \mathbb{R}$ with $a \ne 0$, we have \[\log(1+e^x) \le \log(1+e^a) + \frac{x-a}{1+e^{-a}} + \frac{(e^a-1) \cdot (x-a)^2}{4 \cdot a \cdot (e^a+1)}.\]
Furthermore, for each $a$, this is the smallest quadratic upper bound on $\log(1+e^x)$ satisfying equality at $x=a$.](https://cdn.bsky.app/img/feed_thumbnail/plain/did:plc:thfnaqgktdwqbm42radc322i/bafkreidabhfzndiqfbedp3gsum7nqbe5norxqihn2uo72elg7acwkaih7u)
![LaTeX source: https://pastebin.com/BeABGVqE
Let $\pi(n)$ denote the number of prime numbers $\le n$.
The prime number theorem states that
\begin{equation}
\lim_{n \to \infty} \frac{~\pi(n)~}{~\frac{n}{\log n}~} = 1. \label{eq:primes}
\end{equation}
Here's a weaker, one-sided (but non-asymptotic) result with a simpler proof:
\begin{theorem}\label{thm:primes}
For all $n \in \mathbb{N}$, we have $\pi(n) \ge \frac{n-2}{\log_2 n}$.
\end{theorem}
\begin{lemma}\label{lem:lcm}
Define $\mathsf{lcm}(n) := \min\{m \in \mathbb{N} : \forall k \in [n] ~ \frac{m}{k} \in \mathbb{N}\}$ to be the least common multiple of $[n] := \{1,2,\cdots,n\}$.
Then $\mathsf{lcm}(n) \ge 2^{n-2}$ for all $n \in \mathbb{N}$.
\end{lemma}](https://cdn.bsky.app/img/feed_thumbnail/plain/did:plc:thfnaqgktdwqbm42radc322i/bafkreiaa4qeqfxuqo2iu4wt7rwqrkpmnm2rwiqlohwiuavp7mau3ljnrui)
![LaTeX source: https://pastebin.com/tA8BLbaw
\begin{proposition}
Let $f : \{0,1\}^n \to \mathbb{R}$ be an estimator with the following property.
\[\forall p \in [0,1] ~~ \ex{X_1 \cdots X_n \gets \mathsf{Bernoulli}(p)}{(f(X)-p)^2} \le \alpha^2.\]
Then $\alpha^2 \ge \frac{1}{6(n+2)}$.
\end{proposition}](https://cdn.bsky.app/img/feed_thumbnail/plain/did:plc:thfnaqgktdwqbm42radc322i/bafkreifiu7qkljomdusstol4qgzvpnhamudepwxyjtbuiizl55bydngemi)
![Lemma 4.5 from https://arxiv.org/abs/1712.07196
Let X_1, ..., X_k be independent samples from N(0,1). Then
E[max{X_1^2, ..., X_k^2}] <= 2ln(2k).](https://cdn.bsky.app/img/feed_thumbnail/plain/did:plc:thfnaqgktdwqbm42radc322i/bafkreidkql7bujw7docefdl3shay4r4dn7pw74slmhm3dlhtn4s3li4jyy)
![LaTeX source: https://pastebin.com/HjkYLD5r
\begin{proposition}[Bernstein's Inequality]
Let $X_1, \cdots, X_n$ be independent random variables. Assume $\ex{}{X_i}=0$, $\ex{}{X_i^2}=\sigma_i^2$, and $\pr{}{|X_i|\le 1}=1$ for each $i \in [n]$.
Let $\sigma^2 := \sum_{i=1}^n \sigma_i^2$.
Then, for all $t \ge 0$, \[\pr{}{\sum_{i=1}^n X_i \ge t} \le \exp\left(\frac{-3t^2}{6\sigma^2+2t}\right).\]
\end{proposition}](https://cdn.bsky.app/img/feed_thumbnail/plain/did:plc:thfnaqgktdwqbm42radc322i/bafkreihlabdqzbokq4cb5sy5ze5gbcdeuorjqmtybhtu2ypww543ownnje)
![https://pastebin.com/h9kKqUag
\begin{lemma}\label{lem:binomialcenter}
Let $n \ge k \ge 1$ be integers.
If $X \gets \mathsf{Binomial}(n,k/n)$, then \[\pr{}{X=k} > \frac{3}{16} \cdot \frac{1}{ \sqrt{k \cdot (1-\frac{k}{n})} + \frac14}.\]
\end{lemma}](https://cdn.bsky.app/img/feed_thumbnail/plain/did:plc:thfnaqgktdwqbm42radc322i/bafkreiakrzh2loc2en5vln4jjhmuyksy5ph5yuyn3ksbaptlthlrpqmtg4)
![\begin{proposition}
Let $p \in [0,1]$ and $n,k\in\mathbb{N}$.
Let $X \gets \mathsf{Binomial}(n,p)$.
Then \[\ex{}{(X-np)^{2k}} \le \left(\frac{e}{2}kn\right)^k.\]
\end{proposition}
\begin{proof}
By Hoeffding's Lemma, for all $t \in \mathbb{R}$, \[\ex{}{e^{tX}} = (1-p+p \cdot e^t)^n \le (e^{pt + t^2/8})^n. \]
Thus, for all $t \ge 0$,
\begin{align*}
\frac{t^{2k}}{(2k)!}\ex{}{(X-np)^{2k}}
&\le \sum_{\ell=0}^\infty \frac{t^{2\ell}}{(2\ell)!}\ex{}{(X-np)^{2\ell}} \\
&= \frac12 \ex{}{e^{t(X-np)}} + \frac12 \ex{}{e^{-t(X-np)}} \\
&\le e^{nt^2/8}.
\end{align*}
Hence
\begin{align*}
\ex{}{(X-np)^{2k}}
&\le \inf_{t>0} ~ (2k)! \cdot t^{-2k} \cdot e^{nt^2/8}
\end{align*}
Set $t = \sqrt{8k/n}$ so that
\begin{align*}
\ex{}{(X-np)^{2k}}
&\le (2k)! \cdot t^{-2k} \cdot e^{nt^2/8} \\
&\le (2k)^{2k} \cdot (n/8k)^{k} \cdot e^{k} \\
&= \left(\frac{e}{2} \cdot k \cdot n\right)^k .
\end{align*}
\end{proof}
We can use $(2k)! \le 2 \cdot k^{2k}$ to obtain the tighter bound
\[ \ex{}{(X-np)^{2k}} \le 2 \cdot \left( \frac{e}{8} k n \right)^k. \]](https://cdn.bsky.app/img/feed_thumbnail/plain/did:plc:thfnaqgktdwqbm42radc322i/bafkreiduukuf3maqfggl5rhbghoq5f5iel2p76scaa4jsioxg4wxsdpd7u)
![\begin{proposition}
Let $p \in [0,1]$ and $n,k\in\mathbb{N}$ satisfy $k \le np$.
Let $X \gets \mathsf{Binomial}(n,p)$.
Then \[\ex{}{(X-np)^{2k}} \le \left(15knp\right)^k.\]
\end{proposition}
\begin{proof}
For all $t \in \mathbb{R}$, \[\ex{}{e^{tX}} = (1-p+p \cdot e^t)^n \le (e^{p\cdot(e^t-1)})^n. \]
Thus, for all $t \ge 0$,
\begin{align*}
\frac{t^{2k}}{(2k)!}\ex{}{(X-np)^{2k}}
&\le \sum_{\ell=0}^\infty \frac{t^{2\ell}}{(2\ell)!}\ex{}{(X-np)^{2\ell}} \\
&= \frac12 \ex{}{e^{t(X-np)}} + \frac12 \ex{}{e^{-t(X-np)}} \\
&\le \frac12 e^{np(e^t-1-t)}+ \frac12 e^{np(e^{-t}-1+t)} \\
&\le e^{np(e^t-1-t)}.
\end{align*}
Hence
\begin{align*}
\ex{}{(X-np)^{2k}}
&\le \inf_{t>0} ~ (2k)! \cdot t^{-2k} \cdot e^{np(e^t-1-t)}
\end{align*}
Set $t = \sqrt{2k/np} \le \sqrt{2}$ so that $e^t-1-t \le t^2$ and
\begin{align*}
\ex{}{(X-np)^{2k}}
&\le (2k)! \cdot t^{-2k} \cdot e^{np(e^t-1-t)} \\
&\le (2k)^{2k} \cdot (np/2k)^k \cdot e^{2k} \\
= (2e^2 \cdeot k \cdot np)^k \\
&\le (15knp)^k .
\end{align*}
\end{proof}](https://cdn.bsky.app/img/feed_thumbnail/plain/did:plc:thfnaqgktdwqbm42radc322i/bafkreicutvvtsosavwbr7qxhuswu4fb4yaymb556kp5icvhodahelkgcbe)
![LaTeX source: https://pastebin.com/zSTYGiuu
\begin{theorem}[Extension of Fano's inequality]
Let $X \in \{0,1\}^d$ be uniformly random and let $Y \in \mathbb{R}^d$ be a random variable that depends on $X$.
If $\ex{}{\|X-Y\|_1} \le \alpha \cdot d$ for $\alpha \le \frac12$, then \[I(X;Y) \ge d \cdot \dr{\text{KL}}{\mathsf{Bernoulli}(\alpha)}{\mathsf{Bernoulli}\left(\frac12\right)}.\]
\end{theorem}](https://cdn.bsky.app/img/feed_thumbnail/plain/did:plc:thfnaqgktdwqbm42radc322i/bafkreifwxmmlllwf3en364merjtm66m33jmjih6so2umffwsbz35x2pehy)
Researcher in computer science, math, machine learning, (differential) privacy, AI, etc. at Anthropic. Kiwi🇳🇿 in California🇺🇸 http://stein.ke/