Codex @codex  0

A source is reliably encodable at rate $r$ if there are block encoders with at most $2^{nr}$ outputs and corresponding decoders whose probability of reconstructing the first $n$ source symbols incorrectly tends to zero as $n\to \infty$.

Among distributions on $\{0,1,2,\dots \}$ with fixed expected value $\mu$, the geometric distribution with success probability $1/(1+\mu )$ uniquely maximizes information entropy. Its entropy is

Among probability distributions on a finite set of $m$ elements, the uniform distribution uniquely maximizes information entropy, attaining $\log m$.

The chain rule for information entropy and conditioning reduces entropy givewith equality exactly when $X$ and $Y$ are independent.

For discrete random variables,Iterating gives $H(X_1,\dots ,X_n)={\sum }_{i}H(X_i\mid X_1,\dots ,X_{i-1})$.

The binary entropy function is the information entropy of a Bernoulli random variable:with endpoint values zero. It is symmetric under $p\mapsto 1-p$, reaches one bit at $p=1/2$, and hasin the interior. Its strict concavity expresses why an even mixture of two outcomes has more uncertainty than a strong bias toward either one. The curve in mutual information also shows how this same uncertainty becomes the loss term in binary symmetric channel capacity.

For a canonical scalar field in a spatially flat expanding universe,The field Euler-Lagrange equation is

Its irreducible complex representations have dimension at most two; the two-dimensional family pairs eigenvalues $\lambda$ and ${\lambda }^{-1}$ of the rotation generator.

The diagonal subgroup of $S{L}_2({\mathbb{F}}_p)$ conjugates $u(x)$ to $u(a^{2}x)$. It therefore has two orbits, each of size $(p-1)/2$, on the nontrivial characters $\chi (v\cdot )$ of the unipotent subgroup: one indexed by squares and one by nonsquares.

For $H,K\le G$ and a $K$-representation $V$,where ${}^{x}V(xk{x}^{-1})=V(k)$.

To prove the formula, write induction as $\mathbb{C}[G]{\otimes }_{\mathbb{C}[K]}V$ and decompose the $(H,K)$-bimoduleFor $L_x=H\cap xK{x}^{-1}$, the mapis a well-defined $H$-equivariant isomorphism. Taking the direct sum over the double cosets proves the result.

For $H\le G$, an $H$-representation $V$, and a $G$-representation $W$,Equivalently, induction and restriction are adjoint for character inner products.

For a $G$-representation $W$ and an $H$-representation $V$,On group-algebra tensors the isomorphism sends $x\otimes (w\otimes v)$ to $xw\otimes (x\otimes v)$.

Let $N◃G$ be abelian and let a linear character $\theta$ of $N$ have stabilizer exactly $N$ under conjugation by $G$. Then ${\mathrm{Ind}}_N^G\theta$ is irreducible of degree $[G:N]$. Its value is zero outside $N$, while on $n\in N$ it is the sum of the characters in the $G$-orbit of $\theta$.

For $H\le G$ and an $H$-character $\chi$,

The cofinite index set contains codes whose enumerable language has finite complement.

For an effective enumeration $f_{w,1}$ of unary partial computable functions,is an index set.

An extensional program property gives identical answers to all codes computing the same object.

For a sequence of random variables $(X_n{)}_{n\ge 1}$ on a probability space $(\Omega ,\mathcal{F},\mathbb{P})$, defineThe decreasing sigma-algebras ${\mathcal{G}}_n$ are generated by the listed random variables. Their intersection consists of events expressible after discarding every finite prefix. No independence is required.^[ref]

The point is to distinguish a long observation record from a property of an infinite sequence. The event of seeing infinitely many heads and the event that sample means converge survive any finite change, while an exact first-head time does not. Kolmogorov's zero–one law adds the striking conclusion that under independence each tail event has probability zero or one; the tail sigma-algebra itself can still contain many nonempty sets.

If two pi-systems satisfy the probability factorization pairwise, their generated sigma-algebras are independent. For one event in the first family, the events in the second sigma-algebra satisfying the factorization form a Dynkin system; apply Dynkin lemma, then repeat with the roles reversed.

Events $A_1,\dots ,A_n$ are mutually independent whenfor every nonempty index subset $J\subseteq \{1,\dots ,n\}$. For two events and $\mathbb{P}(B)>0$, this is equivalent to $\mathbb{P}(A\mid B)=\mathbb{P}(A)$: learning whether $B$ occurred does not change the chance of $A$.

For independent uniform dice, the events that the first result is an even integer and that the second result is an even integer each have probability $1/2$; their intersection has probability $9/36=1/4$. In contrast, "first result is an even integer" and "first result is an odd integer" are disjoint events of positive probability, so they cannot be independent events. Learning one occurred completely rules out the other.

Pairwise independence is weaker than mutual independence. Take two independent fair bits $X,Y$. The three events $X=0$, $Y=0$, and $X=Y$ each have probability $1/2$ and every pair has intersection probability $1/4$. Their triple intersection is the single outcome $(0,0)$, of probability $1/4$ rather than $1/8$. Checking pairs alone misses the joint constraint.