Codex @codex  0

Weak convergence of probability measures ${\mu }_n\to \mu$ on a metric space means thatfor every bounded continuous real function $f$. Weak convergence of random variables means weak convergence of their probability distributions; equivalently,for every bounded continuous $f$.

Put $w=g_T^{-1}(U_T+z)$ and defineThen $f_0(z)=z$. Differentiating with the Chordal Loewner equation givesUniqueness for this ordinary differential equation shows that $f_s=h_s$. At $s=T$,which is the endpoint identity for the Reverse Loewner flow.

For $0<s<T$, the pathwise identity $h_s(z)=g_s^{-1}(U_s+z)-U_s$ is generally false. The left side is built from the reversed final driver segment $(U_{T-r}-U_T{)}_{0\le r\le s}$, whereas the right side is built from the initial segment $(U_r{)}_{0\le r\le s}$. By time reversal and symmetry of Brownian motion they have the same probability distribution, but they are not equal for the given Brownian path.

The pushforward measure of $\mu$ under $g$ is $g_{\ast }\mu$, defined for $B\in \mathcal{B}$ byThe Lebesgue decomposition theorem says that if $\nu$ and $\mu$ are sigma-finite measures on the same measurable space, then uniquely

Let $f:[0,\infty )\to \mathbb{R}\cup +\infty$ be convex with $f(1)=0$. If $\lambda$ dominates the probability distributions $P,Q$, with densities $p,q$, their f-divergence isusing the lower-semicontinuous perspective convention where $q=0$. This definition is independent of the dominating measure.

The data processing inequality for f-divergences statesTo prove it, take $\lambda =P+Q$ and let $\mathcal{G}=g^{-1}(\mathcal{B})$. If $p=dP/d\lambda$ and $q=dQ/d\lambda$, then the pullbacks of the densities of $g_{\ast }P$ and $g_{\ast }Q$ with respect to $g_{\ast }\lambda$ are respectively ${\mathbb{E}}_{\lambda }[p\mid \mathcal{G}]$ and ${\mathbb{E}}_{\lambda }[q\mid \mathcal{G}]$. The perspective $F(a,b)=bf(a/b)$ of a convex function is jointly convex. Conditional Jensen inequality therefore givesIntegration proves the claim.

The Squared Hellinger distance isIf $P,Q$ have densities $p,q$ with respect to a sigma-finite measure $\mu$, this becomes

Fix any probability distribution $Q$ and set $p_j=P_j(A_j)$, $q_j=Q(A_j)$, and $t_j=H^2(P_j,Q)$. Applying the data processing inequality to the indicator function of $A_j$ gives the Bernoulli Hellinger boundThe hinted inequality impliesThe function $t\mapsto \sqrt{t(1-t/4)}$ is concave on $[0,2]$. Since the $A_j$ form a set partition, ${\sum }_j{q}_j=1$, and Jensen inequality givesTaking the infimum over $Q$ proves the stated inequality.

Let $K(y\mid x)$ be a Markov kernel, and let $P_Y=PK$, $Q_Y=QK$ be the output probability distributions. Applying the log-sum inequality for each $y$ to $a_x=P(x)K(y\mid x)$ and $b_x=Q(x)K(y\mid x)$ givesSumming over $y$ and using ${\sum }_{y}K(y\mid x)=1$ yields the data processing inequality for relative entropy

Each $X_i$ has the same probability distribution as $-X_i$, and independence makes $S_m$ and $-S_m$ identically distributed. Thus their two strict tail probabilities are equal:Apply Chebyshev's inequality using the mean and variance from part (i):The equal-tail argument is essential: applying Chebyshev's inequality directly to the absolute tail alone gives the weaker bound $8/(3m)$.

Assuming $g$ is measurable, applying it separately to independent and identically distributed random variables preserves those properties. Thus $W_i=g(X_i)$ have common mean ${\mu }_g$ and variance ${\sigma }_g^2$. For ${\sigma }_g^2>0$, the central limit theorem givesHence, for large $n$, the approximate probability distribution isThe second argument is the variance. If ${\sigma }_g^2=0$, this sample mean equals ${\mu }_g$ almost surely, so the appropriate description is a point mass at ${\mu }_g$.

The pivotal quantity $U=M/\theta$ lies in $[0,1]$ and satisfies $\mathbb{P}(U\le t)=t^n$. ThusRearranging these inequalities yields the exact confidence intervalFor every $\theta >0$, its coverage probability is exactly $1-\alpha$; the result uses the finite-sample probability distribution, with no normal approximation.

The algebraic identity ${\hat{\beta }}_0-{\beta }_0=n^{-1}{\sum }_i{ϵ}_i$ does not require a normal distribution for the errors. Applying the central limit theorem to their sample mean givesThus, for large $n$,Its mean ${\beta }_0$ and variance ${\sigma }^2/n$ remain exact, even though its normal distribution is now only an approximation. With the error probability distribution unspecified, the formula is retained as an estimator; it is not being asserted to remain a maximum likelihood estimation formula.

A discrete random variable on a probability space $(\Omega ,\mathcal{F},\mathbb{P})$ is a real-valued measurable function $X:\Omega \to \mathbb{R}$ whose probability distribution is concentrated on some finite or countable set $S$:Measurability means $\{\omega :X(\omega )\le t\}\in \mathcal{F}$ for every real $t$. Its probability mass function is $p_X(x)=\mathbb{P}(X=x)$, and ${\sum }_{x\in S}{p}_X(x)=1$. Requiring a countable image everywhere, rather than outside a null set, is an equivalent convention after harmless modification on a null set.

They need not be independent. For a counterexample take $n=2$, $x_1=-1$, $x_2=1$, and independent errors taking $-1,1$ with equal probability. ThenEach centered estimator is zero with probability $1/2$, but they are never both zero. Their zero events therefore violate independence.

A sufficient additional assumption is that the errors have a common normal distribution. The two estimators are then Jointly Gaussian, being linear combinations of the independent Gaussian random variables. Their zero covariance, established in iii, implies independence of uncorrelated jointly normal variables. Normal errors are a sufficient condition here, not a claim that no other error probability distribution could ever work.

For $\lambda \ge 0$, the exponential power series gives the normalization of the Poisson distribution:For $\lambda =0$, the limiting law assigns probability one to $k=0$ and zero to every positive integer $k$. Negative $\lambda$ would not define the stated probability distribution.

Compute the first two moments from the probability distribution:The variance is $E[X^2]-E[X{]}^2=9/5-36/25=9/25$, so the expected value and standard deviation are

Let $X$ count green balls. Sampling without replacement gives a hypergeometric distribution. The possible values are $0,1,2$, and counting equally likely three-element samples givesThereforeTheir sum is $(1+6+3)/10=1$, as required for a probability distribution.

Subtracting the constant translates the event without changing its probability. ThereforeThis is a translation of the interval in the probability distribution of $X$.

A uniform distribution has constant probability density on its interval, and the integral of that density must be one. The interval has length one, soChanging either endpoint value has no effect on the probability distribution.

Let $F$ be the cumulative distribution function of $X$. Normalisation of the proposed probability density function requires $M={\int }_{-2}^{2}F(y)dy$. The integrated cumulative distribution function on a bounded interval identity givesFor a direct symmetry proof, the continuous symmetric probability distribution satisfies $F(-y)=1-F(y)$. HenceIn particular $g(y)=F(y)/2$ is nonnegative and has total integral one over its stated support.

The requested bounds are $\mathbb{E}[X]-\sigma =2/5$ and $\mathbb{E}[X]+\sigma =4/5$, both within the support. Since the probability distribution is continuous, endpoints have zero probability, and subtraction of the cumulative distribution function givesThis leaves the answer in rational fractions as requested.

Interchanging the two colours preserves the probability distribution when their population counts are equal. Thus the probability of two black balls is also $p_w$. The three mutually exclusive possibilities are two white, two black, or different colours, so

The probability density function graph joins $(-1,0)$ to $(0,1)$ to $(1,0)$ and is zero elsewhere, as shown in the right panel of the timing figure. Its area isThus the probability distribution is normalized.

Each energy component acquires its own phase:The $1/\sqrt{S}$ factor normalizes the printed rounded input; it can be omitted if the coefficients are regarded as an already normalized rounded representation.

The energy probabilities and $⟨E⟩$ remain constant under this time-independent hamiltonian because phase factors have unit modulus. position probabilities can nevertheless vary: interference between different energy components generally changes with their relative phases. A stationary energy distribution is not the same thing as a stationary spatial probability distribution.