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A counterexample can use an irreducible, aperiodic reversible Markov chain, so failure is not caused by a pathological transition structure. Take two binary coordinates, enumerate the four states as $00,01,10,11$, and let $\pi$ be their uniform distribution. Use the Markov kernelIts rows sum to one and it is symmetric, proving detailed balance with $\pi$. Every entry is positive. Let the univariate parametric family be $\tilde{\mathcal{Q}}=\{\mathrm{Bernoulli}(r):0\le r\le 2/5\}$.

For a product initial probability distribution with parameters $r_1,r_2$, additivity of the Kullback-Leibler divergence givesEach summand has derivative $\log (r_i/(1-r_i))<0$ on $(0,2/5]$, so the unique optimum at $t=0$ has $r_1=r_2=2/5$. ThusOn the other hand the allowed initial probability distribution ${\delta }_{00}$, obtained with $r_1=r_2=0$, is sent by one step to $\pi$. Its Kullback-Leibler divergence is zero, the smallest possible value. ConsequentlyThe transition reduces Kullback-Leibler divergence for each input, but may reduce it by different amounts for different inputs. It therefore need not preserve the ordering of competing initial probability distributions, which is exactly why Markov chain variational inference must optimize the initial parameters for its chosen number of steps.

For a discrete random variable with probabilities $p_i$, its Shannon entropy in bits is $H(X)=-{\sum }_i{p}_i{\log }_2{p}_i$, with $0{\log }_{2}0=0$. Gibbs inequality states thatwith equality exactly when $p=q$. If $q_i=0<p_i$, the left side is infinite. Otherwise apply $-\ln u\ge 1-u$ to $u=q_i/p_i$ for $p_i>0$:Equality requires $q_i=p_i$ on the support and no remaining mass outside it. In particular, comparison with the uniform probability distribution on two outcomes gives binary Shannon entropy at most $1$.

Write $q=1-p_1$. For $q>0$, grouping the last two outcomes givesEquality holds if $p_2=p_3$; if $q=0$ it also holds, and then both are zero. Thus equality holds exactly when $p_2=p_3$.

For the discrete memoryless channel, assume $\alpha ,\beta \ge 0$ and $\alpha +\beta \le 1$, and set $w=1-\alpha -\beta$. Both conditional output probability distributions have Shannon entropy $H(w,\alpha ,\beta )$. The third output has probability $\beta$ regardless of the input; the first two have total probability $1-\beta$. Their Shannon entropy is maximized by making them equal, which a uniform input achieves. Hence the channel capacity isand thereforeThe usual zero-times-log convention includes $\beta =1$, when $C=0$. If $w=\alpha$, the rows coincide and every input achieves zero mutual information; otherwise the uniform input is the unique maximizing input.

The orthogonal projections of the Gaussian noise onto the column space of $X$ and its orthogonal complement are independent. Consequently ${\hat{\beta }}_0$ and the residual sum of squares are independent, andThus under the alternative,This is the noncentral t-distribution. The rejection region is fixed by the design, $\nu$ and the chosen level, while this probability distribution depends on $\beta$ only through ${\beta }_0$. Therefore the statistical power does not depend on the other regression coefficients, with the design and $\sigma$ fixed.

The tail sigma-algebra isIts events depend only on arbitrarily late coordinates. The Kolmogorov zero-one law states that for a sequence of independent random variables, every tail event has probability zero or one:Identical probability distributions and integrability are not required for the Kolmogorov zero-one law.

For a stopping time $T$ of a Brownian filtration, define the reflected stochastic processThe Brownian reflection at a stopping time form of the Brownian reflection principle states that $\hat{B}$ is again standard Brownian motion. If $T=\infty$, leave the path unchanged; the second branch is used only when $T<t$.

For an almost surely finite $T$, the Strong Markov property says that $(B_{T+s}-B_T{)}_{s\ge 0}$ is a fresh Brownian motion independent of ${\mathcal{F}}_T$. Its negative has the same probability distribution, so reflecting the future preserves the full path probability distribution. Allowing $T=\infty$ follows by applying this argument to $T\wedge n$ and restricting to each fixed finite time interval.

Let $T_y=\inf \{s\ge 0:B_s=y\}$. Reflect at this stopping time, using the Brownian reflection principle from (a). Path continuity gives $\{S_t\ge y\}=\{T_y\le t\}$, and on this event the reflected endpoint is ${\hat{B}}_t=2y-B_t$. ConsequentlySince $x\le y$, an endpoint ${\hat{B}}_t\ge 2y-x\ge y$ forces the reflected path to have hit $y$ by time $t$. Reflection preserves the first hit of $y$, so the condition $T_y\le t$ is redundant on the right. As $\hat{B}$ has the same probability distribution as $B$, the joint distribution of Brownian motion and its running maximum satisfiesThis argument includes $x=y$; at positive $t$ the boundary endpoint has zero probability under the normal distribution. At $t=0$ both sides vanish because $y>0$.

Taking $x=y=a$ in (c) and using that $B_t$ has no atom at $a$ givesThis is the distribution function of the Brownian first-passage time; (b) ensures that it is a proper probability distribution. Comparing these distribution functions gives the Brownian scaling identity $T_{ca}\stackrel{d}{=}{c}^2{T}_a$ for $c>0$.

For an integer $n\ge 1$, set $T_0=0$ and successively hit the levels $a,2a,\dots ,na$. The Strong Markov property at these finite stopping times and spatial translation imply thatare independent random variables, each with the probability distribution of $T_a$. Thus, for independent copies $T_a^{(j)}$,ThereforeThe paper's definition is the strictly stable distribution convention, with no centering term, a special case of a stable distribution.

The Lévy continuity theorem states that if characteristic functions ${\phi }_n$ of probability distributions ${\mu }_n$ converge pointwise on $\mathbb{R}$ to $\phi$, and $\phi$ is continuous at zero, then $\phi$ is the characteristic function of a unique probability distribution $\mu$ andHere the arrow denotes weak convergence of probability measures, equivalently convergence in distribution of associated random variables. Conversely, weak convergence of probability measures implies pointwise convergence of their characteristic functions to the characteristic function of the limit. The continuity-at-zero condition is essential to ensure that the pointwise limit is a characteristic function.

Suppose $X_n\to X$ with convergence in probability. For real $t$ and $\delta >0$, the elementary inequality $|e^{iu}-e^{iv}|\le \min (2,|u-v|)$ givesFirst let $n\to \infty$, then let $\delta \downarrow 0$. The characteristic functions converge pointwise to that of $X$, which is continuous at zero. The Lévy continuity theorem gives convergence in probability implies convergence in distribution.

For the converse, let $Z$ have the fair Bernoulli distribution, put $X=Z$, and set $X_n=1-Z$ for every $n$. All $X_n$ have the same probability distribution as $X$, so $X_n\stackrel{d}{\to }X$. But $|X_n-X|=1$ always, henceThere is no convergence in probability to $X$. In this example $X_n$ does have convergence in probability to $1-Z$: the counterexample concerns the converse with the specified limit $X$, as usual. If one interprets the question as denying convergence in probability to any limit, take $X_n=Z$ for even $n$ and $X_n=1-Z$ for odd $n$. Its probability distribution is still constant, while the two subsequences have distinct limits from almost sure convergence, ruling out any common limit in probability by uniqueness of a limit in probability.

Let $F_n$ be the distribution function of $X_n$. The limiting constant zero has distribution function $F(x)=0$ for $x<0$ and $F(x)=1$ for $x\ge 0$. For every $\epsilon >0$, both $-\epsilon$ and $\epsilon$ are continuity points of $F$. Thus convergence in distribution implies $F_n(-\epsilon )\to 0$ and $F_n(\epsilon )\to 1$. ThereforeThis provesThe same proof after subtracting a fixed real $c$ proves convergence in distribution to a constant implies convergence in probability. A deterministic limit fixes the coupling automatically; the counterexample in (c) exploits a nonconstant limit whose probability distribution alone does not fix that coupling.

For every $\lambda \in {\mathbb{R}}^2$, the scalar ${\lambda }^{T}X$ is normal with mean ${\lambda }^T\mu$ and variance ${\lambda }^{T}V\lambda$. The moment-generating function of a normal distribution therefore givesThus $\mu$ and $V$ determine the moment-generating function on all of ${\mathbb{R}}^2$. By the uniqueness theorem for moment-generating functions, they uniquely determine the probability distribution of $X$.

Let $\nu$ be an invariant distribution for the continuous-time Markov chain. Using $Q=D_g(P-I)$, its stationarity equation $\nu Q=0$ becomesThus the measure with weights ${\nu }_{i}g(i)$ is invariant for the irreducible positive-recurrent jump chain. Its invariant distribution $\pi$ is unique, so ${\nu }_{i}g(i)=c{\pi }_i$. Normalization gives the invariant-measure transfer between a jump chain and a CTMC formulaThe denominator is finite and nonzero because $\epsilon <g(j)<1/\epsilon$. Thus $\nu$ is a probability distribution and $(X_t)$ is positive recurrent.

For a discrete memoryless channel with input $X$ and output $Y$, its information capacity iswhere $I(X;Y)$ is the mutual information and the maximum is over all input distributions. Capacity is measured in bits when logarithms have base two.