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Applying a Markov kernel to two initial laws cannot increase their total variation distance, even for the distribution of an entire subsequent finite trajectory. If a bounded summand has absolute value at most $B$, the variance of a length-$N$ path sum differs between these laws by at most $6B^2{N}^2$ times their total variation distance. Thus a geometrically long enough burn-in makes this difference negligible relative to $N$.

If $r=dQ/dP$, independence gives $E_{P^{\otimes n}}{\prod }_{i}r(X_i{)}^2=(E_P{r}^2{)}^n$. Since $E_{P}r=1$, this proves the displayed identity. The Cauchy-Schwarz inequality also gives $\Vert Q^{\otimes n}-P^{\otimes n}{\Vert }_1^2\le {\chi }^2(Q^{\otimes n}\Vert P^{\otimes n})$. Thus perturbations of size ${\chi }^2(Q\Vert P)=O(1/n)$ have a uniformly bounded joint total variation distance, a useful input to a metric squared-loss two-point bound.

A sequence $Q_n$ is contiguous with respect to $P_n$ if $P_n(A_n)\to 0$ implies $Q_n(A_n)\to 0$ for every measurable sequence of events. It transfers negligible-event assertions, including consistency, between changing sampling laws. It need not imply small total variation distance.

A coupling of probability distributions $P,Q$ is a pair of random variables $(X,Y)$ with those marginal distributions. On a finite set, and more generally on a standard Borel space, the least achievable mismatch isA maximal coupling attains the minimum; maximal here refers to agreement. An arbitrary coupling can have a much larger mismatch.^[ref]

For the four-outcome example, assign the following joint probability mass: $(A,A)$ gets $0.1$, $(B,B)$ gets $0.1$, $(C,C)$ gets $0.2$, and $(D,D)$ gets $0.3$. The remaining $0.3$ must have $X=A$; place $0.2$ on $(A,B)$ and $0.1$ on $(A,C)$. Row sums give $P$, column sums give $Q$, and the diagonal contains $0.7$. Agreement can reach 70%; mismatch cannot fall below 30%.

To see the lower bound, for any event $E$,The two indicator functions agree whenever $X=Y$. To attain the bound on a finite set, keep mass $\min \{P(x),Q(x)\}$ at $(x,x)$ and couple the remaining excesses. The two residual probability distributions live on disjoint subsets, so their part necessarily mismatches. When the probability distributions are identical, the residual part disappears entirely.

This is a choice of joint construction, not a claim that independently drawn observations are usually equal. Two independent random variables with the same continuous probability distribution can disagree with probability one even though their total variation distance is zero. The construction of a maximal coupling deliberately introduces statistical dependence to reveal how similar the marginal laws are.

A Doeblin condition is a minorization of the whole state space. It implies uniform geometric convergence in total variation distance.

If $r={\max }_{j\ge 2}|{\lambda }_j|$ for a finite irreducible Markov chain with at least two states, then its worst-case total variation distance from its stationary distribution satisfies $d(t)\ge r^t/2$. Choose a possibly complex eigenfunction $f$ for an eigenvalue of modulus $r$, normalize ${\max }_x|f(x)|=1$, and start where the maximum is attained. Since $\pi (f)=0$, $|P^{t}f(x)-\pi (f)|=r^t$; the defining sum for total variation distance bounds this difference by $2d(t)$. For $0<r<1$, this yields $t_{\mathrm{mix}}(\epsilon )\ge (r/(1-r))\log (1/(2\epsilon ))$ for $0<\epsilon <1/2$, because $-r\log r\le 1-r$.

When two density operators are commuting elements, choose a common orthonormal eigenbasis and write $\rho =\mathrm{diag}(p_i)$, $\sigma =\mathrm{diag}(q_i)$. Their quantum fidelity becomes the classical overlap of the two probability distributions, while their trace distance becomes total variation distance:This follows by computing the positive square root of an operator entry by entry in the common eigenbasis. For the qubit mixed states $\rho =\mathrm{diag}(0.9,0.1)$ and $\sigma =\mathrm{diag}(0.6,0.4)$,The two contributions to quantum fidelity are about $0.734847$ and $0.2$. In contrast, the chance that two independent computational basis measurements, one on each preparation, print the same label is $0.9(0.6)+0.1(0.4)=0.58$. That particular event has a specified probability; neither $F$ nor $F_{\mathrm{sq}}$ is that probability.

Here trace distance $0.3$ means an equal-prior quantum state discrimination success of $0.65$ by the Holevo–Helstrom theorem. Measuring in the common computational basis and guessing $\rho$ on outcome zero and $\sigma$ on outcome one achieves $[0.9+0.4]/2=0.65$. Also $\sqrt{1-F^2}\approx 0.3551>D$: the pure-state trace distance and fidelity identity is an equality for pure states, rather than a formula for arbitrary mixed states.

For $n$ independent observations with a normal distribution $N(\mu ,{\sigma }^2)$, $\sigma >0$ fixed, every estimator has ${\sup }_{\mu }{\mathbb{E}}_{\mu }|T-\mu |\ge \sigma /(4\sqrt{2n})$. Compare ${\mu }_0=0$ and ${\mu }_1=\sigma /\sqrt{2n}$: the joint Kullback-Leibler divergence is $1/4$, so the total variation–Hellinger–relative entropy inequality bounds total variation distance by $1/2$. Apply the Le Cam lower bound under absolute-error loss. Degenerate zero-noise laws do not satisfy a positive lower bound.

Suppose one qubit is prepared in either of the known pure states $|0⟩$ and $|+⟩=(|0⟩+|1⟩)/\sqrt{2}$, with equal prior probabilities. Measuring in the computational basis seems natural, but a tilted projective measurement gives better quantum state discrimination. The two density operators and their difference areThe Pauli matrices give ${\Delta }^2=I/2$, and the eigenvalues of $\Delta$ are ${\lambda }_{\pm }=\pm 1/\sqrt{2}$. Their absolute sum gives trace distance $D=1/\sqrt{2}$, agreeing with trace distance between qubit states.

The positive spectral projector points along the unit Bloch vector $\mathbf{n}=(-1,0,1)/\sqrt{2}$:The outcome labels $+$ and $-$ here refer to the sign of the eigenvalue of $\Delta$; they do not denote a Hadamard basis measurement. Guess $|0⟩$ after $P_{+}$ and $|+⟩$ after $P_{-}$. This sign choice is the equal-prior Helstrom measurement for two pure states, a case of the Holevo–Helstrom theorem. ^[ref]

An explicit orthonormal basis for this projective measurement iswith $P_{\pm }=|h_{\pm }⟩⟨h_{\pm }|$. Thus the real coefficient rotation angle is $\beta =-\pi /8$ relative to the computational basis, while the Bloch vector of $P_{+}$ is tilted by $\pi /4$ from the positive $z$ axis toward negative $x$. State-vector and Bloch sphere angles differ by a factor of two.

Write $a=(1+1/\sqrt{2})/2$ and $b=(1-1/\sqrt{2})/2$, so $a+b=1$. The Born rule gives the conditional outcome probabilitiesThe event $P_{+}$ has probability gap $a-b=1/\sqrt{2}=D$, attaining the variational characterization of trace distance. Both preparations are identified correctly with conditional probability $a$, so

For comparison, a $Z$ projective measurement in the computational basis has conditional outcome probabilities $(1,0)$ on ${\rho }_0$ and $(1/2,1/2)$ on ${\rho }_{+}$. Guess zero on outcome zero and plus on outcome one. The average success isand the measured total variation distance is only $1/2$. The optimal tilted projective measurement gains about $0.103553$ in absolute success probability by balancing the two conditional error rates instead of identifying $|0⟩$ perfectly at the expense of $|+⟩$.

These are statistics for alternative measurements on independently prepared copies. Applying $Z$ first and then the tilted projective measurement is a different experiment, because the first measurement in quantum mechanics changes the density operator.

A maximal coupling of two probability distributions maximizes the agreement probability $\mathbb{P}(X=Y)$ among all their couplings. Equivalently, it minimizes $\mathbb{P}(X\ne Y)$. The coupling characterization of total variation distance identifies the minimum mismatch with the total variation distance of the two marginal laws. The term maximal concerns agreement, rather than the size of a correlation coefficient or closeness in a geometric metric.

On a finite set, retain all common probability mass at equal pairs and match the excesses elsewhere. The explicit four-outcome example in coupling characterization of total variation distance realizes this construction and checks its marginal distributions. A maximal coupling need not be unique: different joint arrangements of the residual mass can have the same optimal mismatch.

For estimating the expected value over all continuous probability density functions on $[0,1]$, compare $f=1$ and $g_n(u)=1+(u-1/2)/\sqrt{n}$. Their means differ by $1/(12\sqrt{n})$, and ${\chi }^2(P_{g_n}\Vert P_f)=1/(12n)$. The chi-squared divergence of product measures bounds joint total variation distance by $1/2$. The metric squared-loss two-point bound then gives the displayed uniform minimax risk lower bound.

For any estimator $T$ in a metric parameter space, $d(T,f{)}^2+d(T,g{)}^2\ge d(f,g{)}^2/2$ by the triangle inequality. Average the two risk functions and replace their probability density functions by their minimum. This proves $\max (R_f,R_g)\ge d(f,g{)}^2\int \min (p_f,p_g)/4$. The overlap equals one minus total variation distance. The same proof applies to squared error for a real-valued statistical functional of the parameter, without requiring that functional to be injective.

We prove uniqueness in the sense of at most one Gibbs measure satisfying the printed conditional distributions. No shift-invariance assumption is needed in advance. Write a history as $\eta =({ϵ}_{s-1},{ϵ}_{s-2},\dots )$. The variation hypothesis makes $U$ continuous and bounded on the finite-alphabet sequence space; let $u_{-}\le U\le u_{+}$ and $V=u_{+}-u_{-}$. The conditional distribution of the first spin, obtained by marginalizing the prescribed two-spin law, must beIn particular $q(a\mid \eta )\ge \delta :=e^{-2V}/k>0$ for every symbol and history. Any two candidate measures have this same time-homogeneous one-step kernel, by conditional marginalization and the tower property.

If $\eta ,{\eta }^{\prime }$ agree in their most recent $r$ symbols, each two-spin energy in this formula changes by at mostBoth numerator and denominator therefore change by factors between $e^{-D{\lambda }^{-r}}$ and $e^{D{\lambda }^{-r}}$. Consequently the total variation distance of the two kernels is at most $e^{2D{\lambda }^{-r}}-1\le B{\lambda }^{-r}$, where $B=e^{2D}-1$. Uniform positivity also bounds it by $1-k\delta$. PutThese bounds are decreasing, strictly below one, and summable. We now give the summable-memory coupling argument rather than assuming a small one-step contraction coefficient.

Starting from any two histories at a remote time, use a maximal coupling of the next-spin laws. If the actual histories have agreed for $r$ consecutive symbols, their probability of disagreement is at most $b_r$. Use shared independent uniform random numbers to construct a dominating match-length process $R$: it increases from $r$ to $r+1$ with probability $1-b_r$ and resets to zero with probability $b_r$. Since $b_r$ decreases, the actual consecutive-match length can be kept at least $R$ at every step. When the dominating process does not reset, the actual spins can be coupled to agree.

A new run from zero has probabilityof never resetting. The product is positive because $b_r<1$ and $\sum b_r<\infty$: its tail logarithm converges absolutely, and its finite initial factors are positive. After each reset a fresh run has the same success probability, so the probability of $m$ failed runs is $(1-p_{\ast }{)}^m$. There are almost surely only finitely many resets. Every failed run has finite duration, and thus the time of the final reset is finite almost surely. Its distribution depends only on $(b_r)$, not on the two starting histories. It follows that the probability of any disagreement in a fixed future block tends uniformly to zero as the block's distance from the initial time tends to infinity.

Now take two candidate Gibbs measures, sample their respective histories before time $-N$, and couple their forward kernels as above. Their finite-block laws really are obtained by these kernels: successive conditional expectations use the common conditional distribution $q$. For any fixed coordinate block $[s,t]$, the coupling inequality for total variation bounds the distance between the two candidate block laws by the dominating probability of a reset at or after step $N+s$. That bound tends to zero as $N\to \infty$. Hence all probabilities of finite cylinder sets agree; cylinder sets generate the product Borel sigma-algebra, so the measures agree.

Thus there is at most one measure satisfying these conditional specifications. There is a separate compatibility issue: for an arbitrary unnormalized symbolic potential the finite-horizon conditional distributions printed in the question need not be consistent as the right endpoint changes. The argument proves the requested uniqueness whenever such a law exists; it does not silently infer existence from the variation estimate. A normalized one-sided kernel or the usual compatible Gibbs specification supplies the standard existence formulation. For a concrete incompatibility, take two symbols and $U(a,\eta )=\log W_{{\eta }_1,a}$ with $W=\left(\begin{array}{cc}1& 1\\ 1& 2\end{array}\right)$. This has zero variation once two symbols agree. From a history ending in symbol one, the first-symbol probability obtained from a two-spin block is $2/5$, but from a three-spin block it is $5/13$. Thus the printed finite-horizon specification can indeed have no measure, even though its uniqueness implication remains valid.

Use $I_i=1_{A_i}$ and interpret the printed overlap term as $2\Delta ={\sum }_i{\sum }_{j\in J_i}\mathrm{Pr}(A_i\cap A_j)$; its unindexed $A,B$ are a typographical slip. Take $i\notin J_i$, as usual. If necessary, deleting $i$ from its own neighbourhood gives the proof below with a smaller nonnegative right-hand side, so the stated bound still follows.

If $\lambda =0$, every indicator is zero almost surely and the Poisson distribution with parameter zero is the same point mass. Suppose $\lambda >0$. For any $B\subseteq {\mathbb{Z}}_{\ge 0}$, put $h_B(k)=1_{k\in B}$ and let $Z$ have Poisson distribution with mean $\lambda$. Solve the Poisson Stein equationFor example, for $k\ge 1$ one can takeThe full weighted sum is zero, so using its complementary tail also shows this solution is bounded. The allowed Stein factor is $\Vert \Delta g{\Vert }_{\infty }\le \min (1,{\lambda }^{-1})$, where $\Delta g(k)=g(k+1)-g(k)$.

DefineJoint independence of $A_i$ from all events outside its neighbourhood makes $I_i$ independent of $W_i$, so $p_i\mathbb{E}g(W_i+1)=\mathbb{E}{I}_{i}g(W_i+1)$. Adding and subtracting those terms givesA difference over $m$ integer steps has magnitude at most $m\Vert \Delta g{\Vert }_{\infty }$. In the first sum the step count is $I_i+V_i$; in the second, on $I_i=1$, it is $V_i$. Consequently the absolute value is at mostThe Stein equation identifies the left side with $|\mathrm{Pr}(W\in B)-\mathrm{Pr}(Z\in B)|$. Taking the supremum over $B$ is the definition of total variation distance, proving the Stein-Chen bound with the Poisson Stein factor:

For the random uniform hypergraph, index indicators by four-vertex sets. Each such set is a hypergraph clique exactly when its four triples are all present, so its probability is $p^4$. Distinct sets use disjoint triple indicators unless their intersection has size three; disjoint underlying trials give the required joint independence. A fixed four-set has $d=4(n-4)$ dependent others, and each overlapping pair requires seven distinct triples, with joint probability $p^7$. Thus, with $N=\left(\genfrac{}{}{0ex}{}{n}{4}\right)$ and $\lambda =N{p}^4$, the bracket isSince $\lambda \min (1,{\lambda }^{-1})\le 1$,This calculation applies for $n\ge 4$ and $p>0$; if $p=0$ or $n<4$, the count and its Poisson parameter are zero and the conclusion is immediate.

Let $P$ be the transition matrix of a finite irreducible Markov chain with at least two states, reversible with stationary distribution $\pi$. Its Dirichlet form of a Markov chain isThe Discrete Poincare inequality states that ${\mathrm{Var}}_{\pi }f\le C\mathcal{E}(f,f)$ for every real $f$. Its optimal constant is $C=(1-{\lambda }_2{)}^{-1}$, where ${\lambda }_2$ is the second largest eigenvalue of $P$. Indeed, detailed balance makes $P$ self-adjoint on $L^2(\pi )$. Expand $f-\pi (f)={\sum }_{j\ge 2}{a}_j{v}_j$ in an orthonormal eigenbasis. Then ${\mathrm{Var}}_{\pi }f=\sum a_j^2$ and $\mathcal{E}(f,f)=\sum (1-{\lambda }_j)a_j^2$, proving the inequality and its sharpness.

The useful constructive version is the canonical paths Poincare bound. Choose transition graph paths ${\gamma }_{xy}$ between each ordered pair, write $Q(u,v)=\pi (u)P(u,v)$, and definewhere $N_{uv}$ counts occurrences of the directed transition. The Cauchy-Schwarz inequality along each path in a graph gives $(f(x)-f(y){)}^2\le |{\gamma }_{xy}|{\sum }_{{\gamma }_{xy}}(f(u)-f(v){)}^2$. Multiply by $\pi (x)\pi (y)/2$ and sum. The identity ${\mathrm{Var}}_{\pi }f=\frac{1}{2}{\sum }_{x,y}\pi (x)\pi (y)(f(x)-f(y){)}^2$ proves ${\mathrm{Var}}_{\pi }f\le \rho \mathcal{E}(f,f)$.

For a lazy Markov chain this also proves the mixing time estimateTo see this, the centred initial density has $L^2(\pi )$ norm at most $\pi (x{)}^{-1/2}$; laziness puts all nonconstant eigenvalues between zero and $1-{\rho }^{-1}$. Their powers contract that norm, and the Cauchy-Schwarz inequality bounds total variation distance by half the density's $L^2(\pi )$ norm.

Here is a precise sufficient density condition for the permanent application: every row and every column has more than $n/2$ ones. Interpret the permanent of a matrix as the number $m_n$ of perfect matchings in its bipartite graph, with $n$ vertices per side. Write $m_k$ for the number of $k$-edge matchings in a graph.

First prove the short augmenting paths in a dense bipartite graph bound. Given a nonperfect matching in a graph, choose unmatched $u,v$ on opposite sides. If either has an unmatched neighbour, augment with one edge. Otherwise the matched partners of $N(u)$ and the neighbours $N(v)$ are subsets of the same $n$-element side whose total size exceeds $n$. Their intersection gives an alternating path in a graph $u,b,a,v$ of length three, with $ab$ a matching edge. Repeating proves $m_n>0$. Choose augmentations by a fixed deterministic rule. A reverse augmentation is specified by at most four vertices, so a next-size matching in a graph has at most $C_0=2n^4$ preimages. Therefore

Use the matching generating function $Z(\lambda )={\sum }_{k=0}^n{m}_k{\lambda }^k$ and the weighted matching Markov chain on all matchings in a graph, with stationary weights ${\pi }_{\lambda }(M)={\lambda }^{|M|}/Z(\lambda )$. It stays put with probability $1/2$; otherwise it chooses a graph edge uniformly and proposes its deletion, its addition if both endpoints are unmatched, or its exchange for the unique incident matching edge if precisely one endpoint is matched. Other proposals stay put. Accept with probability $\min (1,{\lambda }^{|M^{\prime }|-|M|})$. Reverse proposals use the opposite exchanged edge, so detailed balance holds. Deletions connect every state to the empty matching in a graph.

For completeness, the canonical paths for the weighted matching chain give a polynomial bound without assuming a mixing theorem. The symmetric difference $I△F$ consists of alternating graph paths and graph cycles. Order its components deterministically, orient open components from their smaller endpoint, and convert them from $I$ to $F$ by moving an unmatched endpoint with exchanges. If the first edge belongs to $I$, delete it first. For a graph cycle, delete one $I$-edge, move the unmatched endpoint around the opened cycle, then insert the last $F$-edge. Each intermediate state is a matching in a graph, and each entire transition path in a graph has length at most $4n$.

Fix a directed transition with initial state $M$. The complementary edge set $K=I△F△M$ agrees with one of the two matchings in a graph outside the active component and has at most two degree-two vertices on that component. Remove at most two offending edges to obtain a matching in a graph $K^{\prime }$. Store those edges and one bit specifying the initial alternating colour of the active component. This encoding is injective for the fixed transition: it restores $K$ and $I△F=M△K$, identifies the active component from the changed edges, and recovers the two colours using the fixed component order. Completed components have $F$'s colour in $M$, pending components have $I$'s colour, and the stored bit resolves the active component. Common edges are $M\cap K$.

There are at most $2(m+1{)}^2$ auxiliary records per $K^{\prime }$, where $m$ is the number of graph edges. If $d=|K|-|K^{\prime }|\le 2$, thenAlso $Q(M,M^{\prime })\ge {\pi }_{\lambda }(M)/(2m\max (\lambda ,{\lambda }^{-1}))$. Sum over the encodings and use the path length bound to obtainThis is polynomial whenever $\lambda$ and ${\lambda }^{-1}$ are polynomially bounded. Starting at the empty matching in a graph, $\log (1/{\pi }_{\lambda }(\varnothing ))=\log Z(\lambda )\le m\log (1+\lambda )$, so the Discrete Poincare inequality gives polynomial-time sampling to any prescribed total variation distance. The addition/deletion/exchange construction is also described in Section 12.4 of Jerrum and Sinclair's survey.

It remains to turn sampling into a relative count; merely sampling perfect matchings would not do that. For $0<\epsilon \le 1/2$, choose ${\lambda }_0=\epsilon /(8m)$ and ${\lambda }_{\ast }=8C_0/\epsilon$. The elementary bounds above giveChoose a schedule ${\lambda }_0<{\lambda }_1<\cdots <{\lambda }_J={\lambda }_{\ast }$ whose successive ratios are at most $1+1/n$. It has $J=O(n\log (n/\epsilon ))$ stages. The annealing ratios for a matching generating function satisfyEach observable lies in $[1,e]$, so the Hoeffding inequality gives relative accuracy $\eta =\epsilon /(10J)$ with $O({\eta }^{-2}\log (J/\delta ))$ independent samples per stage. Run independent copies of the chain from the empty matching in a graph; make the total variation distance per sample at most $\delta /(3JK)$, where $K$ is that sample count. The coupling characterization of total variation distance and the union bound make the entire sample collection coincide with exact stationary samples except with probability at most $\delta /3$. Choose the sample-count constant so all stationary sample means are accurate except with probability at most $\delta /3$.

Let $\hat{R}$ be the product of the sample means, and output $\hat{p}=\hat{R}/{\lambda }_{\ast }^n$. On the successful event its multiplicative error relative to $Z({\lambda }_{\ast })/Z({\lambda }_0)$ lies between $e^{-\epsilon /5}$ and $e^{\epsilon /5}$. Combining the two endpoint bounds provesActivities, sample counts, simulation lengths and rational bit lengths are polynomial in $n,{\epsilon }^{-1},\log ({\delta }^{-1})$. Thus this is a fully polynomial randomized approximation scheme. Small dimensions can be evaluated directly.

Use a fixed logarithm base $b>1$, and write $c={\log }_{b}e=1/\ln b$. All relative entropies below use this same base. Adopt $0\log (0/q)=0$ and $p\log (p/0)=+\infty$ for $p>0$.

For Gibbs inequality, if $q(x)=0<p(x)$ at some point, the relative entropy is infinite and the required conclusion is immediate. Otherwise let $S=\{x:p(x)>0\}$. The elementary inequality $\ln u\le u-1$, with equality only at $u=1$, gives, for every $x\in S$,ConsequentlyThis argument applies to a countable discrete probability distribution too: the difference between the two sides of the pointwise inequality is nonnegative, and the negative part of the entropy summand is bounded by $cq(x)$, hence summable. Thus no subtraction of divergent series is involved. Equality forces both ${\sum }_{S}q=1$ and equality in every pointwise inequality; hence $q(x)=p(x)$ on $S$ and $q=0=p$ off $S$. Conversely $p=q$ plainly gives zero. Therefore $\boxed{D(p\Vert q)\ge 0,D(p\Vert q)=0⟺p=q}$.

For the log-sum inequality, set $F={\sum }_{A}f$, $G={\sum }_{A}g$, and normalize to the probability distributions $p_A=f/F$, $q_A=g/G$. When $A$ is nonempty, positivity ensures $F,G>0$, and direct expansion givesThus normalization reduces the log-sum inequality to Gibbs inequality. Equality holds exactly when $f/F=g/G$. The empty-set case is the zero identity under the usual zero-mass convention.

For the final bound, partition the alphabet using $A=\{x:p(x)>q(x)\}$ and put $a=p(A)$, $d=q(A)$. Since the total signed difference is zero,Apply the log-sum inequality separately to $A$ and its complement. The same normalization argument works for countable sets of finite total mass, using the countable Gibbs inequality just proved. It yields the binary partition bound for relative entropyZero block masses are interpreted by the stated conventions; a positive mass divided by zero makes the bound infinite. The given binary estimate now impliesThis is Pinsker's inequality. Equivalently, for the total variation distance $\Vert p-q{\Vert }_{\mathrm{TV}}=\frac{1}{2}{\sum }_x|p(x)-q(x)|$, the bound is $D(p\Vert q)\ge 2{\log }_{b}e\Vert p-q{\Vert }_{\mathrm{TV}}^2$. The factor ${\log }_{b}e$ keeps the statement valid for either bits or natural-logarithm units.