Kullback's inequality

In information theory and statistics, Kullback's inequality is a lower bound on the Kullback–Leibler divergence expressed in terms of the large deviations rate function.[1] If P and Q are probability distributions on the real line, such that P is absolutely continuous with respect to Q, i.e. P<<Q, and whose first moments exist, then

$D_{KL}(P\|Q) \ge \Psi_Q^*(\mu'_1(P)),$

where $\Psi_Q^*$ is the rate function, i.e. the convex conjugate of the cumulant-generating function, of $Q$, and $\mu'_1(P)$ is the first moment of $P.$

The Cramér–Rao bound is a corollary of this result.

Proof

Let P and Q be probability distributions (measures) on the real line, whose first moments exist, and such that P<<Q. Consider the natural exponential family of Q given by

$Q_\theta(A) = \frac{\int_A e^{\theta x}Q(dx)}{\int_{-\infty}^\infty e^{\theta x}Q(dx)} = \frac{1}{M_Q(\theta)} \int_A e^{\theta x}Q(dx)$

for every measurable set A, where $M_Q$ is the moment-generating function of Q. (Note that Q0=Q.) Then

$D_{KL}(P\|Q) = D_{KL}(P\|Q_\theta) + \int_{\mathrm{supp}P}\left(\log\frac{\mathrm dQ_\theta}{\mathrm dQ}\right)\mathrm dP.$

By Gibbs' inequality we have $D_{KL}(P\|Q_\theta) \ge 0$ so that

$D_{KL}(P\|Q) \ge \int_{\mathrm{supp}P}\left(\log\frac{\mathrm dQ_\theta}{\mathrm dQ}\right)\mathrm dP = \int_{\mathrm{supp}P}\left(\log\frac{e^{\theta x}}{M_Q(\theta)}\right) P(dx)$

Simplifying the right side, we have, for every real θ where $M_Q(\theta) < \infty:$

$D_{KL}(P\|Q) \ge \mu'_1(P) \theta - \Psi_Q(\theta),$

where $\mu'_1(P)$ is the first moment, or mean, of P, and $\Psi_Q = \log M_Q$ is called the cumulant-generating function. Taking the supremum completes the process of convex conjugation and yields the rate function:

$D_{KL}(P\|Q) \ge \sup_\theta \left\{ \mu'_1(P) \theta - \Psi_Q(\theta) \right\} = \Psi_Q^*(\mu'_1(P)).$

Corollary: the Cramér–Rao bound

Main article: Cramér–Rao bound

Let Xθ be a family of probability distributions on the real line indexed by the real parameter θ, and satisfying certain regularity conditions. Then

$\lim_{h\rightarrow 0} \frac {D_{KL}(X_{\theta+h}\|X_\theta)} {h^2} \ge \lim_{h\rightarrow 0} \frac {\Psi^*_\theta (\mu_{\theta+h})}{h^2},$

where $\Psi^*_\theta$ is the convex conjugate of the cumulant-generating function of $X_\theta$ and $\mu_{\theta+h}$ is the first moment of $X_{\theta+h}.$

Left side

The left side of this inequality can be simplified as follows:

$\lim_{h\rightarrow 0} \frac {D_{KL}(X_{\theta+h}\|X_\theta)} {h^2} =\lim_{h\rightarrow 0} \frac 1 {h^2} \int_{-\infty}^\infty \left( \log\frac{\mathrm dX_{\theta+h}}{\mathrm dX_\theta} \right) \mathrm dX_{\theta+h}$
$= \lim_{h\rightarrow 0} \frac 1 {h^2} \int_{-\infty}^\infty \left[ \left( 1 - \frac{\mathrm dX_\theta}{\mathrm dX_{\theta+h}} \right) +\frac 1 2 \left( 1 - \frac{\mathrm dX_\theta}{\mathrm dX_{\theta+h}} \right) ^ 2 + o \left( \left( 1 - \frac{\mathrm dX_\theta}{\mathrm dX_{\theta+h}} \right) ^ 2 \right) \right]\mathrm dX_{\theta+h},$
where we have expanded the logarithm $\log x$ in a Taylor series in $1-1/x$,
$= \lim_{h\rightarrow 0} \frac 1 {h^2} \int_{-\infty}^\infty \left[ \frac 1 2 \left( 1 - \frac{\mathrm dX_\theta}{\mathrm dX_{\theta+h}} \right) ^ 2 \right]\mathrm dX_{\theta+h}$
$= \lim_{h\rightarrow 0} \frac 1 {h^2} \int_{-\infty}^\infty \left[ \frac 1 2 \left( \frac{\mathrm dX_{\theta+h} - \mathrm dX_\theta}{\mathrm dX_{\theta+h}} \right) ^ 2 \right]\mathrm dX_{\theta+h} = \frac 1 2 \mathcal I_X(\theta),$

which is half the Fisher information of the parameter θ.

Right side

The right side of the inequality can be developed as follows:

$\lim_{h\rightarrow 0} \frac {\Psi^*_\theta (\mu_{\theta+h})}{h^2} = \lim_{h\rightarrow 0} \frac 1 {h^2} {\sup_t \{\mu_{\theta+h}t - \Psi_\theta(t)\} }.$

This supremum is attained at a value of t=τ where the first derivative of the cumulant-generating function is $\Psi'_\theta(\tau) = \mu_{\theta+h},$ but we have $\Psi'_\theta(0) = \mu_\theta,$ so that

$\Psi''_\theta(0) = \frac{d\mu_\theta}{d\theta} \lim_{h \rightarrow 0} \frac h \tau.$

Moreover,

$\lim_{h\rightarrow 0} \frac {\Psi^*_\theta (\mu_{\theta+h})}{h^2} = \frac 1 {2\Psi''_\theta(0)}\left(\frac {d\mu_\theta}{d\theta}\right)^2 = \frac 1 {2\mathrm{Var}(X_\theta)}\left(\frac {d\mu_\theta}{d\theta}\right)^2.$

Putting both sides back together

We have:

$\frac 1 2 \mathcal I_X(\theta) \ge \frac 1 {2\mathrm{Var}(X_\theta)}\left(\frac {d\mu_\theta}{d\theta}\right)^2,$

which can be rearranged as:

$\mathrm{Var}(X_\theta) \ge \frac{(d\mu_\theta / d\theta)^2} {\mathcal I_X(\theta)}.$