# Monotonicity, Convexity, and Smoothness of the KL-Divergence between Two Brownian Motions with Different Initializers

We consider the two distributions $$p_t = p_0 * N(0, tI),\quad q_t = q_0 * N(0, t I),$$ where $$*$$ denotes the convolution between two densities, while $$p_0$$ and $$q_0$$ have the same mean and variance. In particular, we assume that $$q_0$$ is $$N(0, I)$$. In other words, we consider two random variables $$X_t = X_0 + N(0, t I)\sim p_t, \quad Y_t = Y_0 + N(0, tI)\sim q_t,$$ where $$X_0\sim p_0$$ and $$Y_0\sim q_0 = N(0, I)$$.

We are interested in characterizing the evolution of $$\text{KL}(p_t, q_t) = \int p_t \log \left(\frac{p_t}{q_t}\right) dx$$ along $$t$$. In particular, we aim to prove:

(i) $$\text{KL}(p_t, q_t)$$ is monotonically decreasing along $$t$$;

(ii) $$\text{KL}(p_t, q_t)$$ is convex along t;

(iii) $$\text{KL}(p_t, q_t)$$ is smooth along t, that is, $$\frac{d^2\text{KL}(p_t, q_t)}{dt^2}$$ is upper bounded.

(Comment: As pointed out by Jon, (i) follows directly from the data processing inequality. As pointed out by Nawaf, (ii) also holds.)

• Point (i) follows from the data processing inequality. – Jon Feb 16 at 0:38
• @Jon You are absolutely right. Thanks a lot for pointing out. – Minkov Feb 16 at 3:33
• Seems better to pose the new part (iii) as a new question, because it was added to the body and the title of the question after the OP received an answer; see, e.g., the nice discussion in meta.mathoverflow.net/questions/3057/splitting-a-question – Nawaf Bou-Rabee Feb 23 at 11:15
• Minkov, I agree with Nawaf: he took the time to answer your question, and to change the question so as to render his answer now obsolete (or out-of-date as you put it) is not good form. I recommend that you accept his answer, and ask your part (iii) as a new question, with a bounty if you like. – Todd Trimble Feb 23 at 20:25
• @NawafBou-Rabee You are absolutely right. I will take your answer and open a new question. Thanks again for your insightful answer! – Minkov Feb 23 at 22:01

Write the KL divergence in terms of the differential entropy of the random variables $$X_t$$ and $$Y_t$$; the result quickly follows. Indeed, since $$Y_t \sim \mathcal{N}(0,1+t)$$, we have \begin{align*} \operatorname{KL}(p_t, q_t) &= - h(p_t) + \frac{1}{2} \int \frac{x^2}{1+t} p_t(x) dx + \frac{1}{2} \log(1+t) + \frac{1}{2} \log(2 \pi) \\ &= - h(p_t) + \frac{1}{2} \mathbb{E}((X_0+\sqrt{t} \mathcal{N}(0,1))^2)\frac{1}{1+t} + \frac{1}{2} \log(1+t) + \frac{1}{2} \log(2 \pi) \\ &= -h(p_t) + h(q_t) \end{align*} where $$h(\cdot)$$ is the differential entropy. By Lemma 2 of Zhang, Anantharam and Geng, subject to $$\operatorname{var}(X_0)=1$$, the minimum of $$-\frac{d^2}{dt^2} h(p_t)$$ is achieved when $$X_0$$ is Gaussian. Thus, $$-\frac{d^2}{dt^2} h(p_t) \ge -\frac{d^2}{dt^2} h(q_t)$$, and hence, $$\frac{d^2}{dt^2}\operatorname{KL}(p_t, q_t) \ge 0$$ which implies that $$\operatorname{KL}(p_t, q_t)$$ is convex with respect to $$t$$.

• @Minkov The answer to the new part (iii) of this question seems to boil down to obtaining a bound on the second derivative of the differential entropy of $X_t$. A formula for this quantity -- attributed to H. McKean, G. Toscani, and C. Villani -- is given in (4) of mdpi.com/1099-4300/20/3/182/htm#B7-entropy-20-00182. This formula involves an expected value over $X_t$, and in order to bound, seems to require additional assumptions on $X_0$. – Nawaf Bou-Rabee Feb 23 at 11:11
• I see. What kind of assumption is needed for this? One possible thing I have in mind is the lower boundedness of the density of $X_0$, which seems a bit strong. – Minkov Feb 23 at 22:23
• @Minkov Agree that seems a bit strong; conjecture that the right assumptions are those that naturally come from expanding the expected value of McKean, Toscani and Villani in terms of the density of $X_0$, and then precisely estimating the terms. – Nawaf Bou-Rabee Feb 23 at 23:33