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For the product law and corrected range of moments, let $E=\mathbf1_{\{Z\le x\}}$. Then

$$H(Y_1,\ldots,Y_n\mid E=1) \ge\mathcal H_n(x)-O_{x_0,\delta}\!\left(\sqrt{n/c}\right).$$

This is the source's route to the lower bound in Lemma 1. The separate finite-window argument is not substituted for it.

Source: published PDF, pp. 4–7. The cutoff is a fixed sufficiently large $D\sqrt{cn}$ in place of the printed 10. The factor $(1+t)/(1+e^t)$ below retains a term dropped in the printed summation. Constants may depend on the fixed positive lower bound $x_0$.

Bears on. Problem 297.

Proof

Write $q=cn$, $a=\Pr(E=1)$, and $b=1-a$. The moment estimates and Berry–Esseen give $a=1/2+O(q^{-1/2})$, so $a,b\ge1/3$ for sufficiently large $n$. For a coordinate $m$, conditional on $Y_m=1$ the variable $Z$ has the law of $Z_m'=Z-Y_m/m+1/m$. Its mean is $x+(1-p_m)/m$ and its variance is $\Theta(q^{-1})$ uniformly by moments. The third-moment estimate holds after the same removal. Since the normal distribution function is Lipschitz, Berry–Esseen gives

$$\Pr(E=0\mid Y_m=1) =\frac12+O\left(q^{-1/2}+\frac{\sqrt q}{m}\right).$$

Bayes' formula, divided by $b$, now yields for $r_m=\Pr(Y_m=1\mid E=0)$,

$$|r_m-p_m|\le C p_m\left(q^{-1/2}+\frac{\sqrt q}{m}\right). \tag{1}$$

Choose a fixed $D$ sufficiently large in terms of these constants. For $m\le D\sqrt q$ use the elementary entropy upper bound 1. For larger $m$, concavity gives $h(r_m)\le h(p_m)+h'(p_m)(r_m-p_m)$. Since $0<p_m<1/2$, $0<h'(p_m)\le\log_2(1/p_m)$; therefore (1) implies

$$h(r_m)\le h(p_m)+ C\left(q^{-1/2}+\frac{\sqrt q}{m}\right)p_m\log(1/p_m). \tag{2}$$

This tangent inequality also handles a conditional parameter above $1/2$. For $t=q/m$ the needed bound is

$$p_m\log(1/p_m) =\frac{\log(1+e^t)}{1+e^t} \le C\frac{1+t}{1+e^t}.$$

The terms with coefficient $q^{-1/2}$ sum to $O(n/\sqrt q)$, because $p\log(1/p)$ is bounded. For the other terms,

$$\sum_{m=1}^n\frac{(1+q/m)e^{-q/m}}m =O\bigl(1+\log_+(1/c)\bigr)$$

by (1) in moments, using $r=1$ and $q$ times the $r=2$ estimate. Subadditivity of conditional entropy and (2) thus give

$$H(Y\mid E=0)\le H(Y)+ O\left(\sqrt q+\frac n{\sqrt q} +\sqrt q(1+\log_+(1/c))\right) =H(Y)+O_{x_0}(n/\sqrt q). \tag{3}$$

The last equality uses $c\le C(x_0)$ and boundedness of $c(1+\log_+(1/c))$ on that range; the finite cutoff's rounding adds at most 1.

Since $E$ is determined by $Y$, the exact chain rule gives

$$H(Y)=h(a)+aH(Y\mid E=1)+bH(Y\mid E=0).$$

Insert (3), use $h(a)\le1$ and $a\ge1/3$, and rearrange: $H(Y\mid E=1)\ge H(Y)-O(n/\sqrt q)$. Finally $n/\sqrt q=\sqrt{n/c}$ and $H(Y)=\mathcal H_n(x)$.