Let $X\_n$ be a random variable distributed on $A\_n:=\{1, \ldots, n\}$ and $g\_n\colon A\_n \to A\_n$ such that $\Pr\big(X\_n \neq g\_n(X\_n)\big) \to 0$. Putting $Y\_n=g\_n(X\_n)$, then by Fano's inequality $$\frac{H(X\_n\mid Y\_n)}{\log n} \to 0,$$ which can be written $$\frac{H(X\_n\mid Y\_n)}{H(X\_n)} \to 0 \qquad (\ast)$$ in the particular case when $X\_n$ is uniformly distributed on $A\_n$. 


As [shown by @AnthonyQuas here](https://mathoverflow.net/questions/128937/order-of-convergence-of-the-conditional-entropy), $(\ast)$ fails in general. Now I'd like to know whether 
$$\limsup \frac{H(X\_n\mid Y\_n)}{H(X\_n)} <1 \qquad (\ast\ast)$$
holds true in general, in the case when $H(X\_n)\to \infty$.



order of convergence of the conditional entropy (2)
I think there's a similar counterexample to the previous one. Let $X\_n$ take the value 0 with probability $1-1/n$ and any value in $\{1,\ldots,N\_n\}$ with probability $1/(nN\_n)$. 
Set $Y\_n=\min(X\_n,1)$. Now $H(X\_n|Y\_n)=(1/n)\log N\_n$ (the additional information in $X\_n$ comes if $Y\_n=1$ which has probability $1/n$; conditioned on $Y\_n=1$, the expected additional information is $\log N\_n$). 


On the other hand, $H(X\_n)=-(1-1/n)\log(1-1/n) - N(1/(nN))\log(1/(nN))\approx 1/n(\log n+\log N\_n)$. Now if $N\_n=2^n$, for example, then the ratio in (\*\*) converges to 1.


