Unbiased estimators of the log odds In the book of Lehmann and Casella (2003) page 83, a random variable $X$ is distributed according to the binomial distribution $Bin(n,p)$, $n$ the number of trials and $p$ the success probability.  
Now they say  suppose $g(p)$ is a function of $p$ which our interest lies, and let $\delta(X)$ be an estimator of $g(p)$ then $\delta$ is unbiased for $g(p)$ if 
\begin{equation}
\sum_{k=0}^n\delta(k)\binom{n}{k}p^k(1-p)^k=g(p) \quad \forall p
\end{equation}
 Now take $g(p)=1/p$, we immediately see that there is no unbiased estimator of $1/p$ for all $p$, for if $p\rightarrow 0$, the left hand side of the equation tends to $\delta(0)$ and the right side to $\infty$. 
My question is, is there an unbiased estimator $\log (p/(1-p))$? maybe not for all $p$?  
 A: For the case $g(p)=1/p$, a more rigorous proof for nonexistence of unbiased estimators is For the binomial distribution, why does no unbiased estimator exist for $1/p$?.  With $g(p)=\log\frac{p}{1-p}$, you could mimick your argument using limits when $p \to 0$, since $\log\frac{p}{1-p} = \log p - \log (1-p)$ which goes to $-\infty$ when p goes to zero. But let us try a more rigorous proof:
Let $x \sim \mathcal{Binom}(n,p)$ with $p \in (0,1)$, and let $\delta()$ be some estimator. Write $\theta= \log\frac{p}{1-p}$ with inverse $p=\frac{e^\theta}{1+e^\theta}$. Then if $\delta$ is unbiased for $\theta$ we must have 
$$
    \sum_{k=0}^n \binom{n}{k}p^k (1-p)^{n-k} \delta(k) - \theta=0
$$
Rewrite this as
$$ \sum \binom{n}{k} e^{k \theta}\delta(k) -\theta (1+e^\theta)^n =0. 
$$ Now use the power series for the exponential function $e^\theta = \sum_{j=0}^\infty \frac{\theta^j}{j!}$, and also expand $(1+e^\theta)^n$ using the binomial theorem, we get
$$
\sum_k \binom{n}{k} \delta(k) \sum_{j=0}^\infty \frac{k^j \theta^j}{j!} =
\sum_k \binom{n}{k} \sum_{j=0}^\infty \frac{k^j \theta^{j+1}}{j!}.
$$ Now use the identity theorem for power series, if two power series are identical on an open interval, then all their coefficients are identical. But for the two power series above, the RHS have a constant term of zero, while the LHS have a constant term $\sum_{k=0}^n \binom{n}{k}\delta(k)$. So for $\delta$ to be an unbiased estimator of $\theta$, this sum must be 0. That would require either $\delta=0$ (which is unbiased only for $\theta=0$), or $\delta$ to take both positive and negative values. This does not give an impossibility yet. 
Take the particular case $n=2$. Then peel off coefficients from the power series corresponding to $\theta^0, \theta^1, \theta^2$. Setting them equal gives a linear system of equations in three unknowns $\delta_0, \delta_1, \delta_2$. Without details, I find the system 
\begin{align}
   \delta_0 + 2\delta_1 + \delta_2&=0 \\
   2\delta_1 +2\delta_2&=2\\
\delta_1+2\delta_2&=4
\end{align} with solution $\delta_0=-2, \delta_1=0, \delta_2=2$. But now it is easy to calculate that this estimator isn't unbiased. 
I will see if I can come back and generalize this argument for all $n$, but it does not seem straightforward. 
