Solve $\frac{1}{2^\theta}\sum_{k=0}^{\theta} {\theta\choose k} \delta(k)=\theta$ for $\delta$

Question

The following arises in unbiased estimation of a parameter for the binomial distribution, but that information is not needed for solving the question. I tried solving this by manipulating the sum to get a recursion, but failed to do so.

Find an explicit function $\delta:\{0,1,\dots\}\to\{1,2,\dots\}$ such that the following holds for all $\theta\in\{1,2,\dots\}$:

$$ \frac{1}{2^\theta}\sum_{k=0}^{\theta} {\theta\choose k} \delta(k)=\theta $$

EDIT

Note that the usual unbiased estimator for this binomial parameter, i.e. the function $\delta(x)=2x$ is not a solution since $\delta$ may not map to $0$. Considering $\theta = 1$ leads to $\delta(0)=\delta(1)=1$, whence considering $\theta=2$ gives that

$$ \delta(0) + 2 \delta(1) + \delta(2) = 8 \iff \delta(2)=5, $$

and so on one may continue to find $\delta(k)$ for every $k$, but I cannot find the pattern and solve to get an explicit function.

Answer 1

Oussama Boussif, in a truly nice use of Pascals's inversion formula, showed that $\delta(k) = 2k $ is a solution. However, the OP said that the solution must have $\delta(k) \in \{1, 2, ...\} $.

To satisfy this, use $\sum_{k=0}^{\theta} (-1)^k \binom{\theta}{k} = 0 $. Therefore, for any real $a$, $\delta(k) =2k+a(-1)^k $ is a solution.

Choosing $a = 1$ gives $\delta(k) =2k+(-1)^k = 1, 1, 5, 5, 9, 9, ... $.

If the results have to be distinct, I would have to think some more.

Answer 2

Let's introduce the following Pascal inversion formula:

$$ {b}_{n}=\sum_{k=0}^{n}{n\choose k}a_k\\ a_n={(-1)}^{p}\sum_{k=0}^{n}(-1)^k {n\choose k}a_k $$

So:

$$ \sum_{k=0}^{\theta}{\theta\choose k}\delta{(k)} = {2}^{\theta}\theta $$

Using the latter, we get:

$$ \delta{(\theta)}={(-1)}^{\theta}\sum_{k=0}^{\theta}(-1)^k {\theta\choose k}k2^k $$

Let's evaluate the latter sum:

We define $y$ such that:

$$ y=\sum_{k=0}^{\theta}x^k {\theta \choose k}={(1+x)}^{\theta}\\ y'=\sum_{k=0}^{\theta}k{x}^{k-1} {\theta \choose k}=\theta {(1+x)}^{\theta-1}\\ \sum_{k=0}^{\theta}k{x}^{k} {\theta \choose k}=xy'=x\theta {(1+x)}^{\theta-1} $$

Now let $x=-2$ and you'll get:

$$ \sum_{k=0}^{\theta}k{(-2)}^{k} {\theta\choose k}=2\theta{(-1)}^{\theta} $$

So:

$$ \delta{(k)}=2k $$