This is a follow-up to this question. The scenario is the same: Somebody rolls $n$ dice (each has $6$ sides, perfectly balanced), and counts the number of $6$s (let's call this number $X_6$). I have to estimate $n$, but this time, instead of knowing $X_6$ directly, I only know an estimate of $X_6$ (let's call it $\hat{X_6}$). This estimator is assumed to be non-biaised and has a constant standard error $c$.
The best estimator I can think of for $n$ is $6\cdot \hat{X_6}$. I can't manage to figure out:
Whether the "non-biaised" property of the estimator $i\rightarrow \hat{i}$ automatically means that my estimator is not biaised.Nevermind, of course it does, the expected value is linear.- What the standard error of my estimator is, or at least a good lower bound for it. If this is too difficult to compute, the same question with "standard error" replaced by "mean/median relative error" is also interesting to me.
(Reason for this question: in my original problem, the estimate comes from a HyperLogLog sketch, and I'm trying to understand what would be the "cost" of computing my sketch only on a sample to estimate global cardinality.)
If I understand your question, given $X_6$, $\hat{X}_6$ is a random variable sampled from a distribution with mean $X_6$ and variance $c^2X_6^2$. Then, we have \begin{align} \mathsf{E}(\hat{X}_6)~&=~\sum_{k=0}^n\mathsf{E}(\hat{X}_6 \mid X_6 = k)\cdot\Pr(X_6 = k) \\ &=~ \sum_{k=0}^n k\cdot \Pr(X_6 = k) \\ &=~ \mathsf{E}(X_6) = \frac{n}{6} \end{align} By the law of total variance, we have \begin{align} \mathsf{Var}(\hat{X}_6)~&=~\mathsf{E}(\mathsf{Var}(\hat{X}_6 \mid X_6)) + \mathsf{Var}(\mathsf{E}(\hat{X}_6 \mid X_6)) \\ &=~\mathsf{E}(c^2X_6^2) + \mathsf{Var}(X_6) \\ &=~c^2\mathsf{E}(X_6^2) + \frac{5}{36}n \\ &=~c^2(\mathsf{Var}(X_6) + \mathsf{E}(X_6)^2) + \frac{5}{36}n \\ &=~c^2(\frac{5}{36}n + \frac{1}{36}n^2) + \frac{5}{36}n \end{align}