Estimator of t rolls of dice given only arithmetic mean

Question

I have trouble with the following:

we roll a six-sided normal dice t-times. I want to find an estimator (honestly I will take any advice) of $t$. But all I have are independent random variables $ X_n $ which are the arithmetic mean of each $t$ rolls.

I will take any advice since I have no idea how can I 'extract' $t$ from arithmetic mean which has the same expected value as one roll of a dice.

Answer 1

If you wanted to do this properly, it would involve a lot of details that would be hard to handle without a computer. The possible values of the $X_i$ depend on $t$ in complicated ways. For instance, any prime factor in the denominator of an $X_i$ must be a prime factor in $t$.

If you want a more tractable approach, you can approximate the distribution by a normal distribution (which essentially means ignoring all information it contains beyond the variance – the mean already being known).

A single fair six-sided die has mean $\mu=\frac{1+6}2=\frac72$ and variance $\sigma^2=\frac16\sum_{k=1}^6\left(k-\frac72\right)^2=\frac{35}{12}$, so the average of $t$ dice has mean $\mu$ and variance $\frac{\sigma^2}n$. Thus, in the normal approximation the log-likelihood of the data $X_i$ is

$$ \frac n2\log t-\frac t{2\sigma^2}\sum_i\left(X_i-\mu\right)^2\;. $$

Setting the derivative with respect to $t$ to zero yields the maximum-likelihood estimator:

$$ \frac n{2\hat t}-\frac1{2\sigma^2}\sum_i\left(X_i-\mu\right)^2\;, $$

and thus

$$ \hat t=\frac{n\sigma^2}{\sum_i\left(X_i-\mu\right)^2} $$

This is essentially the reciprocal of the standard maximum-likelihood estimator of the variance. It obviously has the not very nice feature that it diverges if all the means are exactly $\mu$, but that’s perhaps not suprising, since in this case we don’t really have any information about the size of $t$; the larger $t$, the more likely it is, in the normal approximation, to get so close to the mean so regularly.

Since the estimator is already a function of the minimal sufficient statistic $\sum_i\left(X_i-\mu\right)^2$, it can’t be improved by Rao–Blackwellization. It’s not a very satisfactory result, but to do better I think you’d need to avoid the normal approximation and deal with all the details of which means are possible for which $t$.

Answer 2

I have an idea for a consistent estimator:

Let $Y_i = 1$ when $X_i=1$ and $Y_i=0$ for $X_i \neq 1$. Then we have:

$P(Y_i=1)=(\frac{1}{6})^t$ and so $E(Y_i)=(\frac{1}{6})^t \rightarrow t=-\log_6(E(Y_i))$

Let $S_n=\Sigma_i^nX_i$ and consider $\overline{X}=-\log_6 \frac{S_n}{n}$ Then from the law of large numbers we obtain: $\lim_{n\rightarrow \infty} -log_6\frac{S_n}{n}=-\log_6 (\frac{1}{6})^t = t $

Does anyone have an idea how to improve it to find and unbiased and consistent estimator?

Answer 3

My understanding is that we have a semi-infinite array of independent random variables $(Y_{i,s})_{1 \leq i, \: 1 \leq s \leq t}$, with a uniform distribution in $\{1,\dots, 6\}$. Then, for each $i \geq 1$, we compute $$ X_i = \frac{1}{t} \left ( Y_{i,1} + \cdots + Y_{i,t} \right ), $$ and we want an estimator of $t$ from the i.i.d. sequence $(X_i)$.

Now, we have $$ \mathop{Var} X_i = \frac{1}{t} \mathop{Var} Y_{i,1} = \frac{1}{t} \frac{35}{12}, $$ so $$ t = \frac{35}{12 \mathop{Var} X_i}. $$ To get an estimator of $t$, we could just replace $\mathop{Var} X_i$ by an estimator, and get the estimator $$ \widehat{t_n} = \frac{35n}{12 \sum_{i=1}^n (X_i - \overline{X_n})^2}. $$ This is just a simple approach, and the next step would probably be to talk about bias, asymptotic variance, delta method, etc.