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I'm interested in numerically finding the maximum likelihood estimator of a parameter $\theta$, as well as the confidence interval of this estimator. First I'll describe the method I've been trying, then I'll ask my specific questions about this method.

I perform $N$ binary trials at each possible value of $\theta$ (in my case, $\theta$ is integer-valued and known to fall in a certain range). Given these results, I can choose the maximum likelihood estimator $\hat \theta$ to be the value of $\theta$ that results in the most positive trials.

To find a confidence interval for $\hat \theta$, I can take subsets of my trials and use these to find estimators $\hat\theta_i$. Let the fraction of the trials that I take for each subset be $d$. Let $\hat\theta_\mu$ be the mean of all these estimators $\hat\theta_i$, and $\hat\theta_\sigma$ be the standard deviation.

My questions are: For my estimator, is it better to use the original estimator $\hat\theta$ based on all the data, or $\hat\theta_\mu$ based on my subset estimators? Is the confidence interval just equal to $\hat\theta \pm \hat\theta_\sigma$, or can I take into account the fact that the subset estimators will have more variance than the overall estimator, and use $\hat\theta \pm \sqrt{d} \hspace{3pt}\hat\theta_\sigma$? Is this overall a sound method, or are there other ways to improve it, or another method I should use entirely? I can always run more trials, for example I could run many sets of $N$ trials independently and then get a confidence interval from that, but of course that gets costly.

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I think you want the confidence interval on $\hat p_\theta$, the estimated probability of success for each $\theta$, rather than $\hat \theta$, since I am not sure what the later would mean given that there is a finite discrete set of $\theta$ each with it own independent $p_\theta$. In general I think that if you want find the confidence interval for some statistic you are better off taking bootstrap samples (ie draws of $N$ samples with replacement) rather than subsamples. With bootstrap samples you do not need to worry about choosing a $d$ and your variance estimate is just the variance across the samples. However, in your case sampling seems unnecessary. The $N$ trials for each $\theta$ constitute a draw from a binomial distribution with probability $p_\theta$. So you can estimate the confidence interval for $p_\theta$ analytically. It is fairly easy to see that this is equivalent to the estimate you would get from a large number of bootstrap samples, since the bootstrap samples for each $\theta$ are just draws from the binomial distribution with $p = \hat p_\theta = S/N$ where $S$ is the number of success for $\theta$.

EDIT

If $p_\theta$ looks like "smooth" function of the discrete $\theta$, like the Poisson distribution is a "smooth" function of $k$, then you could try to estimate a CI on that function, but you would need to need to know the form of that function. You would need to guess a function $f$ such that $p_\theta=f(\theta, \phi)$ for some unknown $\phi$ and then you would estimate $\hat \phi$ and its confidence interval. For example when people estimate a Poisson distribution from data, the $k$ of the Poisson corresponds to your $\theta$, but what needs to be estimated the $\lambda$, which comes from the assumption that the form of the distribution is the Poisson.

EDIT 2

If you want to stay non-parametric and want to know how likely any given $\theta$ is to be optimal, then you could take a bootstrap sample from each of the N trials for each $\theta$, or equivalently a size $N$ binomial sample with $p = \hat p_\theta$ and record which $\theta$ sample had the highest success rate. Repeat to get a distribution over $\theta$ for producing the highest success rate sample. That seems like a reasonable proxy for what you are looking for if you do enough repetitions.

Alternatively you could try to estimate the distribution for the regret choosing $\hat \theta_{opt}$ the parameter value that was optimal in your intial trials. Do the same resampling as above, but each time record the difference between the success rate of sample produced by $\hat \theta_{opt}$ and the highest sample success rate of that iteration.

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  • $\begingroup$ What you say is correct, I think. Since there are a large number of possible values for $\theta$, I was thinking of it as continuous when talking about CIs, but this is not technically correct. To give you more background, here is the question that inspired mine. As you can see, the discrete distribution is pretty smooth, so it does make some sense to talk about a CI. $\endgroup$ – hunse Feb 10 '14 at 19:34
  • $\begingroup$ Given the confidence intervals on all the $\hat p_\theta$, is it possible to give a probability that my selected estimator $\hat \theta$ is correct (i.e., $P(\theta = \hat \theta)$), and give probabilities for other values of $\hat \theta$? Essentially, I want estimates of all the probabilities $P(\theta = \theta_i)$, where $\theta_i$ ranges over all the possible values of $\theta$. $\endgroup$ – hunse Feb 10 '14 at 19:38
  • $\begingroup$ If $p_\theta$ looks like "smooth" function of $\theta$, then you could try to estimate a CI on that function, but you would need to need to know the form of that function, $\endgroup$ – Daniel Mahler Feb 10 '14 at 20:42
  • $\begingroup$ Yeah, I was hoping to do it non-parametrically (i.e., not knowing the form of the function/distribution). Obviously in that case a CI wouldn't be possible, but it seems like I should be able to find $P(\theta = \theta_i)$ (of which the top 95% would be a suitable stand-in for a CI). $\endgroup$ – hunse Feb 10 '14 at 20:50
  • $\begingroup$ Knowing $P(\theta = \theta_i)$ will tell me how likely it is that my selected $\hat \theta$ is correct, and whether there are other plausible candidates. If there are other candidates, are they all near the selected $\hat \theta$? Or perhaps the underlying distribution is bimodal, in which case there is another very different possibility for $\hat \theta$. This is what I hope to discover from $P(\theta = \theta_i)$. $\endgroup$ – hunse Feb 11 '14 at 16:15

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