1
$\begingroup$

Revised Version

Situation: We have a sample of size $r$. Then, we count how many out of these $r$ to be "success". Let $X=$ the no. of the "successes" and say $X$ follow Poisson dist.

Goal: To test whether $X$ is really Poisson using Dispersion Test with test statistics given by

$D=\frac{\sum_{i=1}^{n}(X_{i}-\overline{X})^2}{\overline{X}}$ --- As state in here.

Problem:

  1. I have only one sample, so my n=1. So, D=0 if i used $\overline{X}=X_1$. To overcome this, i use bootstrap to generate "n" bootstrap samples and obtain $\overline{X}$. Is this valid?

  2. Should i choose the generated bootstrap samples used for hypothesis testing in such a way the resulting $\overline{X}=Var(X)$? -- I think this rule should be used if i want to estimate the Poisson parameter. I'm confused in using bootstrap for estimation and bootstrap for hypothesis testing.

3.Illustration:

a. The data: 0.42172863 0.28830514 0.66452743 0.01578868 0.02810549. So,$ r=5$.

b.Define “success”: Observation which larger than 0.4. Here, $X=2$.

c.Obtain let say 3 bootstrap samples (from data in Step a):

i) 0.28830514 0.01578868 0.66452743 0.02810549 0.02810549. So, $X=1$.

ii) 0.42172863 0.66452743 0.02810549 0.66452743 0.66452743. So, $X=4$.

iii) 0.02810549 0.66452743 0.01578868 0.66452743 0.42172863. So, $X=3$.

d.So, $\overline{X}= 2.666667$ and $Var(X)= 2.333333$. Note that here, I didn’t control on how the 3 bootstrap samples (in Step c) are generated. That’s why $\overline{X}>Var(X)$.

So, my questions:

1.Is bootstrap a valid technique to obtain $\overline{X}$ in this case?

2.Should I control on how the bootstrap samples generated so that the resulting $\overline{X}=Var(X)$ to be used for the hypothesis testing?

$\endgroup$
  • 1
    $\begingroup$ @user3638, majority of statistical estimation methods are used for one sample. $\endgroup$ – mpiktas Mar 9 '11 at 14:18
  • $\begingroup$ A great book on bootstrapping is Bradley Efron and and Robert Tibshirani's An Introduction to the Bootstrap. It is a very easy read and covers the topics well. $\endgroup$ – Charlie Mar 9 '11 at 16:14
  • $\begingroup$ Maybe you could ask some more specific question, even to get an example of bootstrap use? $\endgroup$ – user88 Mar 9 '11 at 16:51
  • $\begingroup$ The preceding comments addressed an earlier (more general) version of this question (which, being unanswered, I merged with this one). $\endgroup$ – whuber Mar 9 '11 at 17:03
  • $\begingroup$ @mbq: Already add an example in the revised version. $\endgroup$ – user3638 Mar 10 '11 at 4:03
4
$\begingroup$

First of all, I don't think you can estimate (over)dispersion from a sample of size 1. Overdispersion means that the sample variance is more than predicted based on other properties of the sample (mean) and properties of the assumed model (for Poisson models, variance = mean). But for a sample of size one, the population variance estimate is not finite (it involves dividing by $n-1 = 0$).

In any case, your ideas about the bootstrap procedure are not headed in the right direction. When you bootstrap, you are sampling data values from the sample, which in this case won't help. (i.e., say that $X_1 = 1$. Then the possible bootstrap samples are $[1,1], [1,1,1] ...$) Certainly, drawing extra data from a process that constrains the mean to be equal to the variance will invalidate your estimates of dispersion, as that is exactly the hypothesis under test.

Finally, I'm not familiar with the test formula you are using. Perhaps you are trying for the variance to mean ratio aka Index of dispersion, in which case the formula you give for $D$ is not correct?

$\endgroup$
  • $\begingroup$ +1 for being gentle :-). A nit, but maybe a revealing one: a suitable variance estimator in this situation is the number of successes themselves. Thus one can indeed bootstrap from a single observation. But of course that would be self-fulfilling: a Poisson bootstrapped dataset is going to look Poisson! $\endgroup$ – whuber Mar 9 '11 at 16:57
  • $\begingroup$ @whuber: Ah. I think I was confused as to whether the question was asking about bootstrapping from a single success/failure observation, or from a single count observation (composed of several successes and failures). But the point still stands, I think: you can't learn much about the distribution of a random variable with only one sample of it. $\endgroup$ – Aaron Mar 9 '11 at 17:57
  • $\begingroup$ whuber is right. i'm trying to bootstrap the sample to obtain different no. of successes. Anyway, thanks for pointing out (over)dispersion issue with sample size of 1. $\endgroup$ – user3638 Mar 10 '11 at 4:06
3
$\begingroup$

Your question contains a "contradiction". The way you have described it, the variable defined as $X$ cannot have a Poisson distribution. I'll explain why.

So, you choose a cut-off point, $y_{0}$, and define the original variables $Y_{i}\:(i=1,\dotsc,r)$. You then take a new variable $X_{i}$ and define it as:

$$ X_{i} \equiv \Bigg( \begin{matrix} 1 & Y_{i} > y_{0} \\ 0 & Y_{i} \leq y_{0} \end{matrix} $$

Then $X=\sum_{i} X_{i} \sim Poisson(\lambda)$. Raikov's theorem Says that this implies the individual terms are also poisson $X_{i}\sim Poisson(\frac{\lambda}{r})$. But this is impossible, for as $X_{i}$ has been defined, it can only have two values $(0,1)$, a Poisson has support $(0,1,2,\dots)$. Therefore $X$ cannot be poisson.

From the way it has been defined in the question $X$ is binomially distributed $$X \sim Bin(r,\theta)$$ where $$\theta \equiv Pr(Y_{i}>y_{0})=1-F_{Y}(y_{0})$$

What your bootstrap procedure is doing is attempting to estimate $E(X)=r\theta$, but the estimate of $\theta$ is just given by $\hat{\theta}=\frac{X_{obs}}{r}$ and has variance of $\frac{\hat{\theta}(1-\hat{\theta})}{r+I_{B}}$ where $I_{B}$ is what I call the "Bayesian indicator". If you are Bayesian, with haldene prior, $I_{B}=1$, if you are frequentist then $I_{B}=0$.

You could do a bootstrap if you wanted, by because the maths is there, I'm sure you would get the same answers for big enough bootstrap samples (the simulation will only demonstrate the maths). Also note that the binomial is "under-dispersed" with the mean being greater than the variance, as your bootstrap shows. Also your 3 iterations are "approximately" in line with theoretical mean $r\hat{\theta}=2$ and theoretical variance $r\hat{\theta}(1-\hat{\theta})=1.2$. From looking at the bootstrap samples, this is to be expected (your average/variance being below the theoretical average/variance) as you sampled the "high" number disproportionately in 2 of the simulations.

$\endgroup$
  • $\begingroup$ Ok. I get what you mean about "contradiction" in the question. But, binomial distribution can also be approximated by Poisson distribution. So, does that mean $X$ still cant be Poisson? Emm... maybe i should say $X$ is approximately Poisson? And, about the "high" no. disproportionately in the simulations, is anything "wrong" with that? I mean, do i "somehow" need to control it? Sorry for asking so many questions, i cant help because i'm completely confused. $\endgroup$ – user3638 Mar 11 '11 at 4:47
  • $\begingroup$ @tsukiko - Why use an approximate distribution when the exact one is no more difficult to use? You could use a truncated poisson for the $X_{i}$ terms, but this would basically be a substitution of $\theta=\frac{\lambda}{r+\lambda}$ in my equations, and nothing else would change. $\endgroup$ – probabilityislogic Mar 11 '11 at 11:34
  • $\begingroup$ @tsukiko - in regards to the second part of your comment, the "high" number is not "wrong" per se - I was just explaining why your numbers were higher than the theoretical values. If you did 100 bootstraps, you may find the averages closer to the theoretical values. $\endgroup$ – probabilityislogic Mar 11 '11 at 11:36
  • $\begingroup$ @tsukiko - on looking at your question, you want to run a dispersion test for the poisson. The theory says that $X$ is necessarily "under dispersed" because it is binomial $Var(X)<E(X)$. So you know from the start, before doing any bootstraps, that it is not dispersed according to poisson. Further you also know by how much, a factor of $(1-\theta)$. So if $\theta$ is small, then the approximation will be good. But $\theta$ is not small in the illustration. You can run bootstraps, but this will only demonstrate the maths. $\endgroup$ – probabilityislogic Mar 11 '11 at 11:46
  • $\begingroup$ @probabilityislogic I agree with your explanation. Tq for that. I'm using Poisson instead of binomial perhaps for the sake of studying the spatial pattern of "successes" in the original data. $\endgroup$ – user3638 Mar 11 '11 at 12:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy