Statistics 5601 (Geyer, Spring 2006) Examples: Parametric Bootstrap

General Instructions

To do each example, just click the "Submit" button. You do not have to type in any R instructions or specify a dataset. That's already done for you.

Theory

The theory of the parametric bootstrap is quite similar to that of the nonparametric bootstrap, the only difference is that instead of simulating bootstrap samples that are IID from the empirical distribution (the nonparametric estimate of the distribution of the data) we simulate bootstrap samples that are IID from the estimated parametric model.

All the same considerations arise.

• Since the parameter estimate theta hat is not the true parameter value theta. We do not sample from the correct distribution. We should sample from Fθ. We do sample with the same thing with a hat on the θ (which I can't do on a web page).

• Thus the bootstrap does not do the right thing, only close to the right thing when the sample size is large.

• In constructing confidence intervals, it helps to bootstrap pivotal or at least variance-stabilized quantities.

And so forth.

Simulating from a parametric model is not so easy as simulating from the empirical distribution. In fact, it can be arbitrarily complicated. So hard that it is an open research problem how to do it. For some parametric models sampling is easy, others not. In general, it bears no relation to samping from the empirical.

If the observed data are in the vector `x`, then

```x.star <- sample(x, replace = TRUE)
```

makes a nonparametric bootstrap sample.

In contrast, if the observed data are assumed to be IID normal, then

```x.star <- rnorm(length(x), mean = mean(x), sd = sd(x))
```

makes a parametric bootstrap sample. This does not do the right thing because we should specify `mean` to be the true population mean and `sd` to be the true population standard deviation (but since we don't know the population values we must use estimates).

For more contrast, if the observed data are assumed to be IID Cauchy, then

```x.star <- rcauchy(length(x), location = median(x), scale = IQR(x) / 2)
```

makes a parametric bootstrap sample. We can't use `mean(x)` and `sd(x)` as estimators of location and scale because the Cauchy distribution doesn't have moments and hence these aren't consistent estimators (of anything, much less location and scale). Why `median(x)` and `IQR(x) / 2` are consistent (even asymptotically normal) estimators of location and scale would be more theory than we want to go into here. The only point we wanted to make is that the three examples look a lot different from each other.

The Normal Location Problem

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1. We are using a trimmed mean `mean(x, trim = 0.25)` for the location estimator (on-line help). The untrimmed mean is optimal for the normal distribution (assuming we really believe the data are normal), but we can use whatever estimator we want and the bootstrap is still valid.
2. We are using the median absolute deviation `mad(x)` for the scale estimator (on-line help). The standard deviation is optimal for the normal, but . . . .

The Cauchy Location Problem

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1. We are using a trimmed mean `mean(x, trim = 0.25)` for the location estimator (on-line help). No simple estimator is optimal.
2. We are using the median absolute deviation `mad(x)` for the scale estimator (on-line help). No simple estimator is optimal. We change the `constant` argument to be right for the Cauchy distribution. The population MAD being given by `qcauchy(0.75)`, which is 1.0.

Theory II: Multinomial Sampling

To get to some examples with wading through a tremendous amount of theory, we will stick to one parametric model for which the sampling looks fairly similar to the nonparametric bootstrap. This is the multinomial distribution.

The multinomial distribution is the distribution of categorical measurements on IID individuals. The number of individuals in each category make up the data vector `x` and the probabilities of individuals being in each category make up a probability vector `p` (where probability vector means `all(p >= 0)` and `sum(p) == 1`).

Given a probability vector `p` of length `k` and a sample size `n` one creates a multinomial sample with the R statements

```c.star <- sample(1:k, n, prob = p, replace = TRUE)
x.star <- tabulate(c.star, k)
```

(The first statement creates an IID sample of category numbers with the specified probabilities. The second counts the number of individuals in each category. So `x.star` is a vector of length `k`.)

A Chi-Square Test of Goodness of Fit

For example, suppose we observe the multinomial data defined to be `x` in the form below, and we want to test the null hypothesis that the true category probabilities are all equal (to 1 / 6 because there are 6 categories). The R function `chisq.test` does the usual chi-square test that uses the large-sample approximation (that the chi-square test statistic has a chi-square distribution). The remainder of the code does the parametric bootstrap test.

Actually, since the null hypothesis is completely specified here this is, strictly speaking, a Monte Carlo test rather than a parametric bootstrap. The test is exact. (We only say we are doing a parametric bootstrap when the null distribution of the test statistic depends on the true unknown parameter value and we are plugging in an estimate for the unknown truth).

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Goodness of Fit with Estimated Parameters

For another example, suppose we observe the Poisson data defined to be `x` in the form below, and we want to test the null hypothesis that the data is Poisson against the alternative hypothesis that is isn't.

The usual way to do this is to truncate the data, making a category 8 or more for example. The reason for the truncation is that the usual asymptotics of the chi-square test requires it. But this introduces needless complication when we are not using the asymptotics (except that theta hat is near theta).

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• The reason why `tabulate(1 + x, 1 + max(x))` is that `x` contains integers starting at zero but `tabulate` wants integers starting at one.
• The test statistic calculated by the `tstat` function is the likelihood ratio test statistic. By standard likelihood theory, it is asymptotically equivalent to the chi-square test statistic, but is not exactly the same for any finite sample size.
• The reason why `foo[counts == 0] <- 0` is that when `counts[i]` is zero, then so is `p1[i]`, but that makes `foo[i]` zero times minus infinity equals `NaN`, which is not right. Since x log(x) converges to zero as x goes to zero, we want `foo[i]` to be zero in that case.
• Interpretation of the P-value: P = 0.34 means no evidence against H0. Hence we accept the null hypothesis that the data are Poisson.

(A goodness of fit test like this is just like every other hypothesis test except that we want to accept H0 as opposed to the usual situation where we want to reject. Our acceptance of H0 doesn't mean that the data actually have the Poisson distribution, only that they provide no reason to disbelieve they do.)

A Chi-Square Test of Independence

For another example, suppose we observe the contingency table defined to be `x` in the form below, and we want to test the null hypothesis of independence (that the row category labels and column category labels are independent random variables).

For some reason, the R `chisq.test` function does this test for 2-way tables by simulation, but doesn't do the analogous test for 1-way tables (go figure).

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Logistic Regression

Theory

The `glm` function in R (on-line help) does so-called generalized linear models. The theory of these is usually covered in the categorical data course (like Stat 5421), so we won't cover it here.

For those interested, I wrote up some notes about generalized linear models for the other class I am teaching this semester (Stat 5931), but there's no real need to look at it to understand this example.

The response variable in this problem `kyphosis` is categorical with values `present` or `absent` which we model as independent but not identically distributed Bernoulli random variables

Yi = Bernoulli(pi)

where the so-called mean value parameter vector

p = (p1, . . ., pn)

is related to the so-called linear predictor vector

η = (η1, . . ., ηn)

ηi = logit(pi) = log(pi / (1 - pi))

and the linear predictor vector has the usual form of a mean function in ordinary linear regression

η = X β

where X is the so-called design matrix or model matrix for the problem (the rows are cases and the columns are predictor variables) and β is a vector of regression coefficients.

Kyphosis is a misalignment of the spine. The data are on 83 laminectomy (a surgical procedure involving the spine) patients. The predictor variables are `age` and `age^2` (that is, a quadratic function of age), `number` of vertebrae involved in the surgery and `start` the vertebra number of the first vertebra involved.

Practice

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1. The command `pred <- predict(out2, type = "response")` estimates the mean value parameter vector p. If `type = "link"` (the default) it produces the linear predictor η, which is not what we need for the next item.
2. We use predicted values for `out2` because that is the fit for the small model (null hypothesis) and you always simulate under the null hypothesis when doing a test.
3. The command `kyphosis.star <- rbinom(n, 1, pred)` simulates new Bernoulli data with mean value parameter vector p. That's what the parametric bootstrap requires.
4. We save both the bootstrap values of the test statistic `dev.star` and the P-value `pev.star` (which can also be thought of as a test statistic with the proviso that we reject the null for low values of `pev` as opposed to high values of `dev` and other usual test statistics).