4
$\begingroup$

Here is a little brainteaser:

Suppose we know $x|(\mu,\sigma^2) \sim N(\mu, \sigma^2)$ and that $\sigma^2$ can either take on the value 1 or 2 on any particular draw from the distribution (i.e., there is a 50/50 chance that it will be 1 or 2). The goal is to construct a confidence interval with exactly 95% coverage for $\mu$.

There are (at least) two angles of attack here:

  1. We know that the coverage of $x \pm 1.96 \times \sqrt{1}$ is too low and the coverage of $x \pm 1.96 \times \sqrt{2}$ is too high. But on any draw, we could randomly pick one or the other CI (i.e., we use a randomized CI). The problem is then figuring out with what probability we should pick the first or second interval, so that we end up with 95% coverage in the long run.
  2. We know that there is some value $\tilde{\sigma}^2$ between 1 and 2 that will give us the desired 95% coverage if we compute the CI with $x \pm 1.96 \times \sqrt{\tilde{\sigma}^2}$. The problem is then finding this $\tilde{\sigma}^2$ value.

And just in case: This is not homework. If you care to check my profile and go to my website, you'll find that my school days are long over. I actually just constructed this little exercise myself and figured others may enjoy trying to solve it. I'll post an answer in due time.

$\endgroup$
2
$\begingroup$

It's been a while since I've done things like this but I think you would argue as follows.

First, you're distribution is a finite mixture of two normals (for details: What is the variance of the weighted mixture of two gaussians?).

Following the logic there, we have $f(y) = .5f(X_1) + .5f(X_2)$ where Y is the mixture.

That means that $Var(Y) = p_1\sigma_1^2 + p_2\sigma_2^2 + p_1p_2(\mu_1-\mu_2)^2 = .5*1 + .5*2 + .25(\mu_1 - \mu_2)^2 = 1.5 + .25(\mu_1-\mu_2)^2 = 1.5$.

Similarly, $E(Y) = .5E(X_1) + .5E(X_2) = .5\mu + .5\mu = \mu$

So now we have the mean and the variance. However, we can NOT assume that a mixture of two normals is still normal -- and generally this is not the case.

Thus, I don't think your confidence bound for $x$ will be in the form of $x \pm 1.95\sqrt{1.5}$. Although using the central limit theorem, you can calculate the confidence bound for $\bar{x}$ as $\bar{x} \pm 1.96 \sqrt{1.5 /n}$.

As an example of the failure of normality, please see the below R simulation where I simulate data from the process you described. Even though if you hist(data) , the plot will look fairly normal, you can see from the shapiro wilk's test p-value that the data is not normal.

> data <- unlist(lapply(1:5000, function(x){
+   if(rbinom(1,1,.5)){
+     return(rnorm(n = 1, mean = 0, sd = 1))
+   }else{
+     return(rnorm(n = 1, mean = 0, sd = sqrt(2)))
+   }
+ }))
> 
> mean(data)
[1] 0.02801353
> var(data)
[1] 1.480294
> shapiro.test(data)

    Shapiro-Wilk normality test

data:  data
W = 0.99833, p-value = 3.78e-05
$\endgroup$
  • $\begingroup$ You are correct in that the marginal distribution of $x$ is a mixture and it is certainly not normal. Still, one can construct a CI of the form $x \pm 1.96 \sqrt{\tilde{\sigma^2}}$ such that it will have exactly 95% coverage. $\endgroup$ – Wolfgang Nov 18 '16 at 19:11
  • $\begingroup$ @Wolfgang: I believe the 1.96 comes from the fact that you are using the normal distribution. If it's not normal, why would 1.96 still be valid? $\endgroup$ – user1357015 Nov 18 '16 at 19:14
  • $\begingroup$ The mixture is not normal, but conditional on $\sigma^2$, we still have normal distributions. So the trick is to use this fact. Then we can work out the coverage of $x \pm 1.96 \sqrt{1}$ and $x \pm 1.96 \sqrt{2}$ and try to use this for either solution 1) or 2). $\endgroup$ – Wolfgang Nov 18 '16 at 19:20

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.