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So I've been working on a problem in my probability class on which I have become stuck. It involves X1 X2 ... Xn ~ Poisson(lambda)

1 - We were instructed to show that X_bar was sufficient.

Sol'n: I did this using factorization, and by finding T(x) = Sum(x_i)/n, which came from the exponent position above my lambda from the likelihood/pdf. It could've also been achieved via the exponential family method.

2 - We were then instructed to show that X_bar^2 is approximately normal with mean \lambda^2 and a variance depending on \lambda

Sol'n: This was achieved using the central limit theorem and the delta method, which ended up giving me N(0, 4*lambda^3)

In step 3, however, we are asked "to display an approximate pivot based on the result in part #2, and then produce a 95% confidence interval for lambda^2"

We were given a hint, namely that 

sqrt(n)(x_bar^2 - lambda^2) / sqrt (lambda * 4 lambda^2) ~ N(0,1)

I have, up until this point, no concept of what to do with a pivot or with this given info...I am looking for a number of degrees of freedom for a poisson table, which I think is wrong, but I've clearly got an issue with understanding what is being asked of me. If anyone could give me a nudge in the right direction, I'd be really grateful.

Gosset
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1 Answers1

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As opposed to a statistic, a pivotal statistic (or pivotal quantity) can depend on both the sample and parameters, but its distribution does not depend on any parameters. A good example is the standardized mean from a normal distribution:

$\frac{\sqrt n(\bar{X}-\mu)}{\sigma} \sim N(0,1)$

This quantity depends on the sample and parameters, but its distribution does not depend on parameters so it is a pivotal quantity. When you are asked to construct a CI based on pivotal quantity, just use the critical values of the distribution of the pivotal quantity. So for the example I just gave (and using the approximate critical values for the standard normal), the 95% CI based on the pivotal quantity is:

$-2\leq\frac{\sqrt n(\bar{X}-\mu)}{\sigma}\leq2$

In your example, you just need to construct this interval based on the approximate distribution of the pivotal quantity and then solve for $\lambda^2$

BLimkins
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  • thanks that was more insightful than my professor's been all year hahaha thank you so much for taking the time to help me! – Gosset May 02 '14 at 15:34