1 / 29

Confidence Intervals

Confidence Intervals. Lecture XXI. Interval Estimation. As we discussed when we talked about continuous distribution functions, the probability of a specific number under a continuous distribution is zero.

billy
Download Presentation

Confidence Intervals

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Confidence Intervals Lecture XXI

  2. Interval Estimation • As we discussed when we talked about continuous distribution functions, the probability of a specific number under a continuous distribution is zero. • Thus, if we conceptualize any estimator, either a nonparametric estimate of the mean or a parametric estimate of a function, the probability of the true value equal to the estimated value is obviously zero.

  3. Thus, usually talk about estimated values in terms of confidence intervals. Specifically, as in the case when we discussed the probability of a continuous variable, we define some range of outcomes.

  4. Confidence Intervals • Amemiya notes a difference between confidence and probability. Most troubling is our classic definition of probability as “a probabilistic statement involving parameters.” This is troublesome due to our inability without some additional Bayesian structure to state anything concrete about probabilities.

  5. Example 8.2.1. Let Xi be distributed as a Bernoulli distribution, i=1,2,…,n. Then, Therefore, we have

  6. Why? By the Central Limit Theory • Given this distribution, we can ask questions about the probability. Specifically, we know that if Z is distributed N(0,1), then we can define

  7. Table of Normal Probabilities

  8. The values of gk can be derived from the standard normal table as

  9. Assuming that the sample value of T is t, the confidence is defined by

  10. Building on the first term

  11. Using this probability, it is possible to define two numbers h1(T) and h2(T) for which this inequality holds.

  12. Example 8.2.2. Let Xi ~ N(m,s2), i=1,2,…n where m is unknown and s2 is known. We have

  13. Example 8.2.3. Suppose that Xi~N(m,s2), i=1,2,3,…n with both m and s2 unknown. Let be an estimator of m and be the estimator of s2.

  14. Then the probability distribution or the Student’s t distribution with n-1 degrees of freedom is the probability distribution.

  15. Theorem 5.4.1: Let X1, X2,…Xn be a random sample from a N(m,s2) distribution, and let

  16. and S2 are independent random variables • (N-1)S2/s2 has a chi-squared distribution with n-1 degrees of freedom.

  17. The proof of independence is based on the fact that S2 is a function of the deviations from the mean which, by definition, must be independent of the mean.

  18. The chi-square is defined as: • In general, the gamma distribution function is defined through the gamma function:

  19. Substituting X=bt gives the traditional two parameter form of the distribution function

  20. The expected value of the gamma distribution is ab and the variance is ab2. • Lemma 5.4.1. (Facts about chi-squared random variables) We use the notation cp2 to denote a chi-squared random variable with p degrees of freedom. • If Z is a N(0,1) random variable, the Z2~c12, that is, the square of a standard normal random variable is a chi-squared random variable.

  21. If X1, X2,…Xn are independent, and Xi~cpi2, then X1+X2+…Xn~c2p1+p2+…pn, that is, independent chi-squared variables add to a chi-square variable, and the degrees of freedom also add. • The first part of the Lemma follows from the transformation of random variables for Y=X2 which yields:

  22. Returning to the proof at hand, we want to show that (N-1)S2/s2 has a chi-squared distribution with n-1 degrees of freedom. To demonstrate this, note that

  23. If n=2, we get • Given that (X2-X1)/√2 is distributed N(0,1), S22~c12 and by extension for n=k, (k-1)Sk2~ck-12.

  24. Given these results for the chi-square, the distribution of the Student’s t then follows.

  25. Note that this creates a standard normal random variable in the numerator and in the denominator

  26. The complete distribution found by multiplying the standard normal time the chi-squared distribution times the Jacobian of the transformation yields:

More Related