{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

lec24 - .1 Goodness-of-fit test Suppose that we observe an...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Lecture 24 24.1 Goodness-of-fit test. Suppose that we observe an i.i.d. sample X 1 , . . . , X n of random variables that can take a finite number of values B 1 , . . . , B r with some unknown to us probabilities p 1 , . . . , p r . Suppose that we have a theory (or a guess) that these probabilities are equal to some particular p ◦ 1 , . . . , p ◦ r and we want to test it. This means that we want to test the hypotheses H 1 : p i = p ◦ i for all i = 1 , . . . , r, H 2 : otherwise, i.e. for some i, p i 6 = p ◦ i . If the first hypothesis is true than the main result from previous lecture tells us that we have the following convergence in distribution: T = r X i =1 ( ν i- np ◦ i ) 2 np ◦ i → χ 2 r- 1 where ν i = # { X j : X j = B i } . On the other hand, if H 2 holds then for some index i , p i 6 = p ◦ i and the statistics T will behave very differently. If p i is the true probability ( X 1 = B i ) then by CLT (see previous lecture) ν i- np i √ np i → N (0 , 1- p i ) . If we write ν i- np ◦ i √ np ◦ i = ν i- np i + n ( p i- p ◦ i ) √ np ◦ i = ν i- np i √ np i + √ n p i- p ◦ i √ p ◦ i then the first term converges to N (0 , 1- p i ) but the second term converges to plus or minus ∞ since p i 6 = p ◦ i . Therefore, ( ν i- np ◦ i ) 2 np ◦ i → + ∞ 94 LECTURE 24.LECTURE 24....
View Full Document

{[ snackBarMessage ]}

Page1 / 5

lec24 - .1 Goodness-of-fit test Suppose that we observe an...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon bookmark
Ask a homework question - tutors are online