1
$\begingroup$

I have problems to understanding the following discussion.

enter image description here

enter image description here

The questions are:

1)In "Some computation shows that this rule had probability 0.083,..." how $0.083$ calculated?(In a different version wrote $.074$)

2)Why "The stopping rule doesn’t affect the posterior distribution"? and i do not understand why it told ? So what?

3)In the discussion, a stopping rule changed, so likelihood inference changed , but Bayesian not. Is it really a better property? What does that mean? More generally how can i challenge this discussion? Can i find a stopping time rule such that Frequentist Inference has a good property but bayesian Inference does not(opposite of the discussion )? (If it is so finding a rule that bayesian approach has a good property but Frequentist Inference does not, means nothing).

For question (1)

$$P_{H_0}(Z_{30}>1.645 \ or \ Z_{20}>1.645)= P(Z_{30}>1.645)+\mathbb P(Z_{20}>1.645)-P(Z_{30}>1.645 \ and \ Z_{20}>1.645)= 0.1-P(Z_{30}>1.645 \ and \ Z_{20}>1.645)$$ $$P(Z_{30}>1.645 \ and \ Z_{20}>1.645)=P(\frac{\sum_{j=1}^{20}X_j +\sum_{j=21}^{30}X_j}{\sqrt{30}}>1.645 \ and \ Z_{20}>1.645)$$

$$=P(\sqrt{20} Z_{20} +\sum_{j=21}^{30}X_j>1.645\sqrt{30} \ and \ Z_{20}>1.645)$$ But I am stuck here.

Thanks in advance for any help you are able to provide.

Source:casi.pdf, 3.3 Flaws in Frequentist Inference (Page 31).

Source: Another version casi2.pdf

$\endgroup$
3
  • 1
    $\begingroup$ It may not help, but the version I just downloaded states 0.074 where your's says 0.083. The book title states "Corrected November 10, 2017." $\endgroup$ – Bernhard Apr 15 '20 at 7:41
  • $\begingroup$ For me it is 2016(I am not sure!). I see it now for yours. $\endgroup$ – Masoud Apr 15 '20 at 8:39
  • $\begingroup$ The version i use , added at the end of question. $\endgroup$ – Masoud Apr 15 '20 at 8:49
1
$\begingroup$

I can only provide a (partial) answer to 1). This is a topic for example addressed by P. Armitage, C. K. McPherson and B. C. Rowe (1969), Journal of the Royal Statistical Society. Series A (132), 2, 235-244: "Repeated Significance Tests on Accumulating Data".

They consider simulation, and so I will just follow them:

n <- 30

snoop <- function(n){
  x <- rnorm(n)
  t30 <- sum(x)/sqrt(n)
  t20 <- sum(x[1:20])/sqrt(20)

  q <- qnorm(.95)
  return(t20 > q | t30 > q)
}

mean(replicate(1e6, snoop(n)))

This returns values around the 0.074 you quote.

$\endgroup$

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.