# Understanding an example of using Bayesian and Frequentist inference

I have problems to understanding the following discussion.

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.

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

Source: Another version casi2.pdf

• 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." – Bernhard Apr 15 '20 at 7:41
• For me it is 2016(I am not sure!). I see it now for yours. – Masoud Apr 15 '20 at 8:39
• The version i use , added at the end of question. – Masoud Apr 15 '20 at 8:49

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.