Let \(\displaystyle{X}_{{1}},\ldots{X}_{{n}}\) be iid \(\displaystyle{N}{\left(\theta,\ {1}\right)}\).

Adolfo Hebert 2022-03-21 Answered
Let X1,Xn be iid N(θ, 1). A 95% confidence interval for θ is X±1.96n. Let p denote the probability that an additional independent observation, Xn+1, will fall in this interval. Is p greater than, less than, equal to 0.95? Prove your answer.
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Answers (1)

Jaslyn Allison
Answered 2022-03-22 Author has 13 answers

Step 1
Perhaps a little more than a hint but here goes
The question appears to be asking if
p=P{X1.961nXn+1X+1.961n}
=P{1.961nXn+1X1.961n}
is less than, equal to, or greater than 0.95.
Now, it turns out Xn+1X (itself a linear combination of normal random variables) is also normal, with mean 0 and variance 1+1n (due to independence of Xn+1 from the other Xi and hence from X).
Then we have that
P{1.961+1nXn+1X1.961+1n}=0.95 as well.
Now take a look at the interval above and notice that
[1.961+1n,  1.961+1n ]
=[1.961+1n,  1.961n )
1.961n  1.961n 
(1.961n,  1.961+1n ]
Can you deduce where p stands relative to 0.95?

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