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I am reading about confidence intervals and got stuck with this example from L. Wasserman's book titled "All of Statistics". Could anybody explain why PQ(θ ∈ C) = 3/4 in this example? Below is the paragraph from the book:

Let θ be a fixed, known real number and let X1, X2 be independent random variables such that P(Xi = 1) = P(Xi = -1) = 1/2. Now define Yi = θ + Xi and suppose that you only observe Y1 and Y2. Define the following interval that actually contains only one point:

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You can check that, no matter what θ is, we have Pθ(θ ∈ C) = 3/4 so this is a 75 percent confidence interval. Suppose we now do the experiment and we get Y1 = 15 and Y2 = 17. Then our 75 percent confidence interval is {16}. However, we are certain that θ = 16. If you wanted to make a probability statement about θ you would probably say that P(θ ∈ C|Y1, Y2) = 1. There is nothing wrong with saying that {16} is a 75 percent confidence interval. But is it not a probability statement about θ.

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  • $\begingroup$ I didn't fing the definition of $\theta$, is it the same as Q ? $\endgroup$
    – Pohoua
    Mar 31, 2020 at 11:56
  • $\begingroup$ Yes, it's the same. Sorry for that. I'll change. Do you have any ideas? $\endgroup$
    – kirgol
    Mar 31, 2020 at 12:18
  • $\begingroup$ There are only four possible outcomes: make a four-row table of the possibilities, work out their probabilities, and check which rows correspond to the event ${\Pr}_\theta(\theta\in C).$ $\endgroup$
    – whuber
    Mar 31, 2020 at 13:43
  • $\begingroup$ @whuber sorry, can you elaborate on that?. I only see that X1 and X2 can be 1 or -1 and in both cases they have P = 1/2. They are independent, thus P(X,Y) for all 4 possibilities is 1/4. How then we get 3/4 for Q? $\endgroup$
    – kirgol
    Mar 31, 2020 at 16:23
  • $\begingroup$ Because the event comprises three of the rows, whence its chance is three times 1/4. $\endgroup$
    – whuber
    Mar 31, 2020 at 16:25

1 Answer 1

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You can dissociate cases :

  • if $X_1 \neq X_2$, which happens with probability $\frac{1}{2}$, then $X_1 = -X_2$ (since $X$ can only be $1$ or $-1$) and $Y_1 \neq Y_2$. So $C = \{\frac{Y_1 + Y_2}{2}\} = \{\theta\}$. So $\theta \in C$.

  • if $X_1 = X_2 = 1$, which happens with probability $\frac{1}{4}$, then $Y_1 = Y_2 = \theta + 1$ and $C = \{\theta\}$. So $\theta \in C$.

  • if $X_1 = X_2 = -1$, which happens with probability $\frac{1}{4}$, then $Y_1 = Y_2 = \theta - 1$, and $C = \{\theta - 2\}$. So $\theta \notin C$.

So the only possible case in which $\theta \notin C$ is when $X_1 = X_2 = -1$, which happens with probability $\frac{1}{4}$, so $P(\theta \in C) = 1 - \frac{1}{4} = \frac{3}{4}$

I think the point of Wasserman here is that the randomness lies in $C$ and not $\theta$. And indeed in the different cases considered, each time it is the confidence intervall $C$ which changes, not $\theta$.

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  • $\begingroup$ Thank you very much for your explanation. I understood everything $\endgroup$
    – kirgol
    Apr 4, 2020 at 11:56

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