# Question about calculating confidence intervals

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:

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.

• I didn't fing the definition of $\theta$, is it the same as Q ? – Pohoua Mar 31 '20 at 11:56
• Yes, it's the same. Sorry for that. I'll change. Do you have any ideas? – kirgol Mar 31 '20 at 12:18
• 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 ${\Pr}_\theta(\theta\in C).$ – whuber Mar 31 '20 at 13:43
• @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? – kirgol Mar 31 '20 at 16:23
• Because the event comprises three of the rows, whence its chance is three times 1/4. – whuber Mar 31 '20 at 16:25

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 $$\theta$$.

• Thank you very much for your explanation. I understood everything – kirgol Apr 4 '20 at 11:56