# Probability about confidence interval

Let $X_1,...X_n$ be iid $N(\theta,1)$. A 95% confidence interval for $\theta$ is $\overline{X}\pm\frac{1.96}{\sqrt{n}}$.Let p denote the probability that an additional independent observation,$X_{n+1}$, will fall in this interval.Is p greater than, less than, equal to .95?Prove your answer.

Casella & Berger, Chapter 9 - Interval Estimation

First $\overline{X}\approx N(\theta,\frac{1}{n})\Rightarrow Y=\sqrt{n}\overline{X}$~$N(\theta,1)$

I know that $$P(\overline{X}-\frac{1.96}{\sqrt{n}}\leq\theta\leq \overline{X}+\frac{1.96}{\sqrt{n}})=0.95$$

$$P(\theta-\frac{1.96}{\sqrt{n}}\leq\overline{X}\leq\theta+\frac{1.96}{\sqrt{n}})=0.95$$ $$P(\theta-1.96\leq\overline{X}-\theta\leq\theta+1.96)=0.95$$ $$P(\theta-1.96\leq Y\leq\theta+1.96)=0.95$$

That's why I think it's the same, I tried other ways but got nowhere.

Someone give me a hint?

• Smaller (to guess which one is correct, consider $n\to\infty$). What is your try? – Did Jun 6 '15 at 21:06
• @Did It is above what I tried. – Roland Jun 6 '15 at 21:45
• ?? Sorry but where is $X_{n+1}$ in your approach? – Did Jun 6 '15 at 22:22
• I think you are being asked a question about 'prediction intervals' as in regression (but only estimating one parameter). The CI is based on the mean $\bar X = \bar X_n$ of $n$ observations. Now you have a question about an additional individual observation $X_{n+1},$ not before encountered. [By the way, translation note: I think you mean "Justify your answer." Warrant is both a noun and a verb, somewhat different meanings. The verb is rarely used in the imperative, usually as a participle warranted. Slippery idioms on that one.] – BruceET Jun 6 '15 at 22:58
• Clue must be in text somewhere ahead of problem. If I'm right about what's being asked, for prediction interval on next obs., variance is $\sigma^2(1 + 1/n).$ – BruceET Jun 6 '15 at 23:19

Perhaps a little more than a hint but here goes...

The question appears to be asking if

\begin{align} p = P\left\{\bar X - 1.96\sqrt\frac{1}{n} \le X_{n+1} \le \bar X + 1.96\sqrt\frac{1}{n} \right\} \\ = P\left\{-1.96\sqrt\frac{1}{n} \le X_{n+1} - \bar X \le 1.96\sqrt\frac{1}{n} \right\} \\ \end{align}

is less than, equal to, or greater than $0.95$.

Now, it turns out $X_{n+1} - \bar X$ (itself a linear combination of normal random variables) is also normal, with mean $0$ and variance $1+\frac{1}{n}$ (due to independence of $X_{n+1}$ from the other $X_i$ and hence from $\bar X$).

Then we have that $P\left\{ -1.96\sqrt{1+\frac{1}{n}} \le X_{n+1} - \bar X \le 1.96\sqrt{1+\frac{1}{n}} \right\} = 0.95$ as well.

Now take a look at the interval above and notice that

$$\left[ -1.96\sqrt{1+\frac{1}{n}}, \ \ 1.96\sqrt{1+\frac{1}{n}} \ \right] = \left[ -1.96\sqrt{1+\frac{1}{n}}, \ \ -1.96\sqrt{\frac{1}{n}} \ \right) \\ \bigcup \color{red}{ \left[ -1.96\sqrt{\frac{1}{n}}, \ \ 1.96\sqrt{\frac{1}{n}} \ \right]} \\ \bigcup \left( 1.96\sqrt{\frac{1}{n}}, \ \ 1.96\sqrt{1+\frac{1}{n}} \ \right]$$

Can you deduce where $p$ stands relative to 0.95?