I am brushing up on statistical theory to keep my mind active, and I have a question about confidence intervals.

Given a normal distribution with unknown mean $\mu$ and known variance 1, I know that the distribution of the sample mean of $n$ observations follows a $N(\mu, 1/n)$ distribution, so a $95\%$ confidence interval for $\mu$ is $$\bar{X} \pm \frac{1.96}{\sqrt{n}}$$ I also know that if I want to construct a $95\%$ prediction interval for any future observation $X^*$ (assuming the same normal distribution as before), it will be $$\bar{X} \pm 1.96 \sqrt{1+\frac{1}{n}}$$ which is longer.

However, what if I wanted to calculate an upper bound of the probability that a future observation $X^*$ falls in the $95\%$ confidence interval for the mean (i.e., the first interval above)? Since it is not as long as the $95\%$ prediction interval, my intuition is that it cannot be $95\%$ (it will be lower), but I don't know how to prove it to myself, along with deducing exactly what the probability will be that any future observation falls into the confidence interval for the mean of the observed data. Is simply saying that the prediction interval is wider a "proof?"

Any insights would be appreciated.

  • $\begingroup$ This follows rather directly from some of the axioms of probability, of which the most relevant is that when one event $A$ is a subset of another event $B$, then $\Pr(A)\le \Pr(B)$. In fact you can compute this chance exactly with little trouble: simply solve the equation $1.96/\sqrt{n}=Z\sqrt{1+1/n}$ and find what level of coverage $Z$ corresponds to. $\endgroup$
    – whuber
    Commented Apr 6, 2015 at 21:19
  • $\begingroup$ unfortunate that this was deleted; I could see it being useful to other readers. $\endgroup$
    – Glen_b
    Commented Apr 7, 2015 at 6:56

1 Answer 1


Your intuition is correct that it will be lower than 95%. Here is one way to demonstrate it. Let $X_1, \dots , X_n$ be a random sample from a $N(\mu, 1)$ distribution. As you stated the 95% confidence interval, we can write

\begin{align} 0.95 &= P\Bigg[ \bar{X} - \frac{1.96}{\sqrt{n}} \leq \mu \leq \bar{X} + \frac{1.96}{\sqrt{n}} \Bigg] \\ &= P\Bigg[ -1.96 \leq \frac{\mu - \bar{X}}{1/\sqrt{n}} \leq 1.96 \Bigg] \\ &= P \big[ -1.96 \leq Z \leq 1.96 \big], \tag{1} \end{align} where $Z \sim N(0,1)$.

Now, let $X^*$ be a future observation from $N(\mu, 1)$ independent of the first $n$ observations. We want to calculate the probability that it falls in the confidence interval, i.e.

$$ P \Bigg[ \bar{X} - \frac{1.96}{\sqrt{n}} \leq X^* \leq \bar{X} + \frac{1.96}{\sqrt{n}} \Bigg] = P \Bigg[ - \frac{1.96}{\sqrt{n}} \leq X^* - \bar{X} \leq \frac{1.96}{\sqrt{n}} \Bigg] \tag{2} $$

and compare it to 0.95. I re-wrote it this way so that it's easy to see that it is a probability statement about the random variable $X^* - \bar{X}$. Since $X^* - \bar{X}$ is a linear combination of independent normal random variables, it also has a normal distribution (see e.g. Corrollary 4.6.10 in Casella/Berger), and it is not hard to calculate that it has mean $0$ and variance $1 + \frac{1}{n}$. Continuing from $(2)$, divide through the terms by the standard deviation to obtain

\begin{align} P \Bigg[ - \frac{1.96}{\sqrt{n}} \leq X^* - \bar{X} \leq \frac{1.96}{\sqrt{n}} \Bigg] &= P \Bigg[ - \frac{1.96}{\sqrt{n}\sqrt{1+\frac{1}{n}}} \leq \frac{X^* - \bar{X}}{\sqrt{1+\frac{1}{n}}} \leq \frac{1.96}{\sqrt{n}\sqrt{1+\frac{1}{n}}} \Bigg] \\ &= P \Bigg[ - \frac{1.96}{\sqrt{n+1}} \leq Z^* \leq \frac{1.96}{\sqrt{n+1}} \Bigg], \tag{3} \end{align} where $Z^* \sim N(0,1)$. Now $(1)$ and $(3)$ are directly comparable since they are both probability statements about a standard normal random variable. Since the interval in $(3)$ is a smaller interval contained in the interval in $(1)$, it must have a smaller probability. In summary,

\begin{align} P \Bigg[ \bar{X} - \frac{1.96}{\sqrt{n}} \leq X^* \leq \bar{X} + \frac{1.96}{\sqrt{n}} \Bigg] &= P \Bigg[ - \frac{1.96}{\sqrt{n+1}} \leq Z^* \leq \frac{1.96}{\sqrt{n+1}} \Bigg] \\ &< P \big[ -1.96 \leq Z \leq 1.96 \big] \\ &= 0.95. \end{align}

Notice that this argument used exactly what was suggested in whuber's comment. It is interesting to plug in some values of $n$ and see what we get. For example, using $n=20$, calculating in R I get

> n <- 20
> pnorm(q=1.96/sqrt(n+1)) - pnorm(q=-1.96/sqrt(n+1))
[1] 0.3311356

By the way, your question is similar to Exercise 9.2 in Casella/Berger.


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