# Paradoxical result when computing a confidence interval when population standard deviation is not known

I have been trying to construct a $$1-\alpha$$ confidence interval for the mean. The distribution from which a sample is drawn is exponential distribution with density:

$$p(x) = \frac{1}{\beta}e^{-x/\beta}$$

The exponential distribution has mean $$\mu=\beta$$ and standard deviation $$\sigma=\beta$$. As far as my knowledge is concerned, for a sample $$X_1, X_2, \cdots, X_n$$ of size $$n$$, the confidence interval :

$$\left(\bar{X}_n-z_{\alpha/2}\frac{\sigma}{\sqrt{n}}, \bar{X}_n+z_{\alpha/2}\frac{\sigma}{\sqrt{n}}\right)$$

should trap the true $$\mu$$ with $$1-\alpha$$ chance (I am not saying that this interval contains $$\mu$$ with probability $$1-\alpha$$; that will be wrong since $$\mu$$ is not a random variable).

However, this assumes that population standard deviation $$\sigma$$ is known. Since I want to simulate a situation where that too needs to be estimated, I constructed the forllowing interval:

$$\left(\bar{X}_n-t_{\alpha/2}\frac{\hat{\sigma}}{\sqrt{n}}, \bar{X}_n+t_{\alpha/2}\frac{\hat{\sigma}}{\sqrt{n}}\right)$$

where, $$\hat{\sigma}=\sqrt{\frac{1}{n-1}\sum_{i=1}^n (X_i-\bar{X}_n)^2}$$

and $$t_{\alpha/2}$$ is the critical t-value obtained for the Student's t-distribution for $$n-1$$ degrees of freedom.

I wrote a small python script to verify that with this modification, the constructed interval is indeed $$95\%$$ confidence interval when $$\alpha=0.05$$. For my simulations, I chose $$n=10$$, so that we have $$9$$ degrees of freedom, $$t_{\alpha/2}=2.262$$. I could verify that the first interval where population variance is known indeed contains true $$\mu$$ around $$95\%$$ of the time. (I generate $$10000$$ intervals and count how many contain the known $$\mu$$). However, the second one seems to contain $$\mu$$ only $$90\%$$ of the time. I am not sure why even after correctly choosing the correct critical $$t$$ this is happening. If I increase $$t_{\alpha/2}=3.182$$ corresponding to 3 degrees of freedom, I do get $$95\%$$ interval again! But of course this makes no sense because the degrees of freedom is $$9$$ in this case. Any idea what is happening?

• Your confidence interval formula is inappropriate for your probability model. It will be approximately correct for very large samples -- much larger than $n=10$ -- due to the Central Limit Theorem. Try your script with, say, $n=500.$
– whuber
Apr 18, 2021 at 15:28
• The way you’ve written your exponential PDF, the mean is $\beta$, and the variance is $\beta^2$. You can’t know one without knowing the other. You seem to get this, but others might not pick up on this fact.)
– Dave
Apr 18, 2021 at 15:40
• @whuber : I didn't get you. What is the appropriate way in this case? Apr 18, 2021 at 15:52

If you have a random sample of size $$n$$ from $$\mathsf{Exp}(\mathrm{rate}=1/\beta),$$ (with mean $$\beta),$$ then $$\bar X/\beta \sim \mathsf{Gamma}(\mathrm{shape}=n,\mathrm{rate} = n).$$ Thus a 95% CI for $$\beta$$ is of the form $$\left(\frac{\bar X}{U},\, \frac{\bar X}{L}\right),$$ where $$L$$ and $$U$$ cut probability $$0.025$$ from lower and upper tails, respectively of $$\mathsf{Gamma}(n,n).$$

For example, let's take a random sample of size $$n=30$$ from an exponential distribution with mean $$\mu = 15.$$ In R below, I got $$\bar X =14.08$$ and 95% CI $$(10.14,\, 20.86),$$ which does happen to contain $$\mu = 15.$$ Of course, with real data from a real population, one never knows for sure if the 95% CI covers the unknown $$\mu.$$

set.seed(1234)
x = rexp(30, 1/15)
summary(x)
Min.  1st Qu.   Median     Mean  3rd Qu.     Max.
0.05992  4.10425 11.88384 14.07736 24.64535 45.78687
mean(x)/qgamma(c(.975,.025), 30, 30)
[1] 10.14004 20.86475


Three additional runs with no seed specified, gave intervals $$(12.89,\, 26.52),$$ $$(7.36,\, 15.15),$$ and $$(11.02,\, 22.67).$$ But a run using todays date as set.seed(418) missed the mark with CI $$(15.29,\, 31.46).$$

• @BrudeET: So do you mean that my assumption of t-distribution is violated for exponential? Apr 18, 2021 at 16:11
• You wrongly assume normal data when you use a z-interval. In my method you can start with $P(L \le \bar X/\beta \le U) = 0.95,$ with $L$ and $U$ chosen as stated. Then 'pivot' to isolate $\beta$ in the inequality to get $P(\bar X/U \le \beta \le \bar X/L) = 0.95,$ which provides the basis for the form of CI that I suggested. Apr 18, 2021 at 16:14
• Thanks! I should also pay attention to how fast $\bar{X}_n$ approaches normal distribution. For different underlying distributions, convergence is different. Apr 18, 2021 at 16:20
• Better yet, use exactly the right CI for the situation at hand, if you can find it. No use needlessly relying of the CLT or other methods of approximation. // The CLT works OK when $\sigma$ is known because $Z = \frac{\bar X -\mu}{\sigma/\sqrt{n}}$ converges to standard normal as $n$ becomes large. But you have to be somewhat more cautious assuming that the dist'n of $T = \frac{\bar X -\mu}{S/\sqrt{n}}$ is approximated by Student's dist'n. // Some people seem to think t intervals are automatically OK if $n > 30,$ but that is not really correct. Apr 18, 2021 at 16:33
• That's a very useful advice. Thanks. So when exactly is t-distribution helpful? When n is large, we anyway don't need it. When n is small, you seem to imply that only when the underlying distribution is normal, t-distribution is of any use. Apr 19, 2021 at 2:41