Suppose we have $\alpha_1,..\alpha_n,\beta_1,..\beta_n$~$N(0,1)$.

Define $x_i = u_i + \sigma\alpha_i$, $y_i = u_i + \sigma\beta_i$, for $(\mu_1,..,\mu_n) \in \mathbb{R}^n$, $\sigma > 0$.

Consider the model $(x_1,y_1),..,(x_n,y_n)$, for it we get the likelihood function which will yield:

$\mu_i* = \frac{x_i+y_i}{2}$ as the MLE estimator for $\mu_i$, and $\sigma* = \frac{1}{4n}\sum^{23}_{i=1}(x_i-y_i)^2$.

It follows the $\mu_i*$ is unbiased for $\mu_i$, and $2\sigma*$ for $\sigma$.

I want to build a $0.99$ confidence interval for $\mu_i$, supposing $n=23$;

I'm unsure which is the correct way to calculate this -

either $[\mu_i* \pm \frac{\sqrt{\sigma*}}{\sqrt{23}}t_{22,.975}]$

or $[\mu_i* \pm \frac{\sqrt{\sigma*}}{\sqrt{2}}t_{22,.975}]$. I think the latter is correct since we would base this CI on the Cental Limit Theorem using the mean of $x_i$ and $y_i$.

  • $\begingroup$ Is $u_i$ indeed dependent on $i$? If so, why is $u*$ without an index - shouldn't it depend on $i$ too? Also, where did $u$ come from, and what is its relationship to the $u_i$? $\endgroup$
    – Ami Tavory
    Jul 1, 2017 at 17:23
  • $\begingroup$ You still have "It follows the $\mu*$ is unbiased for $\mu$, and $2\sigma*$ for $\sigma$.". Also, I don't quite get the relevance of $n$. If indeed $\mu_i$ and $\sigma_i$ are dependent on $i$ and unrelated, why does having 23 of them make a difference? $\endgroup$
    – Ami Tavory
    Jul 1, 2017 at 17:45
  • $\begingroup$ @AmiTavory edited again :). It makes a difference only because our estimate of $\sigma$ is dependent upon $23$ samples of pairs, not $2$. $\endgroup$
    – Mariah
    Jul 1, 2017 at 17:59
  • $\begingroup$ Ah, much clearer now. I was guessing that the $\sigma$ was a typo too, and it was actually $\sigma_i$. $\endgroup$
    – Ami Tavory
    Jul 1, 2017 at 18:17

1 Answer 1


Suppose we have $\alpha_1,..\alpha_n,\beta_1,..\beta_n$~$N(0,1)$.... Define $x_i = u_i + \sigma\alpha_i$, $y_i = u_i + \sigma\beta_i$, for $(\mu_1,..,\mu_n) \in \mathbb{R}^n$, $\sigma > 0$... $\sigma* = \frac{1}{4n}\sum^{23}_{i=1}(x_i-y_i)^2$... I want to build a $0.99$ confidence interval for $\mu_i$, supposing $n=23$;

When considering the confidence interval for $\mu_i$, there are two things that need to be taken into account:

  • The estimate of $\sigma$ might be off.

  • For the true $\sigma$, the outcome of $x_i - y_i$ might be off.

Let's consider the first item. $ x_i - y_i \sim \mathcal{N}(0, 2\sigma^2)$. As such, $$ \frac{1}{2n} \sum_{i = 1}^n \left( x_i - y_i \right)^2 $$ is a known-mean unbiased estimator for $\sigma^2$ (why $4n$ in your question?). However, $ \frac{1}{2n} \sum_{i = 1}^n \left( x_i - y_i \right)^2 $ is itself a random variable, and so has variance. Using the properties of chi-square distributions (see this question), an $\epsilon$ confidence interval for $2 \sigma^2$ is

$$ \left(\frac{S^2 n}{\chi^2_{1-\epsilon2}}, \frac{S^2 n}{\chi^2_{\epsilon/2}}\right). $$

Obviously, the worst case is the right boundary of this interval. If you plug in this value for $2 \sigma^2$, you can calculate a $\delta$ interval for the normal distribution using the usual method.

However, note that you need to take both uncertainties into account. One way of upper-bounding the probability of error would be to upper-bound it by taking the probability of error as $(1 - \epsilon) ( 1 - \delta)$. If you're looking for a 99% CI, you need to use $\epsilon, \delta$, s.t. $(1 - \epsilon) ( 1 - \delta) \geq 0.99$. One way of doing so (not necessarily the optimal) would be taking $\epsilon = \delta = 1 - \sqrt{0.99}$.

  • $\begingroup$ This is a good answer thanks. I'm wondering specifically whether, assuming I keep the format of the CI's as I wrote them, to divide in the denominator by $\sqrt(2)$ or by $\sqrt(23)$; as I said, I think it's by $\sqrt(2)$ $\endgroup$
    – Mariah
    Jul 1, 2017 at 20:43
  • $\begingroup$ @Mariah I'll update that shortly. However, I think you need to take into account the confidence of the estimator for $2 \sigma^2$. IINM, you're not doing that in the calculation outlined in your answer, no? There's also the factor $\frac{1}{4n}$ where I don't follow your reasoning. $\endgroup$
    – Ami Tavory
    Jul 1, 2017 at 20:46
  • $\begingroup$ @Mariah Regarding your original question, I agree with what you wrote there - you should use the second form. I disagree with the two other points above, though. $\endgroup$
    – Ami Tavory
    Jul 1, 2017 at 20:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.