Let $X_1,\ldots,X_n$ be iid from $\chi^2_v$, where $v\in \mathbb{N}$. So $E(X_i) = v$ and $Var(X_i) = 2v$.

  1. Find a statistic $Y_n$ such that $$\frac{\sqrt{n}(\bar{X}_n-v)}{Y_n} \overset{D}{\to} N(0,1)$$
  2. Suppose $n = 100$ and $\bar{x_n} = 10$. Use the asymptotic result in part 1 to obtain an approximate 95% confidence interval of $v$.

Attempt: Since we have a random sample with common mean and variance we can use the central limit theorem. So we know that $$\frac{\sqrt{n}(\bar{X}_n-v)}{\sqrt{2v}} \overset{D}{\to} N(0,1) \text{ in distribution}$$

Since we need a statistic, let $Y_n$ be the sample standard deviation. Then we get the result using Slutsky's theorem because we know that the sample standard deviation converges in probability to the population standard deviation.

Since we are trying to construct a confidence interval for $v$, we know that the population variance is unknown. We could use the sample variance instead provided the sample size is large enough. But how can I do so without having the realized sample variance of the data? I can't deduce the sample variance using only the sample mean.

Any hints would be appreciated.


2 Answers 2


Since you only know $n$ and $\bar{X}_n$ why not choose a $Y_n$ that isn't a function of anything else but (at most) those two things?

You chose a $Y_n$ without considering what you have to work with and then struck a problem ... but you didn't reconsider whether you might have chosen a different $Y_n$ that avoided the problem.

  • $\begingroup$ The better choice wasn't so obvious to me for some reason... The choice in distribution seemed a bit odd to me, until I read your comments. We can use the sample mean to estimate the variance. $\endgroup$ Commented Apr 13, 2017 at 19:24
  • $\begingroup$ So we should let $Y_n = \sqrt{2*\bar{X}_n}$ as this would not provide a problem. $\endgroup$ Commented Apr 13, 2017 at 19:25
  • $\begingroup$ That would seem to be the most obvious choice. $\endgroup$
    – Glen_b
    Commented Apr 13, 2017 at 21:13

As noted in the OP, we run into issues when trying to use the sample standard deviation. We know that $$S_n^2 = \frac{n}{n-1}\left( \frac{\sum X_i^2}{n} - \bar{X}_n^2\right)\overset{p}{\to} \text{Var}(X_i),$$ but as we see in the second part of the problem, we are not given the realized value of $\sum X_i^2.$

To get around this problem note that in this problem we have the following relationship between the variance and the mean: $$\text{Var}(X_i) = 2v = 2\text{E}(X_i).$$

This allows us to see that we can estimate the variance by doubling our estimate for the mean. WLLN tells us that $$\bar{X}_n \overset{p}{\to}\text{E}(X_i).$$ Thus, by Slutsky's we know that $$2\bar{X}_n \overset{p}{\to}\text{Var}(X_i).$$

As mentioned in the OP, CLT tells us that $$\frac{\sqrt{n}(\bar{X}_n-v)}{\sqrt{2v}} \overset{D}{\to} N(0,1).$$ Therefore, the best choice for $Y_n$ to complete the problem would be $$Y_n = \sqrt{2\bar{X}_n}.$$

This will yield no issues in calculating a confidence interval for $v$.


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