6
$\begingroup$

Find the limiting distribution of $\sqrt{n} \left(\sqrt{\bar{X}} -1 \right) $ if $\sqrt{n} \left( \bar{X}-1 \right) \to N(0,1)$. Can you please check my work below?

In principle, the Delta method should be of use here. We know that according to that particular method if $$ \sqrt{n} \left( X_n -\theta \right) \to N \left( 0,\sigma^2 \right)$$ then $$\sqrt{n} \left( g\left(X_n \right) -g \left(\theta \right) \right) \to N\left( 0,\sigma^2 g \prime \left(\theta \right)^2 \right)$$

We have $g \left(t \right) =\sqrt{t}$ and $g(1)=1$ which implies that $g \prime \left( 1 \right)= \frac{1}{2}$ and then the limiting distribution is $N \left(0, 1/4 \right)$

Is everything alright here? Thanks.

$\endgroup$
7
  • 3
    $\begingroup$ A quick simulation might set you at ease concerning the correctness of your result. In R, for example, you could compute n <- 10^6; var(sqrt(n)*(sqrt(rnorm(n, 1, 1/sqrt(n)))-1)) in about 1/10 second. $\endgroup$
    – whuber
    Dec 19, 2013 at 16:38
  • $\begingroup$ @whuber By the way if you would like to post your comment as a response that I can mark it as correct, do so. $\endgroup$
    – JohnK
    Dec 20, 2013 at 17:11
  • $\begingroup$ If $g(t) = \sqrt{t}$ and $X_n$ is a random variable taking on negative values with positive probability, what is meant by $g(X_n)$ when $X_n$ happens to have value less than $0$? $\endgroup$ Dec 21, 2013 at 0:12
  • $\begingroup$ @DilipSarwate That is a good point. I cannot say. $\endgroup$
    – JohnK
    Dec 21, 2013 at 15:59
  • 1
    $\begingroup$ If you know that $X_1, \ldots, X_n$ is an i.i.d. sample in addition (as it usually should be), then the identity $\sqrt{n}(\sqrt{\bar{X}} - 1) = \sqrt{n}(\bar{X} - 1)/(\sqrt{\bar{X}} + 1)$ and Slutsky's theorem is another way to approach it. $\endgroup$
    – Zhanxiong
    Jun 10, 2018 at 5:34

1 Answer 1

4
$\begingroup$

The result is correct (up to a factor of $\sigma$, which is an unimportant typographical omission). This answer provides two separate ways to double-check it.

We can in fact obtain the PDF of the transformed variables directly: when the $X_n$ are exactly Normal (and not just asymptotically so), the PDF of $\sqrt{n}\left(g(X_n)-1\right)$ can be found via integration as

$$\frac{1}{(\sigma/2)\sqrt{2\pi n}} \left(\sqrt{n}+x\right) \exp\left({-\frac{x^2 \left(2 \sqrt{n}+x\right)^2}{2 n \sigma ^2}}\right)$$

for $x\gt -\sqrt{n}$ (and equal to $0$ otherwise). For fixed $x$, the limiting value as $n\to\infty$ is

$$\frac{1}{(\sigma/2)\sqrt{2\pi}} \exp\left({-\frac {8x^2}{2 \sigma ^2}}\right) = \frac{1}{(\sigma/2)\sqrt{2\pi}} \exp\left({-\frac {x^2}{2 (\sigma/2) ^2}}\right),$$

the PDF of a Normal$(0, \sigma^2/4)$ distribution.


One can check also check the result with a simulation, such as carried out by this R code:

set.seed(17)
n <- 10^6
x <- sqrt(n)*(sqrt(rnorm(n, 1, 1/sqrt(n)))-1)
m <- mean(x); v <- var(x); k <- mean((x-m)^4)
se.v <- sqrt(((n-1)^2 * k - (n-1)*(n-3)*v^2) / n^3)
print(v)              # Variance
print((v - 1/4)/se.v) # Its standardized standard error

This reports a simulated variance (when $n=10^6$) of $0.25015$ times the (unit) standard deviation, which is just $0.44$ standard errors away from $1/4$: evidence that the computed variance is not incorrect for such a large $n$.

$\endgroup$
3
  • $\begingroup$ Thank you but there is no factor $\sigma$ in the exercise as it is a standard normal density we are concerned with. I appreciate your insight. $\endgroup$
    – JohnK
    Dec 20, 2013 at 19:14
  • $\begingroup$ You might want to edit out the references to $\sigma$ in your question then :-). But it doesn't matter--the result is clear and $\sigma$ only factors through everything as a scale factor. $\endgroup$
    – whuber
    Dec 20, 2013 at 19:29
  • $\begingroup$ Yeah of course. In my question I was merely citing the theorem I would use. I will try to make it less misleading next time. $\endgroup$
    – JohnK
    Dec 20, 2013 at 19:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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