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.

  • 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


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:

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 $n$.

  • $\begingroup$ 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

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