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$X_i \stackrel{iid}{\sim} N(\mu,\sigma^2)$ i=1,2,...,n

$Z_n=\sqrt{n}(\bar{X} - \mu)$

I believe the asymptotic distribution of $Z_n$ is $N(0,\sigma^2)$.

So what would the asymptotic distribution of $5\bar{X}^7$ be?

user158565
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pfmr1995
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1 Answers1

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There is no need to appeal to asymptotic theory here, since the exact distribution of the sample mean is well-known when the underlying data is IID normal. If $X_1,...,X_n \sim \text{N}(\mu, \sigma^2)$ you have:

$$\bar{X} \sim \text{N} \Big( \mu, \frac{\sigma^2}{n} \Big).$$

Hence, using the rules for density transformation the density function for the transformed random variable $Y = 5 \bar{X}^7$ is:

$$\begin{equation} \begin{aligned} f_Y(y) dy &= f_\bar{X}(\bar{x}) \Bigg| \frac{dx}{dy} \Bigg| \\[6pt] &= f_\bar{X} \Big( (y/5)^{1/7} \Big) \cdot \frac{1}{35} (|y|/5)^{-6/7} \\[6pt] &= \frac{(|y|/5)^{-6/7}}{35} \cdot \sqrt{\frac{n}{2 \pi \sigma^2}} \cdot \exp \Big( -\frac{n}{2} \Big( \frac{y^{1/7} - 5 \mu}{5 \sigma} \Big)^2 \Big). \\[6pt] \end{aligned} \end{equation}$$

Ben
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  • Sample size $n$ has no relation with final answer? And if $n$ goes infinity, what happens? – user158565 Nov 19 '18 at 05:23
  • @user158565: Sorry, that was an error --- edited. – Ben Nov 19 '18 at 21:23
  • And if n goes infinity, what happens? – user158565 Nov 20 '18 at 02:38
  • If $n \rightarrow \infty$ then $\bar{X} \rightarrow \mu$ and so $Y \rightarrow 5 \mu^7$. If a simple asymptotic approximation to the distribution is desired then this could be obtained using the delta method. – Ben Nov 20 '18 at 02:42
  • So the answer to OP question is: the asymptotic distribution of $5\bar X^7$ is degenerate distribution with value $5\mu^7$. – user158565 Nov 20 '18 at 03:20
  • No, the answer to the OP's question is that the above distribution is the exact distribution for all $n$, which also serves adequately as an asymptotic distribution for large $n$. (If you would prefer to give an alternative asymptotic approximation to this exact distribution, you should feel free to give an alternative answer.) – Ben Nov 22 '18 at 02:41