How can I use the delta method to show that
$\sqrt n$ (1/$\bar Y_n$ - 1/$\mu)$ $\rightarrow$ N (0, $\sigma^2$/$\mu^4$) ?
We know that:
$\bar Y_n$ = $1/n \sum_{n=1}^n Y_i$
$\ Y_i$ is i.i.d; $E(\ Y_i )= \mu\ne 0$ and $V(\ Y_i)= \sigma^2 \gt$ 0
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1 Answer
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Set $f(X) = \frac{1}{X}$ (for $f\colon W\longrightarrow \mathbb R$). Then $Df(X) = -\frac{1}{X^2}$ ($Df$ denotes the derivative). The delta-method states that $$\sqrt n(f(X) - f(\mu)) \rightarrow N(0,\sigma^2Df(\mu)^2)$$ provided that $\sqrt n(X-\mu)\rightarrow N(0,\sigma^2)$. Now your job is to plug in the right value for $X$ and you are done ;-)
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$\begingroup$ Ok, I was just wondering why you did not apply Slutsky as one of the steps. $\endgroup$ Jan 11, 2017 at 14:06
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$\begingroup$ It's not necessary because we are allowed to use the delta method (however, delta method requires that $\sqrt n(X-\mu)$ converges in distribution. If this is not given, you have to proof it first of course, but that's a new question) $\endgroup$ Jan 11, 2017 at 14:13
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$\begingroup$ Ah, ok! I thought I have to show that as well $\endgroup$ Jan 11, 2017 at 14:38