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When proving the delta method of distributions in my textbook we make the following assumption:

Let $X_{n}$ be a sequence of random variables.

and:

${\sqrt{n}[X_n- c]\,\xrightarrow{D}\,\mathcal{N}(0,1)}$

Bearing this in mind it can be shown that:

${\sqrt{n}[g(X_n)-g(c)]\,\xrightarrow{D}\,\mathcal{N}(0,[g'(c)]^2)}$

What I'm struggling to understand why in this case the assumption to implement the delta method is:

${\sqrt{n}[X_n- c]\,\xrightarrow{D}\,\mathcal{N}(0,1)}$

and not

${\sqrt{n}[X_n- c]\,\xrightarrow{D}\,\mathcal{N}(0,\sigma^{2})}$?

Usually $X_{n}$ is an estimator of a population parameter and $c$ the actual parameter being estimated. If the estimator is a consistent asymptotically normal estimator we can conclude that:

${\frac{\sqrt{n}[X_n- c]}{\sigma}\,\xrightarrow{D}\,\mathcal{N}(0,1)}$

Would it be safe to assume that here $\sigma^{2} = 1$ is assumed here?

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  • $\begingroup$ What is your reference? I'm used to the Taylor Series derivation, which assumes higher order terms can be neglected. The wikipedia entry (terrible reference) has the $N(0,\sigma^2)$ $\endgroup$ Commented Nov 22, 2021 at 13:30
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    $\begingroup$ Don't mix asymptotics with ability to apply the delta method in a given non-huge dataset. You may find that the transformed statistic has a very asymmetric distribution, something ignored by the delta method. $\endgroup$ Commented Nov 22, 2021 at 13:39
  • $\begingroup$ In this context (nondegenerate Normal distributions) you can always take $\sigma^2=1$ because the value of $\sigma$ merely establishes what your unit of measurement is. $\endgroup$
    – whuber
    Commented Nov 22, 2021 at 15:26

1 Answer 1

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The two statements :

$$\left(\sqrt{n}(X_n - c) \overset{D}{\to}\mathcal{N}(0, 1)\right) \Longrightarrow \left(\sqrt{n}(g(X_n) - g(c)) \overset{D}{\to}\mathcal{N}(0, g'(c)^2)\right) $$ and $$\left(\sqrt{n}(X_n - c) \overset{D}{\to}\mathcal{N}(0, \sigma^2)\right) \Longrightarrow \left(\sqrt{n}(g(X_n) - g(c)) \overset{D}{\to}\mathcal{N}(0, g'(c)^2\sigma^2)\right) $$ are equivalent.

The second one clearly implies the first one (take $\sigma = 1$).

Now, assume the first one is true, and suppose that $$\sqrt{n}(X_n - c) \overset{D}{\to}\mathcal{N}(0, \sigma^2). $$

Define $Y_n = \frac{X_n}{\sigma}$, $c' = \frac{c}{\sigma}$ and $g_1 : t \mapsto g(\sigma t)$.

Then you have $$\sqrt{n}(Y_n - c') \overset{D}{\to}\mathcal{N}(0, 1),$$ and thus $$\sqrt{n}\left(g_1(Y_n) - g_1(c')\right) \overset{D}{\to}\mathcal{N}(0, g_1'(c')^2)$$ which gives $$\sqrt{n}(g(X_n) - g(c)) \overset{D}{\to}\mathcal{N}(0, g'(c)^2 \sigma^2).$$

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