# Assumption to apply the delta method

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?

• 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 $N(0,\sigma^2)$ Commented Nov 22, 2021 at 13:30
• 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. Commented Nov 22, 2021 at 13:39
• 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.
– whuber
Commented Nov 22, 2021 at 15:26

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