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I'm trying to learn the delta method in probability but couldn't quite get the hang of it. For example: trying to solve a problem from the book statistical Inference :

Consider a random sample from $\mathrm{Beta}(\alpha ,\beta)$ distribution $\alpha > 0, \beta > 0$, use delta method to find approximate distribution of $T_n = {\bar X_n}(1- {\bar X_n})$ and find the approximate distribution of ${\bar X_n}$ when $n$ is large.

My try:

a. By delta method,

$\sqrt{n} [f(x) - f(\mu)] \stackrel{\mathcal{D}}{\longrightarrow} N(0,\sigma^2[f'(x)]^2) $

Here, $f(x) = x(1-x)$ and $f(\mu) = \mu(1-\mu)$

Updated try(after @EMS comments)

$\implies \sqrt{n} [x(1-x) - \mu(1-\mu)] \stackrel{\mathcal{D}}{\longrightarrow} N(0,[1-2\mu]^2)$

$\implies \sqrt{n} [\log(x(1-x)) - \log(\mu(1-\mu))] \stackrel{\mathcal{D}}{\longrightarrow} N(0,2 \log (1-2\mu))$ $\implies \sqrt{n} [\log(x(1-x)) - \log(\mu(1-\mu))] \stackrel{\mathcal{D}}{\longrightarrow} N(0,\mu))$

b. For the second one I think we should just use the central limit theorem to say the mean follows normal distribution with mean = 0 but what is the variance? Is it 1?

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    @EMS Does the above seem correct?2012-03-19

1 Answers 1

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First, let $X_{i}$ be distributed as a Beta distribution with parameters $\alpha$ and $\beta$. Then it has mean $\mu = \frac{\alpha}{(\alpha+\beta)}$ and variance $\sigma^{2} = \frac{\alpha\beta}{(\alpha+\beta)^{2}(\alpha+\beta+1)}$.

Then $\bar{X}_{n} = \frac{1}{n}\sum_{i=1}^{n}X_{i}$, and this will have the same mean as the common mean of all of the $X_{i}$.

The central limit theorem tells us that $ \sqrt{n}\biggl[ \bar{X}_{n} - \mu\biggr] \to \mathcal{N}(0,\sigma^{2}).$

From this knowledge, we can apply the Delta method with the statistic function of interest being $T_{n} = \bar{X}_{n}(1-\bar{X}_{n})$, or more simply $T(x) = x(1-x)$. The Delta method then tells us that \sqrt{n}\biggl[ T(\bar{X}_{n}) - T(\mu)\biggr] \to \mathcal{N}(0,\sigma^{2}[T'(\mu)]^{2}).

Now, T'(x) = 1-2x, and so [T'(\mu)]^{2} = [1-\frac{2\alpha}{(\alpha + \beta)}]^{2} = 4\biggl[\frac{\alpha}{(\alpha+\beta)}\biggr]^{2} -4\biggl[\frac{\alpha}{(\alpha+\beta)}\biggr] + 1.

So, the variance for the distribution of $\sqrt{n}[T-T(\mu)]$ is given by: $ \sigma_{T}^{2} = \biggl(\frac{\alpha\beta}{(\alpha+\beta)^{2}(\alpha+\beta+1)}\biggr)\biggl(4\biggl[\frac{\alpha}{(\alpha+\beta)}\biggr]^{2} -4\biggl[\frac{\alpha}{(\alpha+\beta)}\biggr] + 1\biggr).$

Hopefully you can take it from there. All that remains is adjusting the given distribution of $\sqrt{n}[T-T(\mu)]$ to get just the distribution of $T$, and this should be discussed anywhere that the CLT is discussed.

As for the 'intuition' behind this method, it is a very similar idea to a transformation of variables. The Wikipedia proof for the univariate case does a good job of showing what happens.

When you expand the function $g(X)$ around the point $\theta$ and you assume only a linear approximation, then what you're left with as a scale factor for the term $(X-\theta)$ is g'(\theta), which just comes from simple Taylor series approximation. Dividing both sides by g'(\theta) leaves you with $X-\theta$ on the right hand side, which is something with a known asymptotic distribution. That means all the stuff on the left hand side has to have that same asymptotic distribution. Multiplying by g'(\theta) then gives the result. This also shows why the assumption that g'(x) is not $0$ at $\theta$ (although this is not strictly necessary if you make higher order arguments).