# Inverse Gaussian Distribution and the Central Limit Theorem

Let the random variables $$Y_1,\ldots,Y_n$$ be independent and identically distributed (i.i.d.) (standard) Inverse Gaussian random variables with parameters $$\mu$$ and $$\lambda$$.

Then, let the random variables $$\tilde{A}$$ and $$A$$ be given, which are defined as follows: $$A = \sqrt n \cdot \tilde{A} = \sqrt n \left(\frac{1}{n} \cdot \sum_{i=1}^n \left[e^{b_i Y_i}\right] - \theta\right)$$ ($$b_i \in \mathbb{R}$$ ($$i=1,\ldots,n$$) are constants (with $$b_i \neq b_j$$ for any $$i \neq j$$) and $$\theta$$ is a parameter).

Goal: Determine what ultimately follows if one uses the (Lindeberg-Lévy) Central Limit Theorem on $$e^{b_i Y_i}$$.

The CLT informs us that $$\sqrt n (\overline{Y} - E(Y_i)) \xrightarrow[]{D} \mathcal{N}(0,Var(Y_i))$$, i.e. $$\sqrt n \left(\frac{1}{n} \cdot \sum_{i=1}^n [Y_i] - \mu \right) \xrightarrow[]{D} \mathcal{N}(0,\frac{\mu^3}{\lambda})$$.

It seems to me that it follows that $$\sqrt n \left(\frac{1}{n} \cdot \sum_{i=1}^n \left[e^{b_i Y_i}\right]- e^{b_i \mu}\right) \xrightarrow[]{D} \ldots$$.

Question: How to proceed from here (i.e. how to determine the limit distribution of $$\sqrt n \cdot \tilde{A}$$)?

## 1 Answer

You may consider the random variable \begin{align} \frac{1}{s_n} \sum_{i=1}^n (e^{b_i Y_i} - e^{b_i \mu}) \end{align} where $$s_n^2 = \sum_{i=1}^n \sigma_i^2$$ and $$\sigma_i$$ is the variance of $$e^{b_iY_i}$$. Then, you can check whether the Lindeberg's condition is satisfied. If it is satisfied, using the Lindeberg CLT \begin{align} \frac{1}{s_n} \sum_{i=1}^n (e^{b_i Y_i} - e^{b_i \mu}) \xrightarrow[]{D} \mathcal{N}(0,1) \end{align} as $$n\to\infty$$.