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Michael Hardy
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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$$\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} \cdots$.

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

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 $\sqrt n \cdot \tilde{A}$)?

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 $\sqrt n \left(\frac{1}{n} \cdot \sum_{i=1}^n \left[e^{b_i Y_i}\right]- e^{b_i \mu}\right) \xrightarrow[]{D} \cdots$.

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

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Anna D.
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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 $\sqrt n \cdot \tilde{A}$)?