0
$\begingroup$

Suppose $X_1, X_2, \dots \sim \text{Exp}(\beta)$ (i.e., exponential distribution with mean $\beta > 0$) are iid. Let $T_k = \sum_{i=1}^{k}X_i$, $k = 1, 2, \dots$. Let $\mathcal{N}(\mu, \sigma^2)$ denote a normally-distributed random variable with mean $\mu$ and variance $\sigma^2$. Show that there exist sequences of constants $a_n, b_n$ such that $$a_n(T_n - b_n) \overset{d}{\to}\mathcal{N}(0, 1)$$ (convergence in distribution) as $n \to \infty$.

Since $X_i \sim \text{Exp}(\beta)$, $\mathbb{E}[X_i] = \beta$ and $\text{Var}(X_i) = \beta^2$. By the Central Limit Theorem, $$\sqrt{n}\left(\dfrac{1}{n}T_n - \beta\right)\overset{d}{\to}\mathcal{N}(0, \beta^2)\text{.}$$ Obviously $\dfrac{1}{\beta}\overset{p}{\to}\dfrac{1}{\beta}$, so by Slutsky's Theorem, $$\dfrac{\sqrt{n}}{\beta}\left(\dfrac{1}{n}T_n - \beta\right)\overset{d}{\to}\dfrac{1}{\beta}\mathcal{N}(0, \beta^2) \overset{d}{=}\mathcal{N}(0, 1)\text{.}$$ Thus, $a_n = \dfrac{1}{\beta\sqrt{n}}$ and $b_n = \sqrt{n}$.

The solution I have matches for $a_n$, but says $b_n = n\beta$. Did I do something wrong?

  • 0
    $b_n=E[T_n]$, while $a_n=(\operatorname{Var}(T_n))^{-1/2}$; this ensures that $Z_n:=a_n(T_n-b_n)$ has mean zero and variance $1$. – 2017-01-03

1 Answers 1

2

You just did the algebra wrong. Note that $$ \dfrac{\sqrt{n}}{\beta}\left(\dfrac{1}{n}T_n - \beta\right)=\frac{T_n-n\beta}{\beta\sqrt{n}}=a_n(T_n-b_n) $$ implies that $$ a_n=\frac{1}{\beta\sqrt{n}} \quad\text{and }\quad a_nb_n=\sqrt{n}, $$ which gives $b_n=n\beta$.

  • 1
    I laughed. Thank you. I shouldn't be doing convergence proofs this late. :P Will mark your answer as soon as this site lets me. – 2017-01-03