Problem: Let $X_{1},\ X_{2},\ \cdots$ be independent and identically distributed (i.i.d.) random variables having the ch.f. of the term $1-c|t|^{a}+o(|t|^{a})$ as $t\to0$, where $0<a\leq2$. Determine the constants $b$ and $c$ so that $\sum_{i=1}^{n}X_{i}/(bn^{c})$ converges in distribution to a random variable having ch.f. $e^{-|t|^{a}}.$
I have completed this problem, but condition $0<a\leq2$ is not used. I don't konw what's wrong in my answer. Anyone can help me? Thank in advance. Here are my answer.
Since $X_{1},\ X_{2},\ \cdots$ are i.i.d., so the ch.f. of $\sum_{i=1}^{n}X_{i}/(bn^{c})$ is \begin{align*} \varphi(t)\stackrel{def}{=} \varphi_{\sum_{i=1}^{n}X_{i}/(bn^{c})}(t) & =Ee^{it\sum_{i=1}^{n}X_{i}/(bn^{c})} =\prod_{i=1}^{n}Ee^{it X_{i}/(bn^{c})} \\ & =\prod_{i=1}^{n}\varphi_{X_{i}}(t/(bn^{c}))\\ &=\left[\varphi_{X_{i}}(\frac{t}{bn^{c}})\right]^{n}\\ &=\left[1-c\left|\frac{t}{bn^{c}}\right|^{a}+o(\left|\frac{t}{bn^{c}}\right|^{a}) \right]^{n} \end{align*} By L$\acute{e}$vy continuity theorem, we know that $ \varphi(t)\longrightarrow e^{-|t|^{a}}$ ,then $$ nlog\left[1-c\left|\frac{t}{bn^{c}}\right|^{a}+o(\left|\frac{t}{bn^{c}}\right|^{a})\right]\longrightarrow -|t|^{a}. $$ By Taylor expansion, $log(1-x)=\sum_{n=1}^{\infty}(-x)^{n}/n$, so $$ \begin{cases} ac=1 & \\ \frac{c}{|b|^{a}}=1 & \end{cases} \Longrightarrow \begin{cases} c=\frac{1}{a} & \\ b=\pm\frac{1}{a}^{\frac{1}{a}} & . \end{cases} $$