# Convergence to the stable law

I am reading the book Kolmogorov A.N., Gnedenko B.V. Limit distributions for sums of independent random variables.

From the general theory there it is known that if $X_i$ are symmetric i.i.d r.v such that $P(|X_1|>x)=x^{-\alpha},\, x \geq 1$, then $(X_1+\ldots+X_n)n^{-1/\alpha}\to Y$, where c.f. of $Y$ equals $\varphi_Y(t)=e^{-c|t|^{\alpha}}, \alpha \in (0,2]$, so $Y$ has stable law of distribution.

I want to check it without using that general theorems. So I start as the following, $X_1$ has density of distribution $f_X(x)=|x|^{-\alpha-1}\alpha/2, |x|>1$. Using Levy theorem one must prove that $\varphi^n_{X_1}(t/n^{1/\alpha})\to \varphi_Y(t),\, n \to \infty$ for all $t\in \mathbb R$. $$\varphi_{X_1}(t/n^{1/\alpha})=\int_{1}^{\infty}\cos(tx/n^{1/\alpha})\alpha x^{-\alpha-1}\,dx,$$ for all it is evident that $t$ $\varphi_{X_1}(t/n^{1/\alpha})\to 1, n \to \infty$ so we have indeterminate form $1^\infty$.

So we are to find $n(\varphi_{X_1}(t/n^{1/\alpha})-1)$, but $\varphi_{X_1}(t/n^{1/\alpha})\sim 1+1/2(2txn^{-1/\alpha})^2$, and I can only say something about $\alpha=2$ and I got stuck here. Perhaps, I made a mistake somewhere.

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(+1) Thanks for showing your thoughts and work. –  cardinal Apr 22 '12 at 17:39
Maybe it would help others to specify which "general theorems" you are trying not to use. For example, you are appearing to admit usage of Levy's continuity theorem, which I would consider a pretty general one. :) –  cardinal Apr 22 '12 at 18:00
@cardinal By 'general theorems' I mean neccessary and sufficient conditions for the convergence of a sum of the form $\dfrac{1}{B_n}(X_1+\ldots+X_n-A_n)$ to the specific stable law. –  Tarasenya Apr 22 '12 at 18:02

The problem is, in fact, not so hard as I thought. Note, that it is enough to prove that $\varphi(t)=1-Ct^{\alpha}+o(t^{\alpha}),t \to 0+$. First, I make substitution $y=tx$ in $\int_{1}^{\infty}\cos(tx)\alpha x^{-\alpha-1}\,dx=t^{\alpha}\int_{t}^{\infty}\cos (y) \alpha y^{-\alpha-1}\,dy$. Integrating by parts gives me that the last equals $\cos t-t^{\alpha}\int_{t}^{\infty}\sin y y^{-\alpha}\,dy=1-C t^{\alpha}+o(t^{\alpha})$, and it is a crucial thing that $\alpha \in (0,2]$ to write the last equality, here $C=\int_{0}^{\infty}\sin y y^{-\alpha}\, dy$. Hence, we are done.
I yr integral make the change of variables $z = \frac y {n^{\frac 1 {\alpha}}}$. This brings a factor $\frac 1n$ out front. The write $cos(tz) = 1 + (cos(tz) -1)$. Integrate the 1 explicitly, and the integral invlving $cos(tz)-1$ converges because it is nice at zero.