# If $X_n$ is Gamma $(n,\lambda)$ distributed then $(\lambda X_n -n)/\sqrt n\to N(0,1)$

Let $$X_n$$ be Gamma $$(n,\lambda)$$ distributed, and $$Y_n = \dfrac{\lambda X_n -n}{\sqrt{n}}$$. Show that $$Y_n \rightarrow N(0,1)$$.

My idea to prove this is to use Lévys theorem with the characteristic functions. If I can show that the characteristic functions $$\phi_{Y_n}$$ of $$Y_n$$ converge to a function $$\phi$$, which is continous in $$0$$, then there exists a random variable $$Y$$ such that $$\phi$$ is the characteristic function of $$Y$$. My aim is to get $$\phi_Y (u) = e^{-u^2/2}$$ since I have to show that the $$Y_n$$ converge to a standard normal random variable and the cF describes the distribution uniquely.

I know that the cF of a Gamma$$(n,\lambda)$$-distributed random variable $$X_n$$ is given by $$\phi_{X_n}(u) =\left(\frac{\lambda}{\lambda - iu}\right)^n$$ The cF rules for the linear transformation $$Y_n = \frac{\lambda X_n}{\sqrt{n}} +\frac{-n}{\sqrt{n}}$$ give $$\phi_{Y_n} (u)= e^{-in \frac{u}{\sqrt{n}}} \phi_{X_n}(\frac{\lambda u}{\sqrt{n}}) = e^{-iu \sqrt{n}} \left(\frac{\lambda}{\lambda - i \frac{\lambda u}{\sqrt{n}}}\right)^n$$ that is, $$\phi_{Y_n} (u) = e^{-iu \sqrt{n}} \left(\frac{1}{1 - i \frac{ u}{\sqrt{n}}}\right)^n = e^{-iu \sqrt{n}} \frac{1}{ (1 - i \frac{u}{\sqrt{n}})^n}$$ and now I don't know how to continue calculating.

Since I need the limit of this, I thought about the representation of the exponential function in terms of the limit of a sequence, e.g. $$e^x = \lim_{n \rightarrow \infty} \left( 1 + \frac{x}{n}\right)^n$$ Some hints or a trick would be very nice.

• See also this question for a different approach. – saz Jan 10 at 13:43

## 1 Answer

To sum up, at this moment, you know that $$\phi_{Y_n} (u)=\psi(v_n)^{-n}$$ where $$\psi(v)=e^{v}(1-v)\qquad v_n=\frac{iu}{\sqrt n}$$ Now, by the expansion of the exponential, when $$v\to0$$, $$\psi(v)=\left(1+v+\frac12v^2+o(v^2)\right)(1-v)=1-\frac12v^2+o(v^2)$$ Thus, for every fixed $$u$$, when $$n\to\infty$$, $$\psi(v_n)=1+\frac12\frac{u^2}n+o\left(\frac1n\right)$$ hence $$\psi(v_n)^n\to e^{u^2/2}$$ That is, $$Y_n$$ converges in distribution to the standard normal distribution, as desired.

• Great answer, thank you! – Myrkuls JayKay Jan 9 at 21:20