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.

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

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.

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

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