# Distributions convergence and standard normal random variable using characteristic functions

Suppose $$\lambda_{1}$$, $$\lambda_{2},\ldots\lambda_{n}\ldots$$ be a monotone sequence of positive real numbers that go to infinity. Let $$X_{n}$$ be sequence of random variables with $$\Gamma(\lambda_{n},1)$$ distribution. The probability density function of $$X_{n}$$ is given by
$$f_{n}(x)=$$ $$\begin{cases} \frac{x^{\lambda_{n}-1}\times e^{-x}}{\Gamma(\lambda(n))},\text{if x\ge 0;}\newline 0, &\text{otherwise} \end{cases}$$ Denote $$Y_{n}=\frac{(X_{n}-\lambda_{n})}{\sqrt{\lambda_{n}}}$$

Prove that $$Y_{n}$$ converges in distribution to standard normal random variable as n goes to infinity.

I first found the characteristic function of $$Y_{n}$$ and I could also do the same for moment generating function and I tried to show that characteristic function of $$Y_{n}$$ converges to the characteristic function of the standard normal random variable which is $$e^{(-1/2)t^{2}}$$. To do that I used $$\Phi_{aX+b}(t)=e^{ibt}\times\Phi_{X}(at)$$ and I found $$\Phi_{X_{n}}(t)=(1-it)^{-\lambda_{n}}$$.From here I found the characteristic function as $$\Phi_{aX_{n}+b}(t)=(1-\frac{it}{\sqrt{\lambda_{n}}})^{-\lambda_{n}}\times e^{it(-\sqrt{\lambda_{n}})}$$.Now I am calculating that this characteristic function converges to $$1$$ which is not the characteristic function of standard normal distribution.

• Look harder at the "now I am calculating" bit. Commented Dec 17, 2023 at 21:26
• Yes I recognized that it may not be 1 but in spite of everything I can not compute that limit appropriately. Is it easy? Commented Dec 17, 2023 at 22:34
• If you write clearly how you compute the last limit I will be really pleased. It becomes too complicated after some steps.Thanks Commented Dec 17, 2023 at 22:55

You want to calculate, for fixed real $$t$$, $$\lim_{\lambda\to\infty} \left(1-\frac{it}{\sqrt\lambda}\right)^{-\lambda}\exp(-\sqrt\lambda i t).$$ Take logarithms, so you now want to know $$\lim_{\lambda\to\infty} -\lambda\log\left(1-\frac{it}{\sqrt\lambda}\right)-\sqrt\lambda it.$$ Use the approximation $$\log(1-x)=-x-x^2/2 + o(x^3)$$, valid for small $$x$$, with the choice $$x=it/\sqrt\lambda$$. It's now just a matter of keeping track of powers of $$\lambda$$ in the approximation: \begin{align*} -\lambda\log\left(1-\frac{it}{\sqrt\lambda}\right)-\sqrt\lambda it&= -\lambda \left(-\frac{it}{\sqrt\lambda} -\frac{(it)^2}{2\lambda} + o(\lambda^{-3/2})\right) - \sqrt\lambda it\\&=-t^2/2+o(1/\sqrt\lambda)\end{align*}