Question about deciding the asymptotic distribution

Question

Let $Y_n \sim \text{Gamma}(n, λ)$ where $n$ is an integer. As $n$ goes to infinity, find the asymptotic distribution of \begin{equation*} \sqrt{n}(\log(Y_n/n) - \log(1/\lambda)) \end{equation*}

I tried to write Gamma as a sum of exponential random variables as a tool but still have no thoughts on this problem... I think this Gamma is the shape and rate form, can someone help with me?

Okay here is an update from me: my guess is that the distribution is approximately normal distribution when $n$ is large. By the CLT and delta method, am I right?

Answer 1

Yes, you are right. Let $X_1,...,X_n$ be and iid $Exp(\lambda)$r.v.s, hence $$ Y_n/n = \frac{1}{n}\sum_{i=1}^n X_i \sim Gamma(n, \lambda) $$ thus by the CLT $$ \sqrt{n} ( Y_n/n - E[X]) \xrightarrow{D}N(0, Var(X)). $$ Note that $E[X] = 1/\lambda$, and $Var(X) = 1/\lambda^2$. Using the delta method with $g(x) = \ln (x)$, you have $$ \sqrt{n} ( \ln( Y_n/n ) - \ln(1/\lambda) ) \xrightarrow{D}N\left(0, (g'(1/\lambda)) ^ 2 \frac{1}{\lambda ^ 2}\right), $$
where $g'(1/\lambda) = \frac{1}{1/\lambda} = \lambda$.

Answer 2

From the central limit theorem $\frac Yn\approx \mathcal N(\frac 1 \lambda, \frac 1 {n\lambda^2})$ so $\sqrt n \lambda\left(\frac Yn-\frac 1 \lambda\right)\approx \mathcal N(0, 1)$. We apply the delta method with $\alpha(\mu)=\log \mu$. Then $\alpha'(\mu)=\frac 1\mu$ and $\alpha'(\frac 1 \lambda)=\lambda$. So the delta method says that $\frac{\sqrt n\lambda}{\lambda}(\log(\frac Y n)-\log \frac1\lambda)\approx\mathcal N(0, 1)$ as well.