Compute the next limit using CLT

Let $\{X_n\}_n$ be a sequence of random independent variables with density function $f_{X_n}=4x^2e^{-2x} \chi_{x>0}$, $\forall n \in \mathbb{N}$ if $S_n=\sum_{i=1}^n X_i$ find $$\lim_{n \to \infty} \mathbb{P}[S_n \leq \frac{3n}{2}+\sqrt{3n}]$$

The strategy I used to solve it is to calculate $E[S_n]$ and $Var[S_n]$ and then manipulate $\mathbb{P}[S_n \leq \frac{3n}{2}+\sqrt{3n}]$ so that $$\mathbb{P}\left[\frac{S_n -\frac{3n}{2}}{\sqrt{3n}} \leq 0 \right]$$ is left and apply CLT directly. Is this procedure correct?

Any comments, suggestions or help would be greatly appreciated.

Solution 1:

If $X_i$ are iid with density $f(x)=4x^2\exp(-2x)$, then it has moments

$$E[X_i]=\int_0^\infty xf(x)dx=\frac{3}{2}\\ \text{Var}[X_i]=\int_0^\infty x^2f(x)dx-\left(\frac{3}{2}\right)^2=\frac{3}{4}. $$

By CLT,

$$\lim _{n\uparrow \infty} P\left(S_n\leq \frac{3n}{2}+\sqrt{3n}\right)=\lim _{n\uparrow \infty} P\left(\frac{\frac{1}{n}S_n-\frac{3}{2}}{\sqrt{\frac{3}{4n}}}\leq 2\right)=\Phi(2),$$

where $\Phi$ is the CDF of a standard normal.