Applying the CLT on a modified Random variable

Question

X1, X2, X3, . . . independent, U(0, 1) distributed random variables. Prove that if n → ∞, then:

$ Y_n= \frac{\sum_{k=1}^{n} kX_k - \frac{n^{2}}{4}}{\frac{1}{6}n^{3/2}} \ $ goes to normal distribution N(0,1) in distribution

so what I tried to do is that I defined a new Random Variable $Y=kX$ such that I would apply CLT to it and therefore the new expectation would be E(Y)= $\frac{n}{2}$ and the new variance V(Y)= $\frac{n^{2}}{12}$ and I tried applying the theory but it seems that I am missing something because I couldn't prove the above, can someone please help ?

Answer 1

The new random variable is the following

$$W=\Sigma_{k=1}^n kX_k$$

Now observe that

$$\mathbb{E}[W]=\Sigma_{k=1}^n k\cdot \mathbb{E}[X_1]=\frac{1}{2}\frac{n(n+1)}{2}\approx\xrightarrow{n\rightarrow +\infty}\approx \frac{n^2}{4}$$

$$\mathbb{V}[W]=\Sigma_{k=1}^n k^2\cdot \mathbb{V}[X_1]=\frac{1}{12}\frac{2n^3+3n^2+n}{6}\approx\xrightarrow{n\rightarrow +\infty}\approx \frac{n^3}{6^2}$$

use the CLT to get your result

$$\frac{W-\mathbb{E}[W]}{\sqrt{\mathbb{V}[W]}}\xrightarrow{\mathcal{L}} \Phi_{\{0;1\}}$$

Answer 2

$$E\left\{\sum\limits_{k=1}^{n}kX_k\right\} = \sum\limits_{k=1}^{n}kE\{X_k\} = \frac{1}{2}\sum\limits_{k=1}^{n}k=\frac{n(n+1)}{4}$$

so,

$$ E\{Y_n\} = \dfrac{\frac{n^2+n-n^2}{4}}{\frac{1}{6}n\sqrt{n}} = \frac{3}{2\sqrt{n}} \rightarrow0, \quad \text{as } n\rightarrow\infty$$

and,

$$ E\left\{ \left( \sum\limits_{k=1}^{n}kX_k \right)^2 \right\} = \sum\limits_{k=1}^{n}\sum\limits_{l=1}^{n}klE\{X_kX_l\} = \sum\limits_{k=1}^{n} k^2 E\{X_k^2\} = \frac{1}{3}\sum\limits_{k=1}^{n} k^2 = \dfrac{n(n+1)(2n+1)}{18}$$

so,

$$ E\{Y_n^2\}\rightarrow \dfrac{\frac{n^3}{9}}{\frac{n^3}{36}} = 4 $$

Either I have made a mistake, or $Y_n$ tends to $\mathcal{N}(0, 4)$.