Let $Y_n = \frac{\sum_{i=1}^{n} X_i(X_i+1)}{n}$ . Find a number $c$ such that $Y_n \to c$

Question

Let $X_1, X_2, . . . , X_n$ be independent and identically distributed random variables, each having a $Uniform[0, 0.5]$ distribution. Let $Y_n = \frac{\sum_{i=1}^{n} X_i(X_i+1)}{n}$ . Find a number $c$ such that $Y_n \to c$ ($p$ above the arrow)

Attempt:

$Y_n = \frac{\sum_{i=0}^{n} X_i(X_i+1)}{n} = \frac{\sum_{i=1}^{n} X_i^2}{n} + \frac{\sum_{i=1}^{n} X_i}{n}$

Not sure how to go about this. I believe I need to find their expectation but not sure how.

Answer 1

The density is $2$ on $(0,0.5)$ and $0$ elsewhere. $EX_1=2 \int _0^ {1/2} x \, dx =\frac 1 4$ and $EX_1^{2}=2\int _0^ {1/2} x ^{2}\, dx =\frac 1 {12}$ so $c=\frac 1 4+\frac 1 {12}= \frac 1 3$.

Answer 2

By the Law of Large Numbers $$\frac{\sum_{i=0}^n X_i}{n} \rightarrow \mu$$ where $$\mu = E(X) = \frac{1}{2}(0 + 0.5) = \frac{1}{4}$$ Also $$\frac{\sum_{i=0}^n X_i^2}{n} \rightarrow \mu^2 + Var(X)$$ where $$Var(X) = \frac{1}{12}(0.5)^2 = \frac{1}{48} $$ So $$\frac{\sum_{i=0}^n X_i^2}{n} \rightarrow \frac{1}{16} + \frac{1}{48}$$ Therefore $$c = \frac{1}{4} + \frac{1}{16} + \frac{1}{48} = \frac{1}{3}$$