Prove $\lim_{n \to \infty} \frac{\mathrm{Var}(X_n^*)}{n} = 0$ Suppose $\lim_{n \to \infty} n\mathbb{P}(|X| > n) = 0$ and let $X_n^* = X\mathbb{1}_{\{|X|\leq n\}}$. Prove that
$$
\lim_{n \to \infty} \frac{\mathrm{Var}(X_n^*)}{n} = 0
$$
My attempt was to first bound $\mathbb{E}X_n^*$
$$
\mathbb{E}X_n^* = \mathbb{E}X\mathbb{1}_{\{|X|\leq n\}}$$
$$\leq \mathbb{E}n\mathbb{1}_{\{|X|\leq n\}}$$
$$= n\mathbb{P}(|X| \leq n)$$
$$= n(1 - \mathbb{P}(|X| > n))$$
$$= n - n\mathbb{P}(|X| > n)$$
$$= n \text{ as n $\to \infty$}$$
Then do the same with $\mathbb{E}X_n^{2*}$:
$$
\mathbb{E}X_n^{2*} = \int (X\mathbb{1}_{\{|X| \leq n\}})^2d\mathbb{P}$$
$$= \int X\cdot X\mathbb{1}_{\{|X| \leq n\}})d\mathbb{P}$$
$$\leq n^2\mathbb{P}(|X| \leq n)$$
$$= n^2 - n(n\mathbb{P}(|X| > n))$$
$$= n^2 \text{ as n $\to \infty$}
$$
and finally:
$$
\lim_{n \to \infty} \frac{\mathrm{Var}(X_n^*)}{n} = \lim_{n \to \infty} \frac{\mathbb{E}X_n^{2*} - (\mathbb{E}X_n^*)^2}{n} \leq \lim_{n \to \infty} \frac{n^2 - n^2}{n} = 0
$$
This however seems wrong as in my last equation there is subtraction involved and we cant easily bound that this way (I think). Could anyone help with this proof?
 A: I suppose that the initial condition should be $n\mathbb P(|X|>n)\to 0$. 
This is the same as (I use W.Feller's notations, vol.2, ch.VII, section 7) 
$$
\tau(n)=n\cdot\bigl(F(-n)+1-F(n)\bigr) \to 0 \, \text{ as } \, n\to\infty
$$
where $F(x)$ is cdf of $X$.
Consider 
$$\text{Var}(X_n^*)\leq \mathbb E X_n^{2*}=\int_{-n}^n x^2d F(x)=\int_{-n}^0 x^2d F(x)-\int_0^n x^2d(1- F(x))$$
Integrate by parts
$$
=x^2F(x)\bigm|_{-n}^0 - \underbrace{2\int_{-n}^0xF(x)dx}_{\text{replace $x$ by $-x$}} - x^2(1-F(x))\bigm|^{n}_0 + 2\int_0^nx(1-F(x))dx 
$$
$$
= -n^2(F(-n)+1-F(n))+2\int_0^n xF(-x)dx+2\int_0^nx(1-F(x))dx
$$
$$
=-n\tau(n) + 2\int_0^n \tau(x)\, dx.
$$
Then 
$$
\frac{\text{Var}(X_n^*)}{n}\leq \frac{-n\tau(n) + 2\int_0^n \tau(x)\, dx}{n}=-\tau(n) +\frac{2}{n} \int_0^n \tau(x)\,dx.
$$
The rest step is to prove that $\frac{2}{n} \int_0^n \tau(x)\,dx\to 0$.
Since $\tau(x)\to 0$ as $x\to\infty$, for any $\varepsilon>0$ there exists $N=N(\varepsilon)$ s.t. $0\leq\tau(x)\leq \varepsilon$ for all $x>N$. And also $0\leq \tau(x)\leq N$ for $x\leq N$. Then 
$$
\frac{2}{n} \int_0^n \tau(x)\,dx = \frac{2}{n}\left( \int_0^N \tau(x)\,dx+\int_N^n \tau(x)\,dx\right) \leq \frac{2}{n} (N^2 + (n-N)\varepsilon) \to 2\varepsilon
$$
for $n\to\infty$. We proved that for any $\varepsilon>0$, 
$$
0\leq \limsup_{n\to\infty}\frac{2}{n} \int_0^n \tau(x)\,dx \leq 2\varepsilon,
$$
which completes the proof. 
Note also that formula (7.7) in W.Feller "Introduction to Probability Theore and its Applications", vol.2, VII.7 (Wiley, 1971) containes misprint that is corrected above.
