Properties of a sequence of iid rv's

Question

I cannot do part a), and Im fairly sure that b) and c) will follow from it. If possible could I please have a solution to part a) and hints if you feel necessary to parts b) and c).

Answer 1

Write $$n \, 1_{(X_1^2>\epsilon^2n)}\leq {X_1^2\over\epsilon^2} \, 1_{(X_1^2>\epsilon^2n)}.$$ Now integrate both sides, and use dominated convergence to conclude $n\mathbb{P}(|X_1|>\epsilon \sqrt{n})\to0.$

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The sequence of random variables ${X_1^2\over\epsilon^2} \, 1_{(X_1^2>\epsilon^2n)}$ converges to zero pointwise as $n\to\infty$. Since they are dominated by the integrable random variable $X_1^2$, the dominated convergence theorem guarantees that $$\mathbb{E}\left({X_1^2\over\epsilon^2} \, 1_{(X_1^2>\epsilon^2n)}\right)\to0 \mbox{ as } n\to\infty. $$

Answer 2

For (a): consider $Y$ a simple random variable, that is,a linear combination of characteristic functions of measurable sets. Then $$n\mathbb P\{|X_1|\gt \varepsilon\sqrt n\}\leqslant n\mathbb P\{|X_1-Y|\gt \varepsilon\sqrt n\}+n\mathbb P\{|Y|\gt \varepsilon\sqrt n\},$$ hence using Markov's inequality, $$n\mathbb P\{|X_1|\gt \varepsilon\sqrt n\}\leqslant\frac 1{\varepsilon^2}\mathbb E[(X_1-Y)^2]+n\mathbb P\{|Y|\gt \varepsilon\sqrt n\}.$$ Since $Y$ is bounded, the second term of the RHS is $0$ for $n$ large enough, hence $$\limsup_{n\to +\infty}n\mathbb P\{|X_1|\gt \varepsilon\sqrt n\}\leqslant\frac 1{\varepsilon^2}\mathbb E[(X_1-Y)^2].$$