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I failed to prove the exercise 27.15. in Billingsley's textbook.

Let $X_1,X_2, \dots$ be i.i.d. positive random variables with expectation $m$ and variance $\sigma^2 \in (0,\infty)$.

If $N_t = \max \{n : S_n \leq t \}$, then $$\frac{S_{N_t} -t}{\sqrt{t}} \Rightarrow 0.$$

I already proved that $\frac{t}{N_t} \rightarrow m$ as $t \rightarrow \infty$ and the following facts (what author gave me as hints):

If $X_i$ are i.i.d. with $\mathbb{E}X^2 < \infty$, then $n \, \mathbb{P}(|X_1| \geq \varepsilon \sqrt{n}) \rightarrow 0$ and $\displaystyle n^{-\frac{1}{2}} \max_{1\leq k \leq n} |X_k| \rightarrow 0$ in probability.

Any help will be appreciated.

Thanks!

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Since $S_{N_t}\le t$ and $S_{N_t+1}>t$, we have $$\mathbb P(|S_{N_t}-t|>\sqrt{t}\varepsilon)=\mathbb P(S_{N_t}<t-\sqrt{t}\varepsilon,S_{N_t+1}>t)\le\mathbb P(X_{N_t+1}>\sqrt{t}\varepsilon).$$ Let $M_t=N_t+1$ and observe that $\frac{t}{M_t}\to m$. For $0<\delta<m$ we have $$\mathbb P(X_{M_t}>\sqrt{t}\varepsilon)\le\mathbb P\left(\left|\frac{t}{M_t}-m\right|\ge\delta\right)+\mathbb P\left(X_{M_t}>\sqrt{M_t}\varepsilon\sqrt{m-\delta},\,\left|\frac{t}{M_t}-m\right|<\delta\right).$$ Let $\gamma=\varepsilon\sqrt{m-\delta}$ and $n=\lfloor\frac{t}{m+\delta}\rfloor$. If we further assume that $\delta<\frac{m}2$, then it follows that $n\le M_t< 3n$ on the event $\{|\frac{t}{M_t}-m|<\delta\}$. This implies that the second term above is bounded by $$\mathbb P(\max_{n\le k<3n}X_k>\sqrt{n}\gamma)\le2n\mathbb P(X_1>\sqrt{n}\gamma).$$ Since $n\to\infty$ as $t\to\infty$, putting everything together shows that $\frac{S_{N_t}-t}{\sqrt{t}}\to0$ in probability. Note that the second hint wasn't explicitly used, but the idea behind it was important in our final estimate of the second term.

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  • $\begingroup$ Thank you very much! I read your answer and get critical ideas to prove this. However, I proved this one by assuming $\delta < \frac{m}{2}$ not $\delta > \frac{m}{2}$. $\endgroup$
    – cdamle
    May 3, 2017 at 14:03
  • $\begingroup$ You're right, that was a typo. I will edit. $\endgroup$
    – Jason
    May 3, 2017 at 14:05

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