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Let $A_1,A_2,\cdots$ and $B_1,B_2,\cdots$ be two independent sequences of i.i.d random variables according to probability density function $p(x)$, with $\mathbb{E}[A_1]=\mathbb{E}[B_1]<\infty$ and $Var[A_1]=Var[B_1]<\infty$. For a fixed constant $c>0$, find the constant $k>0$ such that

$$\lim\limits_{n \to \infty}\mathbb{P}\left(\frac{1}{n}\sum\limits_{i=1}^{i=n}A_i\geq \left(1-\frac{c}{\sqrt{n}}\right)\left(1-\frac{k}{\sqrt{n}}\right)\frac{1}{n+k\sqrt{n}}\sum\limits_{i=1}^{i=n+k\sqrt{n}}B_i \right)=1 $$

My approach: $\frac{1}{n}\sum\limits_{i=1}^{i=n}A_i$ is a random variable and weak law of large numbers yields $$\frac{1}{n}\sum\limits_{i=1}^{i=n}A_i\xrightarrow{p}\mathbb{E}(A_i)\,\,\,\,\,\,\,\,\,\,\,(1)$$

Similarly, $$ \frac{1}{n+k\sqrt{n}}\sum\limits_{i=1}^{i=n+k\sqrt{n}}B_i \xrightarrow{p}\mathbb{E}(B_i)=\mathbb{E}(A_i)\,\,\,\,\,\,\,\,\,\,\,(2)$$

[I am not sure about this step:] Therefore, for all $k$

$$\left(1-\frac{c}{\sqrt{n}}\right)\left(1-\frac{k}{\sqrt{n}}\right)\frac{1}{n+k\sqrt{n}}\sum\limits_{i=1}^{i=n+k\sqrt{n}}B_i \xrightarrow{p}\mathbb{E}(B_i)=\mathbb{E}(A_i)\,\,\,\,\,\,\,\,\,\,\,(3)$$

Now, by (1) and (3)

$$\left[\frac{1}{n}\sum\limits_{i=1}^{i=n}A_i-\left(1-\frac{c}{\sqrt{n}}\right)\left(1-\frac{k}{\sqrt{n}}\right)\frac{1}{n+k\sqrt{n}}\sum\limits_{i=1}^{i=n+k\sqrt{n}}B_i \right]\xrightarrow{p}0$$

Thus, for all $k$

$$\lim\limits_{n \to \infty}\mathbb{P}\left(\frac{1}{n}\sum\limits_{i=1}^{i=n}A_i\geq \left(1-\frac{c}{\sqrt{n}}\right)\left(1-\frac{k}{\sqrt{n}}\right)\frac{1}{n+k\sqrt{n}}\sum\limits_{i=1}^{i=n+k\sqrt{n}}B_i \right)=1 $$

Is my approach correct? Is there a simpler precise method to prove? In the problem statement, if we had a function $f(n,c,k)$ instead of $\left(1-\frac{c}{\sqrt{n}}\right)\left(1-\frac{k}{\sqrt{n}}\right)$, such that $\lim\limits_{n\to\infty}f(n,c,k)=1$, could we use the same approach?

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  • $\begingroup$ I observe that your question has random variables $\{B_i\}$ that are never used. Is there a typo in the question? $\endgroup$ – Michael Dec 21 '16 at 18:29
  • $\begingroup$ @Michael Thanks for informing me that. I fixed the typo $\endgroup$ – Susan_Math123 Dec 21 '16 at 18:31
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    $\begingroup$ Your argument seems similar to the following false claim: “If $X_i \rightarrow 1$ and $Y_i \rightarrow 1$ then $P[X_i \geq Y_i] \rightarrow 1$.” This is not true, for example, if $Y_i=1$ and $X_i=1-e^{-i}$ for all $i$. $\endgroup$ – Michael Dec 21 '16 at 18:37
  • $\begingroup$ There are cases when no such $k$ exists. For example if $Var(A)=0$ and $E[A]=-1$. Also, no such $k$ exists if $Var(A)>0$ and $E[A]=0$. You are likely supposed to assume $E[A]>0$ and then use the central limit theorem. $\endgroup$ – Michael Dec 21 '16 at 18:44
  • $\begingroup$ Actually, even if $Var(A)>0$ and $E[A]>0$, there is still no such $k$. Where did this problem come from? if the factors were $(1-c/n^{.5-\epsilon})$ then it would work. $\endgroup$ – Michael Dec 21 '16 at 19:02
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Assume $\{A_i\}$ and $\{B_i\}$ are independent i.i.d. processes with $Var(A_1)=\sigma^2$. If $0<\sigma^2 < \infty$, then there is no such $k$.

Proof: Define $m=E[A_1]$. Fix $c>0$ and $k>0$ as arbitrary constants. Define the following events:

\begin{align} \mathcal{C}_n &= \left\{\frac{1}{n}\sum_{i=1}^n A_i \geq (1-\frac{c}{\sqrt{n}})(1-\frac{k}{\sqrt{n}})\frac{1}{\lceil n + k\sqrt{n}\rceil}\sum_{i=1}^{\lceil n+k\sqrt{n}\rceil}B_i\right\} \\ \mathcal{D}_n &= \left\{\frac{1}{\lceil n+k\sqrt{n}\rceil}\sum_{i=1}^{\lceil n + k\sqrt{n}\rceil}B_i \geq m\right\} \end{align} Notice that $P[\mathcal{D}_n]\rightarrow 1/2$ by the central limit theorem (CLT). Also, $\mathcal{D}_n$ is independent of the $\{A_i\}$ process. Hence, for all $n$ large enough to ensure $1>(1-\frac{c}{\sqrt{n}})(1-\frac{k}{\sqrt{n}})> 0$, we have: \begin{align} P[\mathcal{C}_n|\mathcal{D}_n] &\leq P\left[\frac{1}{n}\sum_{i=1}^nA_i \geq (1-\frac{c}{\sqrt{n}})(1-\frac{k}{\sqrt{n}})m|\mathcal{D}_n\right]\\ &=P\left[\frac{1}{n}\sum_{i=1}^nA_i \geq (1-\frac{c}{\sqrt{n}})(1-\frac{k}{\sqrt{n}})m \right] \\ &=P\left[\frac{1}{\sqrt{n}}\sum_{i=1}^n(A_i-m) \geq - m(c+k) + \frac{ckm}{\sqrt{n}} \right] \rightarrow q \end{align} where we define $q=P[G \geq -m(c+k)]$ for $G$ a Gaussian with mean 0 and variance $\sigma^2$. Note that $0<q<1$. Then:

\begin{align} P[\mathcal{C}_n] &= P[\mathcal{C}_n|\mathcal{D}_n]P[\mathcal{D}_n] + P[\mathcal{C}_n|\mathcal{D}_n^c]P[\mathcal{D}_n^c] \\ &\leq P[\mathcal{C}_n|\mathcal{D}_n]P[\mathcal{D}_n] + P[\mathcal{D}_n^c] \end{align} and so $$ \limsup_{n\rightarrow\infty} P[\mathcal{C}_n] \leq q(1/2)+(1/2) < 1 \quad \Box$$

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  • $\begingroup$ Thanks. I know $P\left[\frac{1}{\sqrt{n}}\sum_{i=1}^n(A_i-m) \geq - m(c+k) \right] \rightarrow q$, but can you (precisely) clarify why $P\left[\frac{1}{\sqrt{n}}\sum_{i=1}^n(A_i-m) \geq - m(c+k) + \frac{ckm}{\sqrt{n}} \right] \rightarrow q$? This is not obvious to me. $\endgroup$ – Susan_Math123 Dec 23 '16 at 3:21
  • $\begingroup$ And, in the last step, why did you use limsup? I think you could use lim. Right? Can you clarify please? (Your clarification helps me understand and learn these stuff which is one of the goals of stackexchange.) $\endgroup$ – Susan_Math123 Dec 23 '16 at 3:23
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    $\begingroup$ If you want, you can just say $P[\frac{1}{\sqrt{n}}\sum_{i=1}^n (A_i-m) \geq -m(c+k) + \frac{ckm}{\sqrt{n}}] \leq P[]$ and then take a limit after. I used $\limsup$ just because I did not want to bother proving the limit exists. $\endgroup$ – Michael Dec 23 '16 at 14:14
  • $\begingroup$ @That was great! $\endgroup$ – Susan_Math123 Dec 29 '16 at 3:50

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