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The proof of the Kolmogorov inequality is based on the algebra over some defined events. The famous Kolmogorov inequality is known as follows

Let $X_k, \space \space i=1,2,...,n$ be independent random variables with common mean $\mathbb{E}(X_k)=0$ and variance $\mathbb{E}(X^2)=\sigma_{k}^2<\infty, \space \space k=1,2,...,n$. Then for any $\epsilon>0$ and $S_n=\sum_{k=1}^{n}X_k$

$$1-\frac{[\epsilon+2c]^2}{\sum_{k=1}^{n}\sigma_{k}^2}\leq P\Big[\max_{1 \leq k \leq n}|S_k-\mathbb{E}(S_k)|\geq \epsilon\Big]\leq \frac{{\sum_{k=1}^{n}\sigma_{k}^2}}{\epsilon^2}$$

The textbooks have different ways of proving this inequality but I want to understand what do the basic defined events (which are used to prove this inequality) signify here so that I can prove the inequality in different manner.

To prove the upper bound of the inequality, one text defines a random variable $t$ as

\begin{align} t & =\begin {cases} \text{1st }k; \space 1\leq k \leq n\space \space \space \text{ such that } S_k^2 \geq \epsilon^2 \text{ if there is such a } \space \space k\\ n+1 \space \space \text{otherwise}\end{cases}\end{align}

$\bullet$ Here what is meant by $\text{1st } \space k$ and how come defining $n+1$ is useful for the upper bound.

Next, most of them define events as follows

Let $\mathcal{B}_k = [|S_k|\geq \epsilon]\ \text{ for } \ k=1,2,..,n \\ \mathcal{A} = [\max_{1 \leq j \leq n}|S_j|\ge\epsilon] \\ \mathcal{A^c} = [\max_{1 \leq j \leq n}|S_j|< \epsilon] \\ \mathcal{A_k} = [|S_k|\geq \epsilon \ \ \text{ and } \ \ |S_j|< \epsilon \ \ \text{ }\ \ j=0,1,..., k-1] \ \ \text{ here }\ \ S_0=0$

$\bullet$ We see that $j=0,1,...,k-1$ and $k=1,2,..,n$ then excluding $j=0\ \ \text{and} \ \ k=k, k+1, k+2, ..., n$ from the last event $\mathcal{A_k}$ we have for the $k-1$ cases:-

$\text{both } \ \ |S_1|\geq \epsilon \ \text{ and } \ |S_1|<\epsilon \ \ \\ |S_2|\geq \epsilon \ \text{ and } \ \ |S_2|<\epsilon \\ \text{...} \ \ \text{ upto } \ \ j=1,.., k-1 \ \ \text{ and } k=1,2,...,k-1$

Writing $\mathcal{A_k}$ as $[\mathcal{B \ \cap \ A^c}]$ what can we infer from this $[\mathcal{B \ \cap \ A^c}]$ for $j,k=1,2,...,k-1$ and ? There is a term "Disjointify $\mathcal{B_k}$ to get the sets $\mathcal{A_1},...,\mathcal{A_n}$". How can we convert an event to several disjoint events?

The proof of the lower bound is based on some additional assumption such that $|X_k|<c<\infty,\ \ P(|X_k|\leq c)=1 \ \forall \ k$. Modifying the event $\mathcal{A}$ as $\cup_{k=1}^{n}\mathcal{A_k}$ gives $\mathcal{A} = [\max_{1 \leq j \leq n}|S_j|>\epsilon]=\cup_{k=1}^{n} [\mathcal{B \ \cap \ A^c}] $.

$\bullet$ The lower bound can be found easily by the manipulation of $[\mathbb{E}S_{n}^{2}: \mathcal{A}]$ and later summing over $S_n^2$ but cannot figure out for $\epsilon>0$, how to use $var(S_n)\leq \epsilon^2+(\epsilon+2c)^2\frac{P(\mathcal{A})}{P(\mathcal{A^c})}$ for the lower bound.

If I just accept the assumptions without understanding what these signify and why they are used, then there is no problem with the different proofs. Any help or explanation regarding the assumptions will be valuable and highly appreciated.


Patrick Alfred Pierce Moran, in his book 'An Introduction To Probability Theory' has defined events as follows to prove the lower bound of the Kolmogorov's inequality:-

It is obvious that all different settings of the notations carry similar meaning but it is not clear to me.

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Here what is meant by $\text{1st } \space k$

This means that $|S_k|\ge \epsilon$ and $|S_i|<\epsilon$ for $i<k$.

how come defining $n+1$ is useful for the upper bound.

Hard to say outside of context. Actually, further you work with events $\mathcal A_k = \{t = k\}$ for $k=1,\dots,n$, so the value $n+1$ is lost anyway.

Writing $\mathcal{A_k}$ as $[\mathcal{B \ \cap \ A^c}]$ what can we infer from this $[\mathcal{B \ \cap \ A^c}]$ for $j,k=1,2,...,k-1$ and ?

What is $\mathcal{B}$ here? Did you mean $\mathcal{B}_k$? Then $\mathcal{B} \cap {A^c} = \varnothing$, as $\mathcal{B}_k\subset \mathcal A$.

There is a term "Disjointify $\mathcal{B_k}$ to get the sets $\mathcal{A_1},...,\mathcal{A_n}$". How can we convert an event to several disjoint events?

I didn't encounter the word "disjointify" before, but I like it! It means "replace with disjoint sets having the same union". Here we replace $\mathcal{B}_1,\dots, \mathcal{B}_n$ with sets \begin{align} \mathcal{A}_1& = \mathcal{B}_1 ,\\ \mathcal{A}_2& = \mathcal{B}_2\setminus\mathcal{B}_1 ,\\ \mathcal{A}_3& = \mathcal{B}_3\setminus(\mathcal{B}_1\cup \mathcal{B}_2) ,\\ &\dots\dots \dots\\ \mathcal{A_n}& = \mathcal{B}_n\setminus\biggl(\bigcup_{i=1}^{n-1}\mathcal{B}_i\biggr). \end{align}.

$\bullet$ The lower bound can be found easily by the manipulation of $[\mathbb{E}S_{n}^{2}: \mathcal{A}]$ and later summing over $S_n^2$ but cannot figure out for $\epsilon>0$, how to use $var(S_n)\leq \epsilon^2+(\epsilon+2c)^2\frac{P(\mathcal{A})}{P(\mathcal{A^c})}$ for the lower bound.

Where did you get this inequality from? For the lower bound, I usually write, $$ \mathrm{E}[S_n^2 1_{\mathcal A}] = \sum_{k=1}^n \mathrm{E}[(S_n-S_k) + S_k)^2 1_{\mathcal A_k})] \\ = \sum_{k=1}^n \left(\mathrm{E}[(S_n-S_k)^2]\cdot \mathrm{P}(\mathcal A_k) + \mathrm{E}[S_k^2 1_{\mathcal A_k}]\right) \le \left(\mathrm{Var}(S_n) + (c+\epsilon)^2\right) \sum_{k=1}^n \mathrm{P}(\mathcal A_k)\\ = \left(\mathrm{Var}(S_n) + (c+\epsilon)^2\right)\mathrm P(\mathcal A) $$ and on the other hand, $$ \mathrm{E}[S_n^2 1_{\mathcal A}] = \mathrm{E}[S_n^2] - \mathrm{E}[S_n^2 1_{\mathcal A^c}]\le \mathrm{Var}(S_n) - \epsilon^2 \mathrm P(\mathcal A^c), $$ which gives $$ \mathrm P(\mathcal A)\ge \frac{\mathrm{Var}(S_n) -\epsilon^2}{\mathrm{Var}(S_n) + (c+\epsilon)^2-\epsilon^2} \ge 1 - \frac{(c+\epsilon)^2}{\mathrm{Var}(S_n)}. $$

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  • $\begingroup$ +1 for this answer, which is helpful. Just want to know one thing. By the phrase "replace with disjoint sets having the same union" do you mean partitioning $\mathcal{B}_1$ then writing them as the union of those partitions? $\endgroup$
    – vbm
    Apr 10, 2020 at 8:55
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    $\begingroup$ I mean that we write $\mathcal A = \bigcup_{i=1}^n \mathcal B_i = \bigcup_{i=1}^n \mathcal A_i$, and the latter sets are disjoint. $\endgroup$
    – zhoraster
    Apr 10, 2020 at 9:29
  • $\begingroup$ Yes, now I have got it. $\endgroup$
    – vbm
    Apr 10, 2020 at 9:45

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