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The followings are from Kai Lai Chung's A Course in Probability Theory, on page 207

$X_{n,j}, j=1,\dots, k_n, n=1,\dots,$ are random variables, where $k_n \to \infty$ as $n\to \infty$.

For every $\epsilon > 0$:

(a) for each $j$, $\lim_n P(|X_{nj}|>\epsilon) = 0$;

(b) $\lim_n \max_j P(|X_{nj}| > \epsilon) = 0$; (holospoudic)

(c) $\lim_n P(\max_j |X_{nj}| > \epsilon) = 0$;

(d) $\lim_n \sum_j P(|X_{nj}| > \epsilon) = 0$;

It is clear that (d) => (c) => (b) => (a).

I can understand (d) => (c), because $$\sum_j P(|X_{nj}| > \epsilon) \geq P(\max_j |X_{nj}| > \epsilon).$$

I can understand (b) => (a), because $$\max_j P(|X_{nj}| > \epsilon) \geq P(|X_{nj}|>\epsilon).$$

Questions:

  1. I wonder why (c) => (b)? Neither do I understand why $$P(\max_j |X_{nj}| > \epsilon) \geq \max_j P(|X_{nj}| > \epsilon),$$ if it is true.
  2. Also why when $X_{nj}, j=1,\dots,k_n$ are independent for each $n$, (d) $\equiv$ (c)? This is in the exercise 1 on page 214.

Thanks and regards!

Michael Hardy
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Tim
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1 Answers1

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Fix $\varepsilon$, and consider some random variables $(Y_k)_k$ and $Y=\sup\limits_kY_k$. For every $i$, $Y\geqslant Y_i$ hence $[Y_i\gt\varepsilon]\subseteq[Y\gt\varepsilon]$. This implies that $\mathrm P(Y_i\gt\varepsilon)\leqslant\mathrm P(Y\gt\varepsilon)$. This inequality holds for every $i$ and the RHS does not depend on $i$ hence $\sup\limits_i\mathrm P(Y_i\gt\varepsilon)\leqslant\mathrm P(Y\gt\varepsilon)=\mathrm P(\sup\limits_iY_i\gt\varepsilon)$.

Edit: When, furthermore, the random variables $(Y_k)_k$ are assumed independent, the estimation of the distribution of $Y$ proceeds as usual. First, for every $\varepsilon$, $[Y\leqslant\varepsilon]=\bigcap\limits_k[Y_k\leqslant\varepsilon]$ hence $$ \mathrm P(Y\gt\varepsilon)=1-\prod\limits_k(1-\mathrm P(Y_k\gt\varepsilon)). $$ Now, $1-x\leqslant\mathrm e^{-x}$ for every $x$ hence $$ \mathrm P(Y\gt\varepsilon)\geqslant1-\exp\left(-\sum\limits_k\mathrm P(Y_k\gt\varepsilon)\right), $$ which shows that (c) implies (d). On the other hand, (d) implies (c) because $[Y\gt\varepsilon]=\bigcup\limits_k[Y_k\gt\varepsilon]$ and the probability of the union is at most the sum of the probabilities.

Did
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  • Thanks, Didier! I understand the first question now. I wonder why (d) and (c) are equivalent when $X_{nj},j=1,\dots,k_n$ are independent for each $n \in \mathbb{N}$? – Tim Jul 01 '12 at 14:49
  • See Edit. $ $ $ $ – Did Jul 01 '12 at 17:38
  • Thanks! (d) => (c), can it be because $$\sum_k P(|Y_k| > \epsilon) \geq P(Y > \epsilon)$$? – Tim Jul 01 '12 at 17:59
  • Yes. $ $ $ $ $ $ – Did Jul 01 '12 at 18:29
  • $[Y⩽ε]=⋂_k[Y_k⩽ε]$, I think $[Y⩽ε] \supseteq ⋂_k[Y_k⩽ε]$ is more accurate, given $Y = \sup_k Y_k$, isn't it? $Y_k$ being independent doesn't make the inequality equality. – Tim Jul 02 '12 at 04:10
  • Really? The equality holds. – Did Jul 02 '12 at 06:16
  • When $Y=\max_k Y_k$, I agree the equality holds. When $Y=\sup_k Y_k$, the equality does not hold when $\sup$ is not achievable. Am I right? – Tim Jul 02 '12 at 06:39
  • No, you are not. – Did Jul 02 '12 at 06:53