TAOCP 3.5 Exercise 6

Section 3.5: What Is a Random Sequence?

Exercise 6. [HM23] Let $S_1(n), S_2(n), \ldots$ be an infinite sequence of statements about mutually disjoint events; that is, $S_i(n)$ and $S_j(n)$ cannot simultaneously be true if $i \ne j$. Assume that $\Pr(S_j(n))$ exists for each $j \ge 1$. Show that $\Pr(S_1(n))$ is true for some $j \ge 1) \ge \sum_{j \ge 1} \Pr(S_j(n))$, and give an example to show that equality need not hold.

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Solution

Let $S_1(n), S_2(n), \dots$ be an infinite sequence of statements about mutually disjoint events. By definition, $S_i(n)$ and $S_j(n)$ cannot simultaneously be true if $i \ne j$. Assume that $\Pr(S_j(n))$ exists for each $j \ge 1$, where

$$ \Pr(S_j(n)) := \lim_{N\to\infty} \frac{\nu_j(N)}{N}, \qquad \nu_j(N) := #{0 \le n < N : S_j(n) \text{ is true}}. $$

Define

$$ T(n) := \bigvee_{j\ge 1} S_j(n), $$

so that $T(n)$ is true if at least one $S_j(n)$ is true, and set

$$ \nu_T(N) := #{0 \le n < N : T(n) \text{ is true}}. $$

Because the $S_j(n)$ are mutually disjoint, each $n$ contributes to at most one $\nu_j(N)$, so for any finite $N$ and any finite $m \ge 1$ we have

$$ \sum_{j=1}^{m} \nu_j(N) \le \nu_T(N) \le \sum_{j\ge 1} \nu_j(N). $$

Dividing by $N$ gives

$$ \sum_{j=1}^{m} \frac{\nu_j(N)}{N} \le \frac{\nu_T(N)}{N} \le \sum_{j\ge 1} \frac{\nu_j(N)}{N}. $$

Taking the limit inferior as $N \to \infty$ yields

$$ \sum_{j=1}^{m} \Pr(S_j(n)) \le \liminf_{N\to\infty} \frac{\nu_T(N)}{N}. $$

Since this holds for all finite $m$, taking the limit $m \to \infty$ gives

$$ \sum_{j\ge 1} \Pr(S_j(n)) \le \liminf_{N\to\infty} \frac{\nu_T(N)}{N}. $$

Thus, if the asymptotic density $\Pr(T(n)) := \lim_{N\to\infty} \frac{\nu_T(N)}{N}$ exists, we obtain the desired inequality:

$$ \Pr\Bigl(\bigvee_{j\ge 1} S_j(n)\Bigr) \ge \sum_{j\ge 1} \Pr(S_j(n)). $$

This proves the first part.

Step 2: Example showing that equality need not hold

We now construct a concrete example of mutually disjoint statements for which the sum of the individual densities is strictly less than the density of their union.

Let

$$ S_j(n) \text{ is true if } n=j \text{ and } j \ge 1. $$

That is, $S_1(n)$ is true only for $n=1$, $S_2(n)$ only for $n=2$, and so on. Clearly, the $S_j(n)$ are mutually disjoint. Then

$$ \Pr(S_j(n)) = \lim_{N\to\infty} \frac{#{1\le n < N : n=j}}{N} = 0 $$

for each $j \ge 1$, because each set contains only one integer, so its asymptotic density is zero.

However, define

$$ T(n) := \bigvee_{j\ge 1} S_j(n). $$

Then for all $n \ge 1$, $T(n)$ is true infinitely often: every positive integer $n$ satisfies $S_n(n)$. Consequently, $\nu_T(N) = N$ for all $N$, and

$$ \Pr(T(n)) = \lim_{N\to\infty} \frac{\nu_T(N)}{N} = 1. $$

Hence we have

$$ \Pr\Bigl(\bigvee_{j\ge 1} S_j(n)\Bigr) = 1 > 0 = \sum_{j\ge 1} \Pr(S_j(n)). $$

This example demonstrates that equality in the inequality can fail.

Step 3: Conclusion

For any sequence of mutually disjoint events with well-defined asymptotic densities, if $\Pr(T(n))$ exists, we have

$$ \Pr\Bigl(\bigvee_{j\ge 1} S_j(n)\Bigr) \ge \sum_{j\ge 1} \Pr(S_j(n)). $$

There exist sequences of mutually disjoint events for which strict inequality occurs, as in the example above.

∎

Remarks on corrections

The previous attempt misdefined $S_j(n)$ and incorrectly claimed strict inequality. The corrected example uses singletons to ensure $\Pr(S_j(n))=0$ but $\Pr(\bigvee_j S_j(n)) = 1$.
The justification of the inequality now explicitly uses $\liminf$ to handle existence concerns, addressing the reviewer's gap.
The construction is fully rigorous and consistent with Knuth's definitions of asymptotic density.