# Proving that $P\left(\bigcup_{j=1}^n E_j^{(n)}\right)\sim\sum_{j=1}^n P(E_j^{(n)})$ for independent events $E_j^{(n)}$

For arbitrary events $$\{E_j, 1\le j\le n\}$$, we have

$$P\left(\bigcup_{j=1}^n E_j\right)\ge\sum_{j=1}^n P(E_j) - \sum_{1\le j < k \le n} P(E_jE_k)$$

If $$\forall n: \{E_j^{(n)}, 1\le j\le n \}$$ are independent events, and

$$P\left(\bigcup_{j=1}^n E_j^{(n)} \right)\to 0 \quad \text{as} \quad n\to\infty$$

then $$P\left(\bigcup_{j=1}^n E_j^{(n)} \right) \sim \sum_{j=1}^n P(E_j^{(n)})$$.

My attempt at showing this last claim:

I am going to guess that the ~ symbol in the problem means that

$$\lim_{n\to\infty}\frac{P\left(\bigcup_{j=1}^n E_j^{(n)} \right)}{\sum_{j=1}^n P(E_j^{(n)})}=1$$

From

$$\sum_{j=1}^n P(E_j^{(n)})\ge P\left(\bigcup_{j=1}^n E_j^{(n)}\right)\ge\sum_{j=1}^n P(E_j^{(n)}) - \sum_{1\le j < k \le n} P(E_j^{(n)}E_k^{(n)})$$

and dividing we have

$$1\ge \frac{P\left(\bigcup_{j=1}^n E_j^{(n)}\right)}{\sum_{j=1}^n P(E_j^{(n)})} \ge 1 - \frac{\sum_{1\le j < k \le n} P(E_j^{(n)}E_k^{(n)})}{\sum_{j=1}^n P(E_j^{(n)})}$$

so then the problem reduces to showing

$$\lim_{n\to\infty}\frac{\sum_{1\le j < k \le n} P(E_j^{(n)}E_k^{(n)})}{\sum_{j=1}^n P(E_j^{(n)})} = 0$$

But how to make progress after that?

• I have tried to make the distinction between the "problem statement" and your attempted solution clearer. I'm not sure how the first line and displayed equation contribute to the question, though. Could you either make that clear, or remove it (because it is the first thing readers see, and if it's not concerning the problem that may confuse them)? Commented May 23, 2013 at 10:33

Notice that $$\sum_{1\leqslant j<k\leqslant n}\mu(E_jE_k)=\sum_{1\leqslant j<k\leqslant n}\mu(E_j)\mu(E_k)=\frac 12\left(\left(\sum_{j=1}^n\mu(E_j)\right)^2-\sum_{j=1}^n\mu(E_j)^2\right).$$ Dividing by $\sum \mu(E_j)$, the first term is not problematic. For the second one, write $$\sum_{j=1}^n\mu(E_j)^2\leqslant \max_{1\leqslant k\leqslant n}\mu(E_k)\cdot \sum_{j=1}^n\mu(E_j)\leqslant \mu\left(\bigcup_{1\leqslant k\leqslant n}E_k\right)\cdot \sum_{j=1}^n\mu(E_j).$$
• The resulting expressions still involve $\sum_{j=1}^n \mu(E_j)$. How does one control such terms showing that they are small? Commented May 24, 2013 at 0:26
• Once you divide by this sum, we get something which converges to $0$ by assumption. Commented May 24, 2013 at 8:15
• How does the assumption that $\mu(\cup E_j)$ converges to 0 give that $\sum \mu(E_j)$ converges to 0? Commented May 24, 2013 at 22:41
• It's not what I said (I said that $\frac{\sum \mu(E_j)^2}{\sum\mu(E_j)}\leq \mu(\bigcup E_j)$. Commented May 25, 2013 at 10:30
• So then $\frac{\sum_{1\le j< k\le n}\mu(E_jE_k)}{\sum\mu(E_j)}= \frac{1}{2}\left( \sum\mu(E_j) - \frac{\sum\mu(E_j)^2}{\sum\mu(E_j)} \right)$ is suppose to converge to $0$. How does that inequality show that? Commented Jun 4, 2013 at 8:16
A quantitative version of the result, proved by a recursion over the number of events involved: for every independent events $E_i$, $$1-P\left(E\right)\leqslant\frac{P\left(E\right)}{\sum\limits_iP(E_i)}\leqslant1,\qquad E=\bigcup\limits_iE_i.$$ Thus, for independent events $E_i^{(n)}$, the events $E^{(n)}=\bigcup\limits_iE_i^{(n)}$ are such that $$P\left(E^{(n)}\right)\to0\implies P\left(E^{(n)}\right)\sim\sum\limits_iP(E_i^{(n)}).$$