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I am seeing a derivation which I do not understand totally. I would appreciate if someone could help me out.

Let $f(t)$ specify the event probability at moment $t$. Further, assume that $P(1)$ denotes the probability of one event occurring during time $T$ and $P(2)$ denotes the probability that two events occur during time $T$.

$P(1) = \int_{0}^{T}f(t)dt$

$P(2) = \int_{t2}^{T}(\int_{0}^{t2}f(t_1)dt_1)dt_2 = \frac{1}{2}(2*\int_{t2}^{T}(\int_{0}^{t2}f(t_1)dt_1)dt_2 = \frac{1}{2}(\int_{0}^{T}f(t)dt)^2 = \frac{1}{2}P(1)^2$

How were the double integrals converted in the final step?

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You forgot $f(t_2)$. Adding it in the double integral defining $P(2)$ makes the integrand symmetric and yields the formula $2P(2)=P(1)^2$. To see this, note that $$ 2P(2)=2\int_{0}^{T}\int_{0}^{t_2}f(t_1)f(t_2)dt_1dt_2 = 2\int_{0}^{T}\int_{0}^{T}\mathbf 1_{t_1\leqslant t_2}f(t_1)f(t_2)dt_1dt_2, $$ which is also, by symmetry, $$ 2P(2) = \int_{0}^{T}\int_{0}^{T}\mathbf 1_{t_1\leqslant t_2}f(t_1)f(t_2)dt_1dt_2+\int_{0}^{T}\int_{0}^{T}\mathbf 1_{t_2\leqslant t_1}f(t_1)f(t_2)dt_2dt_1, $$ which is also $$ 2P(2) = \int_{0}^{T}\int_{0}^{T}f(t_1)f(t_2)dt_1dt_2=P(1)^2. $$

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Interesting. The source does not do this. You have the inner limit from $0$ to $t_2$ and the outer limit from $0$ to $T$. However, the source has the inner limit as $0$ to $t_2$ and outer limit as $t_2$ to $T$. Am I missing something? Thank you! –  Legend Nov 8 '12 at 9:01
    
The first double integral for P(2) in your post has the form $\int\limits_{t_2}^Tg(t_2)dt_2$, which is absurd. (And which begs for the question: what is the source?) –  Did Nov 8 '12 at 10:32
    
+1 for your time. I wasn't sure if you would be able to access this but let me try: static.usenix.org/event/fast08/tech/full_papers/jiang/jiang.pdf Please look at Section 5.2.1 on Page 12. –  Legend Nov 8 '12 at 17:39
    
It looks like I may have mistaken the limits of the integral: I read $t2$ as $t_2$. I just corrected it in my question. –  Legend Nov 8 '12 at 17:44
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