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I have some difficulty in understanding the concept of equality of events. I am referring to the book "Probability and random variables" by papoulis. In that, the author describes equality of events as follows. "Two events A and B are called equal if they consist of the same elements. They are equal with Probability 1 if: P(A) = P(B) = P(AB) "

So far, I have understood the above equation holds, only if the set A and B overlaps (visualising in terms of Venn diagram). But I am not able to correlate this with a real life example. What example would differentiate "equal events" and "equal events with Probability 1" ? I tried understanding this concept with two different dice. But they become independent events, and it will not be possible to have an intersection between the independent events as they occur in different sample spaces.

Could someone please help me out here? I want to understand it with some real life examples. Thank you so much. I have enclosed a copy of the text for your convenience

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I can think of events such as $A=\{1\}$, $B=\{1\}$, $C=\{1,7\}$, for one roll of a common die. Here, $A$ and $B$ have the same elements and are equal events. One could also define the event (set) $B$ in a different way $\{x \ \epsilon \ \ Z \ \ |\ \ 0<x<2 \}$ or $\{x\ \ \epsilon \ \ N \ \ |\ \ x^2=1 \}$, but they all lead to the same elements. Then, $A$ and $B$ are equal events. However, $A$ and $C$ are not equal events since $C$ also contains $7$. But, we know that roll of a die cannot be $7$, this means probability volumes are equal and, $P(A)=P(C)=P(AC)$. So, $A$ and $C$ are equal with probability 1.

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  • $\begingroup$ Thank you! So even if the events A and C don't contain same elements, they can still be "equal events with Probability 1?" $\endgroup$ – Nishanth Rao Dec 23 '18 at 7:07
  • $\begingroup$ Yes, because of an element with zero probability. $\endgroup$ – gunes Dec 23 '18 at 7:08
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Events in probability theory are just sets of outcomes in an underlying sample space $\Omega$. The definition of equality is the same as that used in set theory --- two events $\mathcal{A}$ and $\mathcal{B}$ are equal if and only if the contain the same elements, which occurs if and only if $\mathcal{A} \supseteq \mathcal{B}$ and $\mathcal{A} \subseteq \mathcal{B}$. Since it is possible to have events with probability zero, it is possible to have cases where the probability of two sets is equal, even though the sets themselves are not equal (i.e., do not contain the same outcomes).

To see this, consider the case where you have sample space $\Omega = \{ \omega_1, \omega_2, \omega_4 \}$ and a probability measure $\mathbb{P}$ on the set of all subsets of $\Omega$. Suppose that this probability measure gives the following probabilities:

$$\mathbb{P}(\{ \omega_1 \}) = 0.7 \quad \quad \quad \mathbb{P}(\{ \omega_2 \}) = 0.3 \quad \quad \quad \mathbb{P}(\{ \omega_3 \}) = 0.$$

In this example we have $\mathbb{P}(\{ \omega_1 \}) = \mathbb{P}(\{ \omega_1, \omega_3 \})$ even though $\{ \omega_1 \} \neq \{ \omega_1, \omega_3 \}$. Here we have a case where two events with different outcomes have the same probability, due to the presence of an event with zero probability.

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