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This question is about probability of union of events in a probability product space. Let’s say a fair die is thrown twice and we’re interested to find out probability of getting face value one in 1st throw or two in 2nd throw. There are many ways to solve this as few of them are listed below.

Option 1)
We could think of both the throws as one experiment with sample space $S$. In such case, each sample point will be a double valued point e.g. $S=${{1,1}, {1,2}…{6,6}}. We could define an event $C$ that either 1st throw has face value one or 2nd throw has face value 2, implying $C=$ {{1,1}, {1,2}…,{1,6}, {2,2}, {3,2}…{6,2}}
Finally applying $P(C)$ = size of {$C$} / size of sample space, we could get $P(C) =$ 11/36

Option 2)
Alternately, we could decompose the previous single experiment into two experiments say $E1$ the 1st throw and $E2$ the 2nd throw. We could define two events $A$ and $B$ that a face value of one and of two respectively. In such case, the event $A$ is scoped within $E1$ with its own sample space $S1=${1,2,..6}. Similarly $B$ is scoped within $E2$ and has a different sample space $S2=${1,2,..6}. We could apply sum-rule of probability augmented by independence formula to derive the probability as follows.
$P(A\cup B)=P(A)+P(B)-P(A \cap B) = P(A)+P(B)-P(A).P(B) = 1/6+1/6 – 1/6.1/6= 11/36$

Option 3)
Here I would like to use the decomposed view of two experiments $E1$ and $E2$ like option-2, but would like to apply the size of event set like option-1 e.g.
$P(A \cup B)= $size of {$A\cup B$}/size of product space {$S1 \times S2$}

The question is how I find the size of {$A \cup B$} in terms of size of individual events given that each event is restricted within its sample space or experiment. More precisely, can I apply the set theoretic formula of {$A \cup B$}={$A$}$+${$B$}$-${$A \cap B$} here? It seems I need to use rather something like {$A \cup B$}$=${$A \times S2$}$+${$B \times S1$}$-${$A \cap B$}. Wondering is this something that is mathematically addressed somewhere!

Further, most of the books proves addition rule of probability using Venn diagram using two events, say $A$ and $B$, of same experiment with same sample space, say $S$. How can we prove that the same rule is applicable even when events $A$ and $B$ belong to different experiments with own sample space.
Will appreciate any help on this! I have not taken up Measure Theory or $\sigma$-algebra yet. So exposition of relevant topic without reference to Measure Theory would be highly appreciated.

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The correct approach is always to consider the common sample space. If the events given are independent then the common sample space can be constructed. If they are not independent then either the dependence structure is given on a common sample space or the common sample space is not defined.

In the case of independent dice rolling the sample space is always

$$\Omega=\{(i,j): 1\le i\le 6, 1\le i\le 6\}$$

with

$$P(\{(i,j)\})=\frac1{36}.$$

Option 1) follows this route by counting the elements of event $C$.

Option 2) is also following this approach if we interpret $A$ and $B$ as

$$A=\{(1,1),(1,2),(1,3),(1,4),(1,5),(1,6)\}$$

and

$$B=\{(1,2),(2,2),(3,2),(4,2),(5,2),(6,2)\}.$$

With this

$$A\cup B=\{(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,2),(3,2),(4,2),(5,2),(6,2)\}$$

and

$$A\cap B=\{(1,2)\}.$$

The result will be the same because in $P(A)+P(B)$ the probability $P(\{(1,2)\})$ is enumerated twice so $$P(A\cap B)=P(\{(1,2)\})$$ has to be subtracted.

Option 3) This approach is wrong at the beginning because the $A$ and $B$ belong to different universes. Then the argumentation sneaks back to the right track.

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  • $\begingroup$ My confusion remains unanswered. Let me rephrase it – For a sequence of experiments E1 and E2, irrespective of their dependency relation, can we solve probability question by defining events within each experiment? The solutions of option-2 (and -1), implicitly assume A and B are defined on same sample space. Whereas I would like to see if any alternate solution exists by decomposing the whole experiments and defining events within each experiment. Further I believe, one of the motivation for product space is to enable such decomposed experiments approach unless there is something I’m missing. $\endgroup$ – KGhatak Jul 28 '15 at 11:24
  • $\begingroup$ @BuckCherry: You can answer questions by doing calculations on the separate (decomposed) sample spaces. But you can use these results in the case of the composition of the sample spaces if the experiments are independent. In such a case you can hide that you work on a product space. But why to hide? Let alone that dependent cases cannot be handled that way. $\endgroup$ – zoli Jul 28 '15 at 11:29
  • $\begingroup$ @ zoli - Well, you essentially saying that decomposing into experiments works iif experiments are pairwise independent and otherwise it won’t. It would be great if you have any logical/mathematical proof or some citation on this. $\endgroup$ – KGhatak Jul 28 '15 at 11:41
  • $\begingroup$ @BuckCherry: The essence of modern probability theory (created by Kolmogorov) is the use of product spaces to handle independent and dependent experiments based on the same methodology. (measure theory cannot be gotten around.) See: en.wikipedia.org/wiki/Kolmogorov_extension_theorem $\endgroup$ – zoli Jul 28 '15 at 12:04
  • $\begingroup$ @ zoli – It seems the extension theorem doesn’t provide answer to my question. However, from W. Feller’s Book, Vol I,ch v.4, I found that for series of independent experiments, instead of looking at the sample space as a whole we can consider it as a Cartesian product of individual spaces; allowing us to work with events/probabilities defined within each experiment. He somehow doesn’t speak about dependent events though. What I’m wondering, to start with, is – is it possible to construct a product space for series of dependent events – say Toss a coin and then roll a dice only if it’s a head ? $\endgroup$ – KGhatak Jul 29 '15 at 12:07

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