# Is it always possible to use the previous values of probability of events after combining two random experiments?

The following text is found in Papoulis' book, page 48:

The Cartesian product of two experiments $$S_{1}$$ and $$S_{2}$$ is a new experiment $$S=S_{1}*S_{2}$$ whose events are all Cartesian products of the form: $$A\times B \qquad \qquad (3,2)$$ where $$A$$ is an event of $$S_{1}$$ and $$B$$ is an event of $$S_{2}$$. and their unions and intersections. In this experiment, the probabilities of the events $$A \times S_{2}$$ and $$S_{1} \times B$$ are such that$$P(A\times S_{2})=P_{1}(A)\qquad \qquad P(S_{1}\times B)=P_{2}(B) \qquad \qquad(3,3)$$ where $$P_{1}(A)$$ is the probability of the event $$A$$ in the experiments $$S_{1}$$ and $$P_{2}(B)$$ is the probability of the event $$B$$ in the experiments $$S_{2}$$. This fact is motivated by the interpretation of $$S$$ as a combined experiment. Indeed, the event $$A \times S_{2}$$ of the experiment $$S$$ occurs if the event $$A$$ of the experiment $$S_{1}$$ occurs no matter what the outcome of $$S_{2}$$ is. Similarly, the event $$S_{1} \times B$$ of the experiment $$S$$ occurs if the event $$B$$ of the experiment $$S_{2}$$ occurs no matter what the outcome of $$S_{1}$$ is. This justifies the two equations in (3-3).

Consider the following experiments:

Random experiment 1

tossing equally likely coin with $$P_{1}(Head)=1/2$$

Random experiment 2

Rolling equally likely dice with $$P_{2}(Six)=1/6$$

Random experiment 3

I define a random combined random experiment as follows: I toss a coin If the coin comes Head, I roll the dice, and if the coin comes Tail, I do not roll the dice.

According to $$(3,3)$$ Papoulis states that the following relationship must be established : $$P(S_{1}\times Six)=P_{2}(Six)=1/6$$ but my calculation says: $$P(S_{1}\times Six)=P(Head\times Six)+P(Tail\times Six)$$ $$P(Tail\times Six)=0$$ $$P(Head\times Six)=P(Six \,|\, Head)\times P_{1}(Head)=1/6 \times 1/2=1/12$$ as you see $$P(S_{1}\times Six)=1/12 \neq P_{2}(Six)$$

Although Papoulis considers (3-3) to be always true , regarding to the above example, I do not think this claim is true for random experiments that are dependent.

Am I wrong?

• You're correct. The definition in the book is basically imposing independence of the two experiments, and the properties would not hold for dependent experiments like yours. Commented Apr 10, 2022 at 20:21
• thanks, but if you look at page 48 of the book in the next paragraph after mentioned paragraph, it is emphasized that (3,3) is always true Commented Apr 10, 2022 at 20:30
• I don't have the book so I can't really comment on this. The author may mean it is always true for independent experiments? Commented Apr 10, 2022 at 21:03

The experiment you are describing is not the Cartesian product of $$S_1$$ and $$S_2$$. The outcomes in the Cartesian product would be $$(H,1),(H,2),(H,3),(H,4),(H,5),(H,6),\\ (T,1),(T,2),(T,3),(T,4),(T,5),(T,6),$$ but the outcomes in your experiment are $$(H,1),(H,2),(H,3),(H,4),(H,5),(H,6),\\ (T,\varnothing),(T,\varnothing),(T,\varnothing),(T,\varnothing),(T,\varnothing),(T,\varnothing),$$ Since your experiment is not a Cartesian product, there is no reason that $$(3,3)$$ should hold.

Edit: With the example in your comment, I now understand the heart of the issue. The idea is that $$(3,3)$$ is the definition of what it means for $$S$$ to be the Cartesian product of $$S_1$$ and $$S_2$$. When $$(3,3)$$ fails, as it does in your comment, you should conclude that the situation is not a Cartesian product. The idea is that if $$S$$ is defined by allowing $$S_1$$ and $$S_2$$ to happen simultaneously, then the probabilities you get from ignoring one of the events should be the original probabilities for the other.

In your example where $$P(\text{six}\mid H)=1/6$$ and $$P(\text{six}\mid T)=1/10$$, you would have $$P(\text{six})=\frac12\cdot \frac16+\frac12\cdot \frac1{10}\neq \frac16$$, so the the outcome of the die is not behaving like standard die roll.

However, it can be the case that $$(3,3)$$ holds, even though the events are dependent. Let $$S_1$$ and $$S_2$$ both be coin flips, with probabilities given by the following table:

$$S_1 \setminus S_2$$ H T
H $$40\%$$ $$10\%$$
T $$10\%$$ $$40\%$$

You can check that $$P(A\times S_2)=P_1(A)$$ and $$P(S_1\times B)=P_2(B)$$ for all subsets $$A$$ and $$B$$, but the events are not independent.

• thanks for answer, let's modify the experiment a bit:"if coin comes Tail I will roll dice but this time suppose that conditional probability of Six given Tail is 1/10"now we have Cartesian product of S1 and S2,in this case result of the calculation will be P(S1*Six)=1/12 + 1/20 which is not equal to P1(Six)=1/6 or P2(Six)=1/10; because of these, I think (3,3) is NOT true for DEPENDENT sub-experiments; is my conclusion correct or what Papoulis has said? Commented Apr 11, 2022 at 14:30
• Ah, I see the issue now. I have edited my answer, take a look. Commented Apr 11, 2022 at 21:19
• Dear Mike, I think it is clear that the sample space of the experiment contains all possible ordered pairs of coin and dice (and is a Cartesian product) so according to Papoulis (3-3) should be true in the case of my example. I don't understand, can you please explain to me why should I conclude that the situation(in my example) is not a Cartesian product? Commented Apr 12, 2022 at 17:55
• This will be my last comment. In my opinion, Papoulis is NOT saying that Cartesian product $\implies$ $(3,3)$. They are saying that $(3,3)$ is part of the definition of Cartesian product, so that if $(3,3)$ is satisfied, you can then say $S$ is a Cartesian product. This is the best I can determine from looking at your post alone. Commented Apr 12, 2022 at 18:01