0
$\begingroup$

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?

$\endgroup$
3
  • $\begingroup$ 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. $\endgroup$
    – angryavian
    Commented Apr 10, 2022 at 20:21
  • $\begingroup$ 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 $\endgroup$ Commented Apr 10, 2022 at 20:30
  • $\begingroup$ 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? $\endgroup$
    – angryavian
    Commented Apr 10, 2022 at 21:03

1 Answer 1

0
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ 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? $\endgroup$ Commented Apr 11, 2022 at 14:30
  • $\begingroup$ Ah, I see the issue now. I have edited my answer, take a look. $\endgroup$ Commented Apr 11, 2022 at 21:19
  • $\begingroup$ 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? $\endgroup$ Commented Apr 12, 2022 at 17:55
  • $\begingroup$ 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. $\endgroup$ Commented Apr 12, 2022 at 18:01

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .