Calculatiion of the joint probabilities

I was refreshing my knowledge regarding the definition of joint probabilities and read a page from the book: 'Econometrics for dummies' which gave the following example.

What I intended to do was to calculate the joint probabilities $$P(Y,X)$$ from the unconditional probabilities $$P(Y)$$ and $$P(X)$$.

I came across the following two definitions of a joint probability:

Definition 1: $$P(Y,X) = P(Y) \times P(X)$$

Definition 2: $$P(Y,X) = P(Y|X) \times P(X) = P(X|Y) \times P(Y)$$

Using Definition 1, I could not obtain the joint probabilities from the tables. For example multiplying $$P(Y=1)=0.35$$ and $$P(X=2)=0.10$$ did not give me $$P(Y=1,X=2) = P(1,2) = 0$$.

I do not understand why Definition 1 does not give me the right answer.

I believe the issue is that I am facing is:

I know that the conditional expectation of $$P(Y|X)$$ should not necesarily be equal to $$P(Y)$$. But does the second definition not imply that $$P(Y|X)$$ and $$P(Y)$$ are equal?

The first identity of the joint probability with the product of marginals

$$P(X,Y) = P(X) P(Y)$$

only holds when $$X$$ and $$Y$$ are independent. Based on the table and the fact that the product of the marginals do not equal the joint probabilities, should simply lead you to suspect a lack of independence.

The second identity is true by the definition of conditional probability where

$$P(X\lvert Y) := \frac{P(X,Y)}{P(Y)}$$ hence

$$P(X\lvert Y) P(Y) = P(X,Y).$$

And this second "definition" does not imply that $$P(Y) = P(Y\lvert X)$$, I do not know why you are lead to that conclusion.

• I marked your answer as the correct one.Thanks for the clarification regarding the property of independence. $P(Y) = P(Y|X)$, given that X and Y are independent! It all makes sense now. I lead to that conclusion if you'd combine definition 1 and 2 (However, using definition 1 assumes independence): $P(Y,X) = P(Y|X) \times P(X) = P(Y) \times P(X)$ – Nadia Merquez Oct 18 '19 at 11:00
• yes that is true $P(X\lvert Y) = P(X)$ and $P(Y\lvert X) = P(Y)$ given independence. So for two independent variables conditional and marginal probabilities are the same - knowing the value of one of the variables does not put you in a better position to predict the other. But of course the key again is that by using definition 1 you have already assumed independence (as you yourself state). – Jesper for President Oct 18 '19 at 11:05
• In order to gain full clarification, given that there is a form of dependence between $X$ and $Y$ and one would like to obtain the joint probability $P(Y,X)$. Then, one would have to use the second identity $P(Y,X) = P(Y|X) \times P(X) = P(X|Y) \times P(Y)$, implying that at least one marginal probability and one conditional probability has to be known, right? – Nadia Merquez Oct 18 '19 at 11:29
• To say that one "would have to" is probably to strong in general. But in this context where the choice is between rule 1 and rule 2, then I would say you are right. But note also that in the table you actually observe joint frequencies that can be used to estimate the joint probabilities directly, but I guess that is another story. – Jesper for President Oct 18 '19 at 11:40
• Thanks for the insightful remarks! – Nadia Merquez Oct 18 '19 at 11:51