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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.

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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?

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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.

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  • $\begingroup$ 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): $P(Y,X) = P(Y|X) \times P(X) = P(Y) \times P(X)$ $\endgroup$ – Nadia Merquez Oct 18 '19 at 11:00
  • $\begingroup$ 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). $\endgroup$ – Jesper for President Oct 18 '19 at 11:05
  • $\begingroup$ 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? $\endgroup$ – Nadia Merquez Oct 18 '19 at 11:29
  • $\begingroup$ 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. $\endgroup$ – Jesper for President Oct 18 '19 at 11:40
  • $\begingroup$ Thanks for the insightful remarks! $\endgroup$ – Nadia Merquez Oct 18 '19 at 11:51

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