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The following joint probability distribution exists for two dependent random variables

           1    3    6   9
       2 0.11 0.05 0.20 0.08
    X  3 0.20 0.02 0.00 0.10
       7 0.00 0.05 0.10 0.09

How do I find $\mathbf{P}(Z = XY)$?

I'm not sure if I'm supposed to sum across all rows/columns, then find the products, or just take them as it is. For example:

Is $\mathbf{P}(Z = 63) = 0.09$, or $\mathbf{P}(X=7) \cdot \mathbf{P}(Y=9)$, that is $0.24 \cdot 0.27 = 0.065$?

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Welcome to MSE! If you plan on further submission, this may come in handy to make them look good:… – gnometorule Feb 13 '13 at 7:11

$P(Z=XY)$ is just a look-up. No multiplication involved. You will have to deal with multiple $X,Y$ pairs which give the same $Z$ in which case, the probabilities are additive.


A good way of testing whether ANYTHING is a valid PDF is to see if it integrates/sums up to 1.

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That's not quite true; there's some computation in adding the probabilities for different factorizations of the same product; it just so happens that there's only one such case here and one of the probabilities for that case is zero. – joriki Feb 13 '13 at 7:21
@joriki. True, what I tried to say is that there is no multiplication involved for finding out the probability of a given pair. I'll edit to remove ambiguity. – Inquest Feb 13 '13 at 7:22

The latter in your first formulation, i.e. the former in your second formulation. You need to take into account that one value of the product can be formed in two different ways, though.

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