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Given the following probability mass function: $$ P_{xy}(x,y) = C\left(\frac{1}{2}\right)^x \cdot\left(\frac{1}{2}\right)^y $$ determine C.

$\textbf{Hint}$ : use the definition for a geometric series :

$$\sum_{n=0}^\infty r^n = \frac{1}{(1-r)}$$

edit: I'm not quite sure what to even do, I would normally integrate to find a constant? but I'm using geometric series now, and I don't really have any values for $x, y$ or $P_{xy}(x,y)$...

x and y can take any integer value equal or bigger then 0

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    You missed the most crucial part of a series i.e. the argument. do you know how to determine if a function is a probability dist. in terms of summing over all the support (x,y)? – 2017-02-14
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    I can't tell if the "teh" is a deliberate meme joke or just a typo. I like it either way. – 2017-02-14
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    What values can $x$ and $y$ take? – 2017-02-14
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    @The Count It's the Canadian coming out of meh, fixed it. – 2017-02-14
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    @Chinny84 I'm taking Calculus2 and Stats at the same time, we just started series this week, so I'm kind of behind the second years at my college that already have a grasp on series. I'm not quite sure what you are referring too. – 2017-02-14
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    @paul Integers equal to 0 or higher – 2017-02-14
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    @cicero866 I am Canadian now! So for a function to be a probability function you need the integral (or summation) over all possible values of $x,y$ to be $1$. This condition alone with the fact that you have a separable mass function we can compute the sums easily as they are geometric in nature (hence the hint). – 2017-02-14

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Because $\sum\limits_{ \{x,y\} = 0}^\infty \left( {1 \over 2} \right)^x \left( {1 \over 2} \right)^y = {1 \over 1 - 1/2}{1 \over 1 - 1/2} = 4$, the normalization constant must be $C = 1/4$.

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    if x and y = 0 then would that not mean that everything in the parentheses would then become 1? – 2017-02-14
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    @cicero866: Probability mass at ${x,y} = {0,0}$ is $C \cdot 1$. Since $C$ must be $1/4$ (for normalization), the mass at the origin is in fact $P(1,1) = 1/4$. – 2017-02-14