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p(x, y) = c(x + y), x = 1, 2, 3, y = 1, . . . , x

I tried solving it by integrating it over y and them summing it up over x. But I have a doubt can I integrate the function since it is a probability mass function. Please help.

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  • $\begingroup$ In this case summation should be used instead of integration since $x, y \in \mathbb{N}_+$. $\endgroup$ – Saad Jan 25 '18 at 2:22
  • $\begingroup$ i tried summation but how can i sum it from 1,.....,x $\endgroup$ – Akshay Patil Jan 25 '18 at 2:26
  • $\begingroup$ If $x\in \mathbb N$ and $y\in \mathbb R$ what you did seems right. $\endgroup$ – Macavity Jan 25 '18 at 2:32
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You need to compute $\sum_{x=1}^3\sum_{y=1}^x (x+y)$ and seem to be having trouble with the inner sum. It splits up into two terms. The first is easy cause the $x$ just comes out: $$ \sum_{y=1}^x x = x\sum_{y=1}^x 1 = x^2.$$ For the other we have the formula $$ \sum_{y=1}^x y = \frac{x(x+1)}{2}.$$

Then you need to sum the resulting expression from $x=1$ to $3,$ which can be done by either looking up / figuring out the formulae for $\sum_{i=1}^n i^2$ or more simply, by hand since it's only three terms.

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Note that $$\sum_{y=1}^x c(x+y) = cx^2 + c\frac{x(x+1)}{2}~.$$ Now, sum this function of $x$ from $1$ to $3$ and equate the resulting expression to $1$ to get $c$.

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Evaluate $c$ when the total probability mass equals 1.$$1{~=~\sum_{x=1}^3\sum_{y=1}^x c(x+y)\\~=~ c\sum_{x=1}^3(x^2+\sum_{y=1}^x y)\\~=~c\sum_{x=1}^3(x^2+x(x+1)/2)\\~=~ \frac c2\sum_{x=1}^3(3x^2+x)\\ ~~\vdots}$$

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