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For the following pdf I need to calculate the constant $c$.

$f(x, y)=c \frac{2^{x+y}}{x ! y !} \quad x=0,1,2, \ldots ; y=0,1,2, \ldots$

If I am not mistaken, in the case of a closed set for the discrete variables I would need to evaluate the outcomes of the pdf for all pairs $(0,0), (0,1),..., (2,2)$. Next, I should sum all these probabilities and equal the sum to 1. From there I can evaluate $c$.

The problem here is that the set of variables is not closed, which implies that there are infinite discrete sets. I could numerically approach this constant, since the pdf converges to 0 real quick, but I want an exact expression.

What is the best approach? Thanks!

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  • $\begingroup$ A pdf must be positive and also add up to $1$ for "all" the possible values that the random variables it describes takes. In your case, it seems that $x$ takes zero and all natural numbers, so does $y$; not the few pairs you have mentioned in your description. $\endgroup$
    – Math-fun
    Sep 8, 2022 at 13:24
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    $\begingroup$ @JRN you are right, thanks! OP knows that a sum over all values is needed. Your comment above helps OP, indeed. $\endgroup$
    – Math-fun
    Sep 8, 2022 at 13:32
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    $\begingroup$ @JRN Thanks for your help! I was not aware of this rule $\endgroup$
    – Tim
    Sep 8, 2022 at 13:34
  • $\begingroup$ @math-fun Thanks! $\endgroup$
    – Tim
    Sep 8, 2022 at 13:35
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    $\begingroup$ You'd probably be best calling this a pmf, although I've seen it called a pdf sometimes. $\endgroup$
    – J.G.
    Sep 8, 2022 at 13:49

2 Answers 2

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Hint: $\sum_{x=0}^\infty 2^x/(x!)=\text{e}^2$

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A Poisson random variable $K$ with parameter $\lambda$ has PMF defined by $$P(K=k)=\dfrac{e^{-\lambda} \lambda^k}{k!},\ k\in \{0,1,...\} $$.

Notice that your expression for $f(x,y)$ is proportional to the joint distribution of 2 independent Poisson random variables with paramenter $\lambda=2$: $$f(x,y)=c\left(\dfrac{2^x}{x!}\right)\left(\dfrac{2^y}{y!}\right)$$ Therefore, if $c=e^{-4}=e^{-2}e^{-2}$ you have: $$f(x,y)=\left(\dfrac{e^{-2} 2^x}{x!}\right)\left(\dfrac{e^{-2} 2^y}{y!}\right)$$ in the domain you defined. This is a valid PMF and therefore $c=e^{-4}$ is the answer.

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  • $\begingroup$ Check your assertion about the pdf of a Poisson random variable. -1 pending correction. $\endgroup$ Sep 8, 2022 at 16:39
  • $\begingroup$ @DilipSarwate... opps, thanks! $\endgroup$
    – bluemaster
    Sep 8, 2022 at 16:49

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