# How do you evaluate a constant in a pdf?

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!

• 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. Sep 8, 2022 at 13:24
• @JRN you are right, thanks! OP knows that a sum over all values is needed. Your comment above helps OP, indeed. Sep 8, 2022 at 13:32
• @JRN Thanks for your help! I was not aware of this rule
– Tim
Sep 8, 2022 at 13:34
• @math-fun Thanks!
– Tim
Sep 8, 2022 at 13:35
• You'd probably be best calling this a pmf, although I've seen it called a pdf sometimes.
– J.G.
Sep 8, 2022 at 13:49

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