Unknown in discrete variable problem

[buddingscientist]

Let X be a random variable with probability function:
fx(x) = c/x!, x = 0, 1, 2, ...

Find c.

By first guess was to form the sum:
\sum_{i=0}^{x} c/i! = 1
But I have no idea if that's the right approach or how to proceed.

 

[Timbuqtu]

Probably you have to normalize the probability function, in other words the total probability should be 1:

\sum_{x=0}^{\infty} f(x) = 1

This is easy because:

\sum_{x=0}^{\infty} \frac{1}{x!} = e

 

[buddingscientist]

ahhh!
stupid me wasn't aware of that result, thanks heaps for that.




thanks, c = 1/e for anyone whos interested. I was able to complete the other problems relating to this question.

however I have one small problem, in my studies I've learned "a random variable X will be defined to be discrete if the range of X is countable" - introduction to theory of statistics (mood). but since the values of X was unbounded in the question (X = 0, 1, 2, ...) i.e: Z+ that is uncountable. ?

 

[honestrosewater]

however I have one small problem, in my studies I've learned "a random variable X will be defined to be discrete if the range of X is countable" - introduction to theory of statistics (mood). but since the values of X was unbounded in the question (X = 0, 1, 2, ...) i.e: Z+ that is uncountable. ?

Does countable mean finite or countably infinite? It almost surely means countably infinite. The nonnegative integers are easily seen to be countable:
{1, 2, 3, ...}
{0, 1, 2, ...}
I can't read your original question, so if you meant something else by "unbounded", sorry, but the nonnegative integers are bounded below by 0.