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I have been struggling with the following problem in probability:

Assume X is a random variable with the following probability density function:

$$ P(X = k) = \frac{A}{k!}, k=0,1,2,... $$

How to find the coefficient A ?

My first thought was to integrate this function to get its cumulative distribution function, and then make it equal to 1. However, I find out that integrating such a function is not so simple.

Is it possible to solve it in a simple way?

Thanks!

Pablo Mello
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  • I know that $$\mathrm{e}^z = \sum_{k=0}^\infty \frac{z^k}{k!}$$ can you see how we can apply this? – Chinny84 Apr 09 '17 at 15:06
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    @Chinny84 Perhaps in future you could provide a reference or a name to formulas for the OP to look up and use. E.g. its the Taylor Expansion of $e^z$. – Ian Miller Apr 09 '17 at 15:25
  • @IanMiller Yes, I should of provide a link or at least the name of the expansion - but it was quicker to write the sum. +1 – Chinny84 Apr 09 '17 at 15:27
  • @Chinny84 Oh that's right, this is the Taylor's expansion of $e^z$, which after setting z=1, is just what I am looking for! Nice! Thanks! – Pablo Mello Apr 09 '17 at 15:32
  • @Chinny84 No problem. I had a mental blank when I first saw it and couldn't put my finger on the name then 5 minutes late it hit me. – Ian Miller Apr 09 '17 at 15:35

1 Answers1

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This is a discrete distribution, so instead of integrating, try summing over $k$:

$$\sum_{k=0}^\infty P(X=k) = 1$$

Argon
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  • This isn't much of an answer. More of a comment. Perhaps you could show the OP how to do the sum. From their question they understand that summing (integrating) is required. – Ian Miller Apr 09 '17 at 15:19
  • I used Taylor's expansion of $e^z$ indicated by @Chinny84, but indeed it's better to think of it as a summation than an integral. Now I can solve it. Thanks for the answer! – Pablo Mello Apr 09 '17 at 15:38
  • @IanMiller I don't think that is true; a sum is not an integral, and if OP knew to sum, he would have indicated so ("integrating such a function is not so simple" indeed!). As per his comment here, this was the source of the confusion. – Argon Apr 10 '17 at 04:01