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For a random variable $x$, define a probability distribution $p[x=n]=c (3^n/n!)$ when $x=0, 1, 2, \dots$ and $p(x)=0$ otherwise. Find the value of $c$.

My professor provided the solution $$ \sum_{x=0}^\infty \ c\frac{3^n}{n!}=1 $$ so $c\;e^3 = 1$.

I can't understand why the summation has value $1$.

Rick Decker
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Mr Flint
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    Please see here and here for how to format your mathematics expressions with LaTeX, and see here for how to use Markdown formatting. If you need to format more advanced math, there are many excellent LaTeX references on the internet, including StackExchange's own TeX.SE site. If you see a piece of LaTeX you want to know the code for on the site, you can right click on it, go to "Show Math As", then choose "TeX Commands". – Zev Chonoles Jun 09 '12 at 16:15
  • @Mr Flint : You should try to make sure that you do not mix up variables. There is a high probability that the two $x$ in you first sentence should either be uppercase and lowercase or $x$ and $k$. – Phira Jun 09 '12 at 16:31

1 Answers1

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I will change notation a little to make things look more familiar. For $n=0$, $1$, $2$, $3$, and so on, we have $$P(X=n)=c\frac{3^n}{n!}.$$

The probabilities must add to $1$, so $$\sum_{n=0}^\infty c\frac{3^n}{n!}=c\left(1+\frac{3}{1!}+\frac{3^2}{2!}+\frac{3^3}{3!}+\frac{3^4}{4!}+\cdots\right)=1.\tag{$1$}$$ Recall from calculus the power series (Taylor series, Maclaurin series) for the exponential function. $$e^t=\sum_{n=0}^\infty \frac{t^n}{n!}=1+\frac{t}{1!}+\frac{t^2}{2!}+\frac{t^3}{3!}+\frac{t^4}{4!}+\cdots.$$ So (taking $t=3$) we may rewrite $(1)$ as $$ce^3=1,$$ from which we conclude that $c=\frac{1}{e^3}=e^{-3}$.

André Nicolas
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