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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 $1$.

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    @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 $k$.2012-06-09

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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 $c=\frac{1}{e^3}=e^{-3}$.

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    heya, thanks for the answer :)2012-06-09