# Gamma Densities proof question

I understand the first portion of the problem but then how do I conclude that $f_n(x)$ gives us a probability density function on $[0,\infty)$.

If someone could point me in the right direction that would be a tremendous help!

Suppose that $t \geq 0$ and $n$ is a non-negative integer. For $x \geq 0$ define \begin{eqnarray*} f_n(x) & = & \frac{x^n}{n!}\exp(-x)\\ S_n(t) & = & \int_t^\infty f_n(x)\;dx \end{eqnarray*}

Conclude that \begin{eqnarray*}S_{n+1}(0) = S_n(0) \end{eqnarray*} for each non-negative integer $n$, and use this observation to conclude that $f_n(x)$ gives us a probability density function on $[0,\infty)$.

These densities are known as Gamma densities and $S_n$ is called a survival function. Conclude that\begin{eqnarray*} S_{n}(t) = \sum_{k=0}^{n} \frac{t^k}{k!}\exp(-t)\end{eqnarray*} Thus the survival function of a Gamma distribution is the distribution function of a Poisson distribution.

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Hint: What is $S_0(0)$? What are the conditions on an arbitrary $f$ such that it is a valid probability density function? Now, conclude. – cardinal Jan 29 '12 at 1:03
1 - and a continuous distribution would integrate up to 1 – Taylor Zwick Jan 29 '12 at 1:07
Good. (You also need nonnegativity, of course, but this should be easy to see for your functions of interest.) – cardinal Jan 29 '12 at 1:08
Ok I understand what you mean. Thanks!! – Taylor Zwick Jan 29 '12 at 1:28