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Gamma density

Consider the distribution of the sum of two independent Exponential($\lambda$) random variables. I showed that it has a density of the form:

\begin{displaymath}f_S(s)=\int_0^s\lambda^2 e^{-\lambda s} dy= \lambda^2 s e^{-\lambda s}
\end{displaymath}

This density is called the $Gamma(2,\lambda)$ density. In general the gamma density is defined with 2 parameters $(t,\lambda)$ (both positive reals, most often t is actually integer) as being non zero on the positive reals and defined as:

\begin{displaymath}f(x)=\frac{\lambda e^{-\lambda x} (\lambda x)^{t-1}}{\Gamma(t)}\end{displaymath}

where $\Gamma(t)$ is the constant that makes the integral of the density sum to one:

\begin{displaymath}\Gamma(t)=\int^{+\infty}_0 e^{-y} y^{t-1} dy\end{displaymath}

By integration by parts we showed the important recurrence relation:

\begin{displaymath}\Gamma(t)=(t-1)\Gamma(t-1)\end{displaymath}

Because $\Gamma(1)=\int_0^{+\infty}e^{-y}dy=1$, we have for integer t=m

\begin{displaymath}\Gamma(m)=(m-1)(m-2)\cdots \Gamma(1)=(m-1)!\end{displaymath}

The particular case of the integer t can be compared to the sum of n independent exponentials, it is the waiting time to the nth event, it is the twin of the negative binomial.

From this we can guess what the expected value and the variance are going to be: If all the Xi's are independent $Exponential(\lambda)$, then if we sum n of them we have $E(\sum_{1}^n
X_i)=\frac{n}{\lambda}$ and if they are independent: $var(\sum_{1}^n
X_i)=\frac{n}{\lambda^2}$

This generalizes to the non integer t case:

\begin{displaymath}E(X)=\frac{1}{\Gamma(t)} \int_0^{+\infty}
\lambdax e^{-\lambd...
...da
x)^t=\frac{\Gamma(t+1)}{\lambda\Gamma(t)}=\frac{t}{\lambda}
\end{displaymath}


next up previous index
Next: Sum of two independent Up: Sums of Continuous Random Previous: Sums of Continuous Random
Susan Holmes
1998-12-07