1
$\begingroup$

Problem: A random variable $X$ has a gamma distribution and is called a gamma random variable, if its probability density function is given by $f(x) = \begin{cases}\frac 1{b^\alpha\cdot \Gamma(\alpha)} x^{\alpha-1}\cdot e^{-x/b}&\mbox{ for }x>0,\\\ 0&\mbox{ otherwise}, \end{cases}$ where $\alpha > 0$, and $b > 0$.

($\Gamma$ is the gamma function defined by $\Gamma(\alpha) = \int_0^{+\infty}y^{\alpha-1} \cdot e^{-y}dy$ for $\alpha > 0$). When $x$ is a positive integer $\Gamma(\alpha) = (\alpha-1)!$ .

Show that the exponential distribution is the gamma distribution with $\alpha = 1$ and $b = 0$.

Answer(not sure if correct): Exponential distribution is given by $f(x) = \frac 1{\theta}e^{-x/\theta}$ for $ x > 0$ and $0$ otherwise. If we integrate this function from lets say $0$ to $+\infty$, we get the integral $\frac 1{\theta^2}\cdot e^{-x/\theta}$. This looks nothing like the gamma distribution! Does anyone have any information on this particular kind of distribution? Thanks in advance

  • 0
    Thanks!! Thanks for the fixing the text, and thanks for the hel$p$... Awesome!2012-05-01

0 Answers 0