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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

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I fixed tex on your post. Check I didn't modify anything. – Davide Giraudo May 1 '12 at 16:28
The exponential distribution is (with your notation) the Gamma distribution with $\alpha=1$ and $b=\theta.$ This follows directly by substitution on the expression of $f(x);$ you don't have to integrate anything. It seems you have mistaken $\theta$ for $0.$ – Xabier Domínguez May 1 '12 at 16:34
Thanks!! Thanks for the fixing the text, and thanks for the help... Awesome! – jay May 1 '12 at 16:45

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