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When we talk about how a random variable, say X, is exponential($\lambda$) or some other value what is that saying? Does this have to do with its distribution function and density function and the relationship between the two?

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It means that $X$ is exponentially distributed with parameter $\lambda$, hence has density $\chi_{[0,\infty)}\lambda\exp(-\lambda\cdot)$. –  martini Sep 26 '12 at 13:53
@martini-So if you had something like if X is exponential($\lambda$), then show the random variable Y=$\lambda$X is exponential(1) you would use the density function to help prove this? –  Sprock Sep 26 '12 at 14:25
You can also use the distribution function, which will be $\chi_{[0,\infty)} \bigl(1- \exp(-\lambda\cdot)\bigr)$. You have $F_Y(x) = P(Y \le x) = P(X \le \lambda^{-1}x) = F_X(\lambda^{-1}x) = \chi_{[0,\infty)}(\lambda^{-1}x)(1 - \exp(-x)) = \chi_{[0,\infty)}(x)(1-\exp(-x))$. Hence $Y$ is exponential(1). –  martini Sep 26 '12 at 14:33
But you also need to know there are two different parametrizations for the exponential distribution in use, tyhe one assumed in above comments and the one replacing $\lambda$ above with $\lambda^{-1}$. –  kjetil b halvorsen Sep 26 '12 at 19:10
Closely related –  Jacob Akkerboom Mar 10 at 14:17

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