I've encountered a question: How is $e^x$ related to its probability density function $\lambda e^{-\lambda x} (x>0)$?
How is $e^x$ related to its probability density function \lambda e^{-\lambda x} (x>0)?
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probability
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0@DilipSarwate Inquest I'm asking that because i'd have to deal with $exp(x)$ and the pdf of an exponential function in this problem. (http://math.stackexchange.com/questions/244564/moment-generating-function-from-independent-exponential-lambda-to-gamman) Others have mentioned that I'm getting confusing the density of an exponential function. If you could shine some light on this question? really appreciated!! – 2012-11-26
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Let $\lambda$ be a positive real number. The random variable $X$ has exponential distribution with parameter $\lambda$ if the density function $f_X(x)$ is given by $f_X(x)=\lambda e^{-\lambda x}$ if $x\ge 0$, and $f_X(x)=0$ for $x\lt 0$.
It turns out that for such a random variable $X$, we have $\Pr(X\gt x)=e^{-\lambda x}$ if $x\ge 0$. So the probability that $X \gt x$ decays exponentially as $x$ increases. That is probably the motivation for calling the distribution exponential.
Sometimes, one says that $X$ has exponential distribution with parameter $\lambda$ by writing $X$ ~ $\text{Exp}(\lambda)$. One should not confuse this with the exponential function $\exp(x)$, which is just another name for $e^x$.
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0@user133466: You are right, $\lambda$ got left out. I am typo-prone. Easy to correct in an answer, but not in a comment. – 2012-11-26