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I've encountered a question: How is $e^x$ related to its probability density function $\lambda e^{-\lambda x} (x>0)$?

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    What do you mean by "related"? Are you asking this question because the PDF is called "Exponential Distribution"?2012-11-26
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    Do you mean what is $E[e^X]$ when $X$ is an exponential random variable with parameter $\lambda$?2012-11-26
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    @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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    so Exp($\lambda$) = $\lambda e^{-\lambda x}$ if $x\ge 0$ where as exp$(\lambda)$ = $e^\lambda$?2012-11-26
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    The second is fine, the first definitely not. It is best to think of the terms as entirely unrelated. We had $X$~$\text{Exp}(\lambda)$. No equality sign. Here ~ is an abbreviation for "has distribution." And $\text{Exp}(\lambda)$ just means exponential distribution with parameter $\lambda$, just like $\text{Binom}(n,p)$ means binomial distribution with parameter $n$ and $p$.2012-11-26
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    so $X \sim \text{Exp}(\lambda)$ = $\lambda e^{-\lambda x}$ if $x\ge0$ like this?2012-11-26
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    No. If $X$ ~ $\text{Exp(\lambda)}$, that means that the **probability density function** of the random variable $X$ is $e^{-\lambda x}$ (when $x\gt0$). $\text{Exp}(\lambda)$ is **not** a function. It is just a way of identifying (naming) a distribution. There are similar abbreviations for normal, uniform, Gamma, many others.2012-11-26
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    i got it, i believe you meant to write "that the **probability density function** of the random variable $X$ is $\lambda e^{−λx}$ [instead of $ e^{−λx}$] when $x>0$" correct?2012-11-26
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    @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