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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"? – Inquest Nov 26 '12 at 0:05
Do you mean what is $E[e^X]$ when $X$ is an exponential random variable with parameter $\lambda$? – Dilip Sarwate Nov 26 '12 at 0:06
@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. (…) 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!! – user133466 Nov 26 '12 at 0:09
up vote 1 down vote accepted

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$(\lambda)$ = $e^\lambda$? – user133466 Nov 26 '12 at 0:32
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 $p$. – André Nicolas Nov 26 '12 at 0:39
so $X \sim \text{Exp}(\lambda)$ = $\lambda e^{-\lambda x}$ if $x\ge0$ like this? – user133466 Nov 26 '12 at 0:41
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. – André Nicolas Nov 26 '12 at 0:46
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? – user133466 Nov 26 '12 at 0:49

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