# How is $e^x$ related to its probability density function $\lambda e^{-\lambda x} (x>0)$?

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. (math.stackexchange.com/questions/244564/…) 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

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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