Differences in Two PDFs of the Exponential Distribution As always, I hope this email finds you well and warm today.
I am having an issue understanding the exponential distribution with rate parameter $\lambda$. I am seeing two different pdfs for this distribution; the one in R says $\lambda\exp(-\lambda x)$, while the one in my textbook says $\frac{1}{\lambda}\exp(\frac{-x}{\lambda})$.   I understand that the formula in R calculates $\lambda=\frac{1}{rate}$, but I am totally confused as to what I should use.   If $P \sim Exponential(4)$,  does this mean its pdf is $(4)\exp(-4x)$, or does it mean its pdf is $\frac{1}{4}\exp(\frac{-x}{4})$?  It gives different answers in R as well, as the second argument in pexp()is $rate=\frac{1}{\lambda}$.
My question in particular comes from a homework question:
Let $Z \sim Exponential(4)$. Compute $Pr(Z \geq 5)$.
Ans: $\frac{1}{e^{20}}$
Whereas, I did:
$=>\int_5^\infty \frac{1}{4} e^{\frac{x}{4}} dx$
$=>-1\int_5^\infty e^u du$
$=>e^{\frac{5}{4}}$
Or by using R:
> pexp(5, 1/4, lower.tail=F)
I feel like every week there's a new issue that I'm having with this textbook :P as always please help!
--
 A: The notation "$\text{Exponential}(b)$" on its own is ambiguous and will depend on how your source defines it. The way I usually prefer to communicate about the exponential distribution is "Suppose $X$ has an exponential distribution with mean ___," for example, to a more general audience. The geometric and negative binomial distributions have similar problems as well (i.e., trials or number of failures).
If your source says that $\text{Exponential}(\lambda)$ corresponds to a PDF of $f_{X}(x) = \dfrac{1}{\lambda}e^{-x/\lambda}$ for $x, \lambda > 0$, that should be the notation you use. Unfortunately, you will need to know how to translate between different notations.
Judging by the prior questions you've asked on MSE, you seem to be using Evans and Rosenthal's text.
You have $X \sim \text{Exponential}(4)$. In Example 2.4.5, they define an $\text{Exponential}(\lambda)$ distribution to have PDF of $\lambda e^{-\lambda x}$, so your PDF should be $4e^{-4x}$. This matches the form in R. So, you need to use rate = 4.
pexp(q = 5, rate = 4, lower.tail = FALSE)

[1] 2.061154e-09

