I am doing an exercise where I'm supposed to calculate the quartiles of the exponentially distributed function $f_\mathbb{X}(x)=\lambda e^{-\lambda x}$. So, first I calculate the distribution function, $F_\mathbb{X}(x)$, to be

$F_\mathbb{X}(x) = 1 - e^{-\lambda x}$

I know this is correct. Then, to calculate the first quartile $x_{0.25}$, I set

$F_\mathbb{X}(x_{0.25}) = 1-0.25$

I then perform the following calculations:

$F_\mathbb{X}(x_{0.25}) = 1-e^{-\lambda x} = 1-0.25$

$-e^{-\lambda x} = -0.25$

$e^{-\lambda x} = 0.25$

$-\lambda x = ln(0.25)$

$x = -{ln(0.25)\over\lambda}$

The same calculations are made for the other two quartiles, resulting in the following three quartiles:

$F_\mathbb{X}(x_{0.25}) = 1-0.25 => x = -{ln(0.25)\over\lambda}$

$F_\mathbb{X}(x_{0.5}) = 1-0.5 => x_{0.5} = -{ln(0.5)\over\lambda}$

$F_\mathbb{X}(x_{0.75}) = 1-0.75 => x_{0.75} = -{ln(0.75)\over\lambda}$

which, to me, seems pretty reasonable. The book, however, provides these answers:

$x_{0.25} = -{ln(0.75)\over\lambda}$

$x_{0.5} = -{ln(0.75)\over\lambda}$

$x_{0.75} = -{ln(0.25)\over\lambda}$

These answers does, on the other hand, not seem reasonable at all. Is the book wrong? And, if it's not, where am I wrong?

  • $\begingroup$ The book is right. $\endgroup$ Jul 20, 2013 at 14:43
  • $\begingroup$ Ok. Could you please explain where I went wrong? $\endgroup$ Jul 20, 2013 at 14:44
  • $\begingroup$ To put it briefly, for first quartile you want to set $F_X$ equal to $0.25$. $\endgroup$ Jul 20, 2013 at 15:05

1 Answer 1


The usual definition of the first quartile is the place $q_1$ such that $\Pr(X\le q_1)=0.25$.

In our case, $F_X(x)=1-e^{-\lambda x}$ and therefore we want $1-e^{-\lambda q_1}=0.25$.

This manipulates to $e^{-\lambda q_1}=1-0.25=0.75$. Taking logarihms, we get the book's answer of $-\frac{\ln (0.75)}{\lambda}$.

For the second quartile, $\ln(0.75)$ is replaced by $\ln(0.5)$, and for the third by $\ln(0.25)$.

Remark: Note that $-\ln(0.75)\lt -\ln(0.5)\lt -\ln(0.25)$. This feels as if it is going the wrong way. It isn't. For the logarithms are all negative.

I think the book's answer is (though correct) not presented in a good way. Better would be the equivalent $q_1=\ln(1/0.75)\cdot \frac{1}{\lambda}$. Then everything is positive. Shouldn't we all be positive?

  • $\begingroup$ I see what you mean, and that all makes sense. I'd just like to add that my book says that "The number $x_p$ that is a solution to $F_\mathbb{X}(x_p)=1-p$ is called the $p^{th} quantile$". Is this faulty then, for this gives that $Pr(X\le q_1) = 1-0.25 = 0.75$, or is this misinterpreted? $\endgroup$ Jul 20, 2013 at 15:24
  • $\begingroup$ There are some inconsistencies in usage, unfortunately. I was using the most standard definition, see for example this Wikipedia definition. $\endgroup$ Jul 20, 2013 at 15:53
  • $\begingroup$ I see. I'll take a look at this! Thanks a lot! $\endgroup$ Jul 20, 2013 at 16:04

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