Question Given: For the exponential distribution with CDF:

$$F(y) = 1- e^{-\lambda y} $$

show that the (i/n+1)-th theoretical quantile is given by:

$$ F^{-1}\bigg(\frac{i}{n+1}\bigg) = \frac{1}{\lambda} \ln \bigg(\frac{n+1}{n+1-i}\bigg) $$

My attempt: I set $ \frac{i}{n+1} = x $, then solved for $y$ in the equation $ x = 1-e^{-\lambda y} $

and I arrived at the answer:

$$ F^{-1}\bigg(\frac{i}{n+1}\bigg) = -\frac{1}{\lambda} \ln \bigg(\frac{n+1-i}{n+1}\bigg) $$

which seems correct to me, but it doesn't match the answer given in the question. I don't understand how to get from my answer to the one given.

  • $\begingroup$ Are you familiar with the rule $n\ln x = \ln x^n$. What does it say when $n = -1$? $\endgroup$ – N. F. Taussig Apr 15 '15 at 18:48

In general, if $x > 0$ then $\ln x = -\ln \left(\frac1x\right)$. The quantity you have in parentheses after $\ln$ is just $1$ over the quantity that the "official" answer has, so your answer is in fact equal to that answer.

The answer in the question is in slightly simpler terms than yours (no negative sign required), which is one good reason why it would be stated that way.


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