# Exponential Distribution Theoretical Quantile

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.

• Are you familiar with the rule $n\ln x = \ln x^n$. What does it say when $n = -1$? – 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.