I'm stuck with an exercise that I'll just write down first:

Based on a sample $x_1,...,x_{47}$, that can be considered coming from an unknown distribution, we study a qq-plot where the empirical quantiles for the logarithmic sample are being compared to the corresponding quantile-values for the exponential distribution with parameter 1. The quantile-pairs approximately describes a straight line with intercept 0. Further we have $\frac{1}{47} \sum_{k=1}^{47} ln(x_k) \approx 0.4$. Estimate the 95%-quantile $F_X^{-1}(0.95)$ for the unknown distribution.

I've tried several ways, but somehow I seem to have missed something theoretical important. First I need to estimate the distribution function, but I can't use the empirical distribution-function since I don't know what $x_1,...,x_{47}$ are. Then I thought that it should be an exponential function because of the straight line in the qq-plot, but that doesn't add up with the solution-manual.

According to the solution-manual I should arrive at $F_{ln(X)}^{-1}(p)=-c \cdot ln(1-p)$, where $c$ is estimated to $0.4$.

It feels like an easy problem but I just can't figure it out right now, maybe I've stared at it too long... Anyway, I'm most grateful for all the help I can get. Thank you in advance!

  • $\begingroup$ Hint: for an Exponential distribution and $0\lt p \lt 1,$ what is the formula for quantile $p$? Use that to obtain a quantitative description of the "straight line with intercept 0" in terms of the quantiles of $\log(X).$ $\endgroup$ – whuber Apr 12 '18 at 16:57
  • $\begingroup$ Thank you for your answer! I have done that, but I get $F_{ln(X)}^{-1}(p)=-ln(-p)$, and that doesn't add up with the solution-manual. I don't understand why the "c" is in the quantile function, to start with... $\endgroup$ – AnnieFrannie Apr 13 '18 at 5:33
  • $\begingroup$ Try working this in reverse: given $F^{-1}_{\log(X)} = -c \log(-p),$ what does that imply about the expectation of $\log(X)$? Compare that to the value of $0.4$ observed in the data. $\endgroup$ – whuber Apr 13 '18 at 14:35

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