Solved – Exponential distribution Q-Q plot homework question

Time to put this one to bed, I think

For an exponential qq plot, we fix the theoretical distribution to have λ=1.

Sure, because in a QQ plot we care about how far it is from exponential, not what the parameter value is. But in any case, you can see approximately what it is from the slope of the line the points should lie along in the QQ plot.

In order to compare the plotted points with the line y=x, what kind of rescaling should be applied to the data and why?

While you could rescale the data by dividing by their mean (yielding data with a mean of 1), I think that you should not do so.

If you do this with exponential data, the result will be data that no longer has an exponential distribution. In small samples the effect may be strong. Further, if the data aren't exponential, that adjustment may be badly impacted by large outliers.

It may do okay with very large samples, but there's really no need.

“Hint“ given with this problem: If X follows an exponential distribution with parameter λ, then λX follows an exponential distribution with parameter 1.

Yeah, as I thought — my first guess is they probably want you to estimate $lambda$ by $hat{lambda}=1/bar{x}$, and then scale to unit mean — but as I already explained, I think that's actually not a great idea, because it alters the very distribution you're checking for.

Better I think to plot the the data, and if you want a line to compare to, put some moderately robust estimate of the trend as a line on the graph … without putting too much stock in the position of the line (it's more a guide for the eye, to help you spot where the thing curves).

So far, I have considered the graph of ln (x) but not sure what to do with that and the hint.

Okay, that's another way to go. If you do it that way, consider $Zsim text{Exp}(1)$, and $X=Z/lambda$. Then $ln(X)=ln(Z)-lnlambda$, implying that a plot of quantiles of $ln(X)$ should be parallel to expected quantiles of $ln(Z)$, and the distance to the $y=x$ line tells you about $lambda$. But to be honest I don't think this is the intent of the question.