## Independence of the Range and Minimum of a Sample from an Exponential Distribution

A few years ago I answered a question on math.SE about the distribution of the sample range from an exponential (1) distribution.  In my answer I claim that the range and the minimum of the sample are independent, thanks to the memoryless property of the exponential distribution.  Here’s a direct derivation of the independence of the range and minimum, for those (including the graduate student who contacted me about this last month!) for whom the memoryless justification is not sufficient.

Let $Y_{(1)}, Y_{(2)}, \ldots, Y_{(n)}$ be the order statistics from a sample of size n from an exponential (1) distribution.  The joint distribution of $Y_{(1)}$ and $Y_{(n)}$  is given by

\begin{aligned} f_{Y_{(1)}, Y_{(2)}}(y,z) &= \frac{n!}{(n-2)!} e^{-y} e^{-z} \left( (1 - e^{-z}) - (1-e^{-y}) \right)^{n-2} \\ &= n(n-1) e^{-y} e^{-z} (e^{-y} - e^{-z})^{n-2}, \text{ for } 0 < y < z. \end{aligned}

Now, let’s do a double variable transformation.  Let $R = Y_{(n)} - Y_{(1)}$; i.e., R is the sample range.  Let $S = Y_{(1)}$, the minimum.  Then $Y_{(n)} = R + S$ and $Y_{(1)} = S$.  The Jacobian of the transformation is $\begin{vmatrix} 1 & 1 \\ 0 & 1 \end{vmatrix} = 1$.  Substituting, we have

\begin{aligned} f_{R,S}(r,s) &= n(n-1) e^{-s} e^{-r-s} (e^{-s} - e^{-r-s})^{n-2} \\ &= n(n-1) e^{-2s} e^{-r} (e^{-s})^{n-2} (1 - e^{-r})^{n-2} \\ &= (n-1) e^{-r} (1-e^{-r})^{n-2} n e^{-ns}, \text{ for } r, s > 0.\end{aligned}

Since $f_{R,S}(r,s)$ factors into a function of r times a function of sR and S are independent.  (We have $f_R(r) = (n-1) e^{-r} (1-e^{-r}), r > 0$; and $f_{Y_{(1)}}(y) = n e^{-ny}, y > 0$.)