Self-study: Order statistics and distributions I'm currently working my way through _Probability for Statistics and Machine Learning by Anirban DasGupta_. I've come across the following question and after a bit of effort I could not prove it. Let $X_1,...,X_n$ be i.i.d observations from a continuous distribution $f$ with cdf $F$. Show that the density of $Y = \frac{X_{(1)} + X_{(n)}}{2}$ (mid-range order statistic) is given by $$ F_Y(y) = n \int_{-\infty}^{y} [F(2y-x) - F(x)]^{n-1} f(x) dx $$ and hence find the cdf for the mid-range order statistic for $f(x) = e^{-x}, x>0$ and $n=5$. **Progress:** I know I should be using the convolution formula to prove the first part. Even with the result, I'm having trouble with the second part as my integral has a $y$ term in it which diverges to $\infty$.

Consider: If the mid range order statistic is at most $y$, then one the $n$ samples will be the least order statistic whose value, call it $x$, lies somewhere below $y$, and the remaining $n-1$ samples will lie somewhere between that and at most $2y-x$.

Then the _cumulative distribution function_ of the mid-range order statistic is given by the integral:

$$F_Y(y)= \int_{-\infty}^y \bbox[gainsboro,0.1ex]{\color{gainsboro}{n f_X(x)~\big(F_X(2y-x)-F_X(x)\big)^{n-1}}}\operatorname d x$$

Now substitute $n=5\\\f_X(x)= e^{-x}\mathbf 1_{x\in[0;\infty)}\\\F_X(x)=(1-e^{-x})\mathbf 1_{x\in[0;\infty)}\\\F_X(2y-x) = (1-e^{-2y}e^x)\mathbf 1_{(2x-y)\in[0;\infty)}$

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Note: Always pay attention to the supports.