How to prove the test statistic of the Wilcoxon signed rank test is symmetric about mean How to prove the test statistic of the Wilcoxon signed rank test is symmetric about its mean?
I know that if I want to prove a distribution is symmetric, I need to show
$f(m-t)= f(m+t)$ but $W$ doesn't have a "easy" pmf so I don't know how to do it.
I tried a small number of $n$, and I can easily see that (by calculating all the possibilities) it is symmetric about its mean.
 A: Recall how this test works.  The data consist of $n$ ordered pairs $(x_i,y_i),i=1,2,\ldots,n$.  A procedure--the details will not matter--is followed to assign an integer $R_i$ to each pair according to the size of its difference, $|x_i-y_i|$. The test statistic $W$ is the sum of the $R_i$ corresponding to indexes $i$ with a positive difference minus the sum of all $R_i$ corresponding to $i$ with a negative difference.
The null hypothesis is that these differences are realizations of independent variables having a common symmetric distribution about zero.  That's our point of departure, so let's be clear what it means.  Letting $(X,Y)$ be a random variable of which each $(x_i,y_i)$ is an independent realization, "symmetric" means the distributions of $X-Y$ and $-(X-Y) = Y-X$ are the same.
The conclusion is now immediate: the symmetry under the null implies we could equally well compute $W$ by reversing each ordered pair; that is, by replacing the data $(x_i,y_i)$ with $(y_i,x_i)$.  This changes all the signs of the $x_i-y_i$, but leaves their sizes the same (whence the $R_i$ do not change), and therefore it merely negates $W$.  Consequently $W$ and its $-W$ have the same distribution, QED.
