# $(M,T), (-M,T), (M,-T), (-M,-T)$ all are equal in distribution when the original distribution is symmetric

Suppose $X_1, X_2, \ldots , X_n$ are iid random variables coming from a symmetric distribution $F$. Let $S_n=X_1+X_2+\cdots+X_n$. Let $M$ be the first term among $X_1, X_2, \ldots , X_n$ which has highest absolute value. Let $T=S_n-M$. Then $(M,T), (-M,T), (M,-T), (-M,-T)$ all are equal in distribution.

I can show that $(M,T), (-M,-T)$ are equal in distribution.

Edit: $(X_1,\ldots,X_n)$ is equal in distribution with $(-X_1,\ldots,-X_n)$ as $F$ is symmetric. We note that $(M,T)(-X_1,\ldots,-X_n)$ (that is the $M,T$ computed from $-X_1,-X_2,\ldots,-X_n$ as observations) is same as $(-M,-T)(X_1,\ldots,X_n)$. So we have the above equal in distribution.

But I am finding difficulties in showing the rest. Any help/suggestion would be appreciated.

This is a fact that I found in Feller volume 2, page 150. He uses it without any justification.

• How did you show that $(M,T), (-M,-T)$ are equal in distribution ? Maybe it gives someone an idea to show the other cases. – callculus Apr 13 '17 at 6:09
• @callculus $(X_1,\ldots,X_n)$ is equal in distribution with $(-X_1,\ldots,-X_n)$. We note that $(M,T)(-X_1,\ldots,-X_n)$ is same as $(-M,-T)(X_1,\ldots,X_n)$. So it follows. – Sayan Apr 13 '17 at 8:01