I got a confusion regarding the following distribution conversion. From wikipedia I got that If $Xi$~$Laplace$($0,1/b$)
Then $|Xi|$~$exp$ with mean $b$
so 2*$sum_{i=1}^n |X_i|$/b~ Chisquare $(2n)$
Now again from wikipedia I got If $Xi$~$Laplace$($a,1/b$) then $2bsum_{i=1}^n |Xi-a|$~ Chisquare (2n) Then according to this the above turns out to be $2bsum_{i=1}^n |Xi|$~ Chisquare (2n) which is contradictory,can anyone please help me out to reach a firm conclusion about it
I understand: the results are both correct; the difference you see is due to a different parametrization of the Laplace.
In the first statement they consider the following density:
$$f_X(x)=\frac{1}{2b}e^{-\frac{|x|}{b}}$$
which transformed get $Y=|X|\sim Exp(\frac{1}{b})$ (neg exp with mean $b$, as they state). In this situation (see proof of the second statement) the result is correct:
$$\frac{2}{b}\Sigma_i Y_i \sim Gamma\Big(n;\frac{1}{2}\Big)=\chi_{(2n)}^2$$
In fact, using the same parametrization for the Laplace, you get, for the second statement:
$$f_X(x)=\frac{1}{2b}e^{-\frac{|x-a|}{b}}$$
$$Y=|X-a|$$
with standard transformation you get
$$F_Y(y)=F_X(a+y)-F_X(a-y)=1-\frac{1}{2}e^{-\frac{a+y-a}{b}}-\frac{1}{2}e^{\frac{a-y-a}{b}}=1-e^{-\frac{y}{b}}$$
Thus
$$Y\sim Exp\Big(\frac{1}{b}\Big)=Gamma\Big(1;\frac{1}{b}\Big)$$
As known (easy to prove with MGF and independence) you get $\Sigma_i Y_i \sim Gamma\Big(n;\frac{1}{b}\Big)$
and concluding (very easy to be proved with a simple monotonic transformation)
$$\frac{2}{b}\Sigma_i Y_i \sim Gamma\Big(n;\frac{1}{2}\Big)=\chi_{(2n)}^2$$
Now the two statement are coherent
I understand that they are ambiguous because saying Laplace(0;1/b) I mean the following density:
$$f_X(x)=\frac{b}{2}e^{-b|x|}$$
With this parametrization (that is the one they used in the second statement and correctly, in my opinion), the correct statement is
$$2b\Sigma_i Y_i \sim Gamma\Big(n;\frac{1}{2}\Big)=\chi_{(2n)}^2$$
Concluding: Case by case, do yourself all the calculations, expliciting well the parametrization you use and get your own result. It is a good exercise, not difficult but very useful