Characteristic function of triangular distribution over $[0,2]$.

859 Views Asked by At

Here it is shown that if $X_1,\dots ,X_n$ are iid random variables, then the characteristic function of $S=\sum_{i=1}^nX_i$ is the product of the respective characteristic functions of the $X_i$.

Now if I have a symmetric triangular distribution on $[0,2]$ (peak at $1$), its characteristic function should be the same as the one of $U_{[0,2]}+U_{[0,2]}$: $$\Big(\frac{e^{2it}-1}{it}\Big)^2=\frac{-e^{4it}+2e^{2it}-1}{4t^2}$$

But if I derive the CF by integration I get $$\frac{2e^{it}-e^{2it}-1}{t^2}$$ which—according to Wiki—seems to be the correct result. Where did I missapply the linked "rule"?