Let X and Y two independent random variables with exponential distribution of parameter a>0. Proof using characteristic functions, that U = X+ Y and V = X- Y are not independent.
1) I calculate the characteristic function of X and Y (is the same)
$\varphi_X(t)$ = $\varphi_Y(t)$=$\left( 1- \frac{it}{a} \right)^\left( -1 \right)$
2) I know that $\varphi_\left( X+Y \right) (t)=\varphi_X(t).\varphi_Y(t)$, if X and Y are independent, that is:
$\varphi_U(t)$=$\varphi_\left( X+Y \right) (t)=\left( 1- \frac{it}{a} \right)^\left( -2 \right)$
3) I know that $\varphi_\left(-Y \right) (t)=\bar\varphi_Y(t))$, where $\bar\varphi_Y(t))$, is the complex conjugate of $\varphi_Y(t))$
4) I calculate $\varphi_V(t)$=$\varphi_\left( X-Y \right) (t)=\varphi_X(t).\bar\varphi_Y(t)$
5) How can I proof that U and V are not independent?
You show that they are not independent by showing that $$ \varphi_{(X+Y) + (X-Y)}(t) \neq \varphi_{(X+Y)}(t) \varphi_{(X-Y)}(t) $$ To see this, assume $a>0$ and note that $$ \varphi_{(X+Y) + (X-Y)}(t) = \varphi_{2X}(t) = \int_{0}^\infty e^{itx} f_{2a}(x)\,dx = \frac{2a}{2a-it} $$ and the characteristic functions of $X\pm Y$ are $$\varphi_{(X+Y)}(t) = \int_{x=0}^\infty \int_{y=0}^\infty e^{it(x+y)} f_a(y) f_a(x)\,dy\,dx = \frac{a^2}{(a-it)^2} \\ \varphi_{(X-Y)}(t) = \int_{x=0}^\infty \int_{y=0}^\infty e^{it(x-y)} f_a(y) f_a(x)\,dy\,dx = \frac{a^2}{a^2+t^2} $$ whence $$ \varphi_{(X+Y)}(t) \varphi_{(X-Y)}(t) = \frac{a^4}{(a-it)^3(a+it)} \neq \frac{2a}{2a-it} = \varphi_{(X+Y) + (X-Y)}(t) $$ Contrast this with what you get doing the same steps for Gaussian-distributed variables sharing the same mean and $\sigma$, where that last equality does turn out to be true.