# Let X and Y two independent random variables with exponential distribution of parameter a>0. U = X+ Y and V = X- Y are not independent

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 need to work out a formula for the joint characteristic function of $(U,V)$, that is, $\phi(s,t)=E\exp( sU +tV)$. Apr 2, 2019 at 18:16
• I never work with a join characteristic function. The teacher has not taught us how to do it, has not done exercises about it and it is not in the given notes. Is there another way to do it? Apr 2, 2019 at 18:35
• I never work with a join characteristic function with characteristic functions. Apr 2, 2019 at 18:37

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.
• It's $2a/(2a-it)$. Note that $$\left( 1-\frac{it}{a} \right)^{-1} = \left(\frac{a-it}{a}\right)^{-1} = \frac{a}{a-it}$$ and the $-it$ in the denominator of $\varphi_{2X}$ comes about in the same way, replacing $a$ by $2X$. Apr 4, 2019 at 6:48
• I do not think that $\phi_{2X}(t)=\frac{2a}{2a-it}$. If $X$ is exponential with parameter $a$ and mean $1/a$, then $cX$ is exponential with parameter $a/c$ and mean $c/a$, provided $c>0$. The CF of $X$ is $\frac{a}{a-it}$ and the CF of $cX$ is $\frac{a/c}{a/c-it}=\frac{a}{a-ict}$. This is also seen by $\phi_{cX}(t)=\phi_X(ct)$. Can you correct me or verify this? Apr 7, 2019 at 18:04