$\varphi_{X+Y}(t)=\varphi_X(t) \cdot \varphi_Y(t)$, but X and Y are not independent Consider $X,Y$ random variables with joint distribution:
$$f_{X,Y}(x,y)=\begin{cases}
\frac14\left[ 1+xy(x^2-y^2)\right] &  |x|\leq 1,\;|y|\leq 1  \\
0 & \text{otherwise}
\end{cases}$$

Proof $\varphi_{X+Y}(t)=\varphi_X(t) \cdot \varphi_Y(t)$, but $X$ and $Y$  are not independent.

Here $\varphi_V(t)$ denotes a characteristic function of random variable $V$.

My step-by-step calculation is shown below, but I think something is wrong.
1.$f_X(x)=\frac{1}{2}$ if $|x|\leq 1$
2.$f_Y(y)=\frac{1}{2}$ if $|y|\leq 1$
3.$f_{X,Y}(x,y)\neq f_X(x)f_Y(y)$ implies X and Y are not independent
4.$\varphi_X(t)=\frac{1}{2it}(e^{it}-e^{-it})$
5.$\varphi_Y(t)=\frac{1}{2it}(e^{it}-e^{-it})$
6.$\varphi_{X+Y}(t)=\frac{1}{2it}(e^{it}-e^{-it})$
Obviously I got $\varphi_{X+Y}(t) \neq \varphi_X(t) \cdot \varphi_Y(t)$. What went wrong?
 A: Maria's discovery can be expressed as follows. Let $U$ and $V$ be iid uniform on $[-1,1]$.  The distribution of $X$ is the same as that of $U$. the distribution of $Y$ is the same as that of $Y$, and the distribution of $X+Y$ is the same as that of $U+V$.  To see this note these symmetry properties of the joint density function $f(x,y)$ for $X$ and $Y$:
$$ \frac {f(x,y)+f(x,-y)} 2 = g(x,y),$$
$$ \frac {f(x,y)+f(-x,y)} 2 = g(x,y),$$
and
$$ \frac {f(x,y)+f(y,x)} 2 = g(x,y)$$
where $g$ is the joint density function  for $U$ and $V$.  Then, for example, $$P(X\le t)=\iint_{x\le t} f(x,y)\,dxdy = \iint_{x\le t} g(x,y)\,dxdy=P(U\le t)$$ because the set of $(x,y)$ for which $x\le t$ is the same as the set of $(x,-y)$ for which $x\le t$, and similarly for $y\le t$.  The set of $(x,y)$ such that $x+y\le t$ is the same as the set of $(y,x)$ such that $x+y\le t$.
In effect, if $\varphi$ is the joint characteristic function of $(X,Y)$ and $\psi$ that of $(U,V)$, Maria has discovered that $\varphi(t,0)=\psi(t,0)$, that $\varphi(0,t)=\psi(0,t)$, and that $\varphi(t,t)=\psi(t,t)$ for all real $t$.  This is, of course, far from showing that $\varphi(t,u)=\psi(t,u)$ for all $(t,u)$.   This is well known in medical circles.  For CAT scanners to work they have to, in effect, measure all the one dimensional margins of $\cos\theta X + \sin\theta Y$, once per projection angle $\theta\in[0,\pi)$.  Maria has found, in effect, a mathematical tumor that is invisible when CAT scanned only at the 3 particular $\theta$ angles $0,\pi,\pi/2.$
The "obviously something went wrong" statement is a mistake, however: nothing (other than the typo in equation 6) is wrong up to that point.  Just because the formula $\varphi_{X+Y}=\varphi_X\varphi_Y$ holds for all values of the arguments only when $X$ and $Y$ are independent does not mean  it cannot hold for some  values of the arguments  when $X$ and $Y$ are dependent.
