# $\varphi_{X+Y}(t)=\varphi_X(t) \cdot \varphi_Y(t)$, 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?

• Why do you think $(1)$ and $(2)$ give the right expressions for $f_X$ and $f_Y$? Apr 11, 2019 at 22:28
• Why do you think $\varphi_{X+Y}=\varphi_X\varphi_Y$ is a problem? Apr 11, 2019 at 23:09
• @RhysSteele (1) and (2) are correct. For (1), when you integrate out over $y$, the $y^1$ and $y^3$ terms are odd functions of $y$ and hence vanish, etc. Maria, (6) is wrong: it should be the square of what you state. (It contradicts the problem title and the high-lit sentence, too.) Apr 11, 2019 at 23:23

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.