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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?

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    $\begingroup$ Why do you think $(1)$ and $(2)$ give the right expressions for $f_X$ and $f_Y$? $\endgroup$ Apr 11, 2019 at 22:28
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    $\begingroup$ Why do you think $\varphi_{X+Y}=\varphi_X\varphi_Y$ is a problem? $\endgroup$ Apr 11, 2019 at 23:09
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    $\begingroup$ @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.) $\endgroup$ Apr 11, 2019 at 23:23

1 Answer 1

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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.

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