# Does $E[e^{it(aX + bY)}]=E[e^{itaX}]E[e^{itbY}]$ for every $a,b\in\mathbb{R}$ imply that $X$ and $Y$ are independent?

Let $X, Y$ be two random variables such that for every $\alpha, \beta \in \mathbb{R}$, $$E[e^{it(\alpha X + \beta Y)}]=E[e^{it\alpha X}]E[e^{it\beta Y}]$$ for all $t\in\mathbb{R}$. Does it follow that $X$ and $Y$ are independent?

The answer is YES. Let me introduce some notations first.

Suppose $X=(X_1,\cdots,X_n)$ is an $\mathbb{R}^n$-valued random variable (i.e., a random vector). The characteristic function for $X$, denotes as $\varphi_X(u)$, is a function from $\mathbb{R}^n$ to $\mathbb{R}$: $$\varphi_X(u):=E(e^{iu\cdot X}),\quad u\in\mathbb{R}^n$$ where $u\cdot X$ is the inner product in Euclidean space. In particular, when $n=1$, we have the characteristic function for random variables. Note also that $$\varphi_X(u)=\varphi_{u\cdot X}(1)\tag{1}$$

The answer to your question follows directly from the following theorem (Probability Essentials by Jacod and Protter, Chapter 14):

To see how your question fits in this setting, let $Z=(X,Y)$ (these $X$ and $Y$ are random variables in OP) and $t=1$. Let $u=(\alpha,\beta)\in\mathbb{R}^2$. Then $$E(e^{iZ\cdot u})=E(e^{i\alpha X})E(e^{i\beta Y}).$$

If $$\color{blue}{\forall (a,b)\in\mathbb{R}^2\ \forall t\in\mathbb{R}}\ \quad \varphi_{(a,b)\cdot (X,Y)}(t)=\varphi_{aX}(t)\varphi_{bY}(t)\tag{2}$$ then do we have that $X$ and $Y$ are independent?
Note carefully that the condition $(2)$ implies in particular that $$\color{blue}{\forall (a,b)\in\mathbb{R}^2} \quad \varphi_{(a,b)\cdot (X,Y)}(1)=\varphi_{aX}(1)\varphi_{bY}(1)\tag{3}$$ which, by $(1)$, is equivalent to say that $$\color{blue}{\forall (a,b)\in\mathbb{R}^2}\quad \varphi_{(X,Y)}(a,b)=\varphi_X(a)\varphi_Y(b).\tag{4}$$ In order to apply the theorem, note that $a,b$ here play the role of those $u_i$'s.
Note also that in $(2)$, $\varphi_{(a,b)\cdot(X,Y)}(t)$ is the characteristic function of the random variable $aX+bY$ when $a,b$ are fixed (and $t$ is the variable of the characteristic function). While in $(4)$, $\varphi_{(X,Y)}(a,b)$ is the characteristic function of the random vector $(X,Y)$ evaluated at $(a,b)$. $(1)$ tells you the relationship between these two objects.
• $\alpha X + \beta Y$ is just one random variable, not a random vector. – Carlos Mendoza May 27 '16 at 16:26
• @CarlosMendoza : $E[e^{i a X + i b Y}] = E[e^{iaX}]E[e^{ibY}]$ for every reals $a,b$ implies (by inverse Fourier transform of the characteristic function) that $f_{(X,Y)}(x,y) = f_X(x)f_Y(y)$ where $f_{(X,Y)}$ is the pdf of the random vector $(X,Y)$ (and when the pdf is not a function, you'll need some work but $f_{(X,Y)}(x,y) = f_X(x)f_Y(y)$ will still be true in the sense of distributions). finally this is exactly what means "$X,Y$ are independent" – reuns May 27 '16 at 17:36
• @CarlosMendoza : I'm explaining to you why $E[e^{iaX+ibY}] = E[e^{iaX}]E[e^{ibY}]$ for every reals $a,b$ implies $X,Y$ are independent. finally set $t=1$ in the OP question and voila. – reuns May 27 '16 at 19:13
• @CarlosMendoza: Note that if you are told $\phi_{X+Y}(t)=\phi_X(t)\phi_Y(t)$, this does not mean $X,Y$ are independent. The condition in OPs question is considerably stronger. Check this out: math.stackexchange.com/questions/376511/… A trivial example would be letting $X=Y$ where $X$ is Cauchy (the means add) – Alex R. May 27 '16 at 19:44
• @CarlosMendoza: The logic is as the following. Let $A$ be the condition in the question of OP, where $t$, in your language, is the input variable. Let $B$ be the condition of the theorem, namely, the particular case $t=1$. And let $C$ be the conclusion that $X$ and $Y$ are independent. By the theorem, $B$ implies $C$. Now $A$ implies $B$. (I'm not say that $A$ is equivalent to $B$.) Therefore, we conclude that $A$ implies $C$. – Jack May 27 '16 at 20:22