A random variable is symmetric if and only if its characteristic function is real-valued

Question

Quick summary: I am stuck on the implication: $\phi_X$ real-valued $\rightarrow$ $X$ symmetric.

Assume you have a probability space $(\Omega, \mathcal{F},P)$, and a random varaiable $X: \Omega \rightarrow \mathbb{R^d}$, it is measurable with respect to $\mathcal{B}(\mathbb{R^d})$.

We define the characteristic function $\phi_X: \mathbb{R^d}\rightarrow\mathbb{C}$. With $\phi_X(u)=\int_\Omega e^{i<u,X(\omega)>}dP=\int_{\mathbb{R^d}}e^{i<u,x>}d\,u_x$, the measure $u_x$ is the distribution function:$u_x(B)=P(X^{-1}(B)), B \in \mathcal{B}(\mathbb{R^d}))$.

We say that X is symmetric iff $P(X\in B)=P(X\in-B)), B \in \mathcal{B}(\mathbb{R^d})$. Or equivalently $\mu_X(B)=\mu_X(-B)$.

I am supposed to show that X is symmetric if and only if the characteristic function of X: $\phi_X$ is real valued.

One part I seem to have managed: If we start with that X is symmetric, by then splitting $\mathbb{R^d}$ in all its hyperdimensional "octants" or "quadrants" and $\{0\}$, the result follows(but there is some work involved, but I think it is ok this way).

The problem is the other way: If I start with the knowledge that the characteristic function is real valued, how should I then work to show that X is symmetric? I tried assumeing that there is one set B such that $P(X \in B)\ne P(X \in -B)$, and then try to arrive at a contradiction:

If I split $\mathbb{R}^d$ in: $\{0\}, L_1=(+,+,+..,+), L_2=(-,-,-,,-), L_3=(+,+,+,..,-)L_4=(-,-,-,..,+)$ etc.

So that I get $\{0\}$ and $2^d$ $L_k$s. Where $L_{2i-1}=-L_{2i}$. Then I must have that $P(X \in B)\ne P(X \in -B)$, implies that there is atleast one i, such that $P(X \in B\cap L_{2i-1})\ne P(X \in -B\cap L_{2i})$ or $P(X \in B\cap L_{2i})\ne P(X \in -B\cap L_{2i-1})$, because if all these were equal, and since they are a partition of $\mathbb{R}^d$(with $\{0\}$) we would end up with: $P(X \in B)= P(X \in -B)$.

But this is as far as I got, beacuse even if I have that: $P(X \in B\cap L_{2i-1})\ne P(X \in -B\cap L_{2i})$ I can't see how this would contradict the characteristic function beeing real valued. My only idea is to chose one u, such that we get a complex number when we integrate, my only idea is choosing u for instance 1 in those parts that is a + in $L_{2_i-1}$, but I can't really see this leading anywere, since 0 in a exponential funciton, doesnt lead to a 0 in the function itself.

Any tips or hints or help on how to solve this?

Answer 1

Here is another approach wich might be more intuitive:

Recall that for any random vector $X$ defined on some probability space $(\Omega,\mathcal{F},\mu)$, that the characteristic function $\phi:(\mathbb{R}^d,\mathcal{B}(\mathbb{R})^d)\to(\mathbb{C},\sigma(\mathbb{C}))$ is defined to be $$ \phi_X(\xi)=\mathbb{E}\mathrm{e}^{i\xi^\top X} $$ for any $\xi\in\mathbb{R}^d$. But from complex analysis, we may equivalently write $\mathbb{E}\mathrm{e}^{i\xi^\top X}=\mathbb{E}\cos\xi^\top X+i\mathbb{E}\sin\xi^\top X$, from which the claim would follow.

Namely, as $\sin$ is an asymmetric functional, any probability measure $\nu$ inducing a random variable $X$ that is symmetric around zero trivially satisfies $\mathbb{E}_\nu\sin\xi^\top X=0$, hence a symmetric random variable has a real characteristic function for any $\xi\in\mathbb{R}^d$.

For the other direction, you would need to argue that if $\phi_X(\xi)\in\mathbb{R}$ for all $\xi\in\mathbb{R}^d$, then it must logically follow that $\mu=\mu_X$ is symmetric around zero. To do so, observe that if $\phi_X:\mathbb{R}^d\to\mathbb{R}$, then for any $\xi\in\mathbb{R}^d$, it must be that $\mathbb{E}\sin\xi^\top X=0$. In particular, $\mathbb{E}\sin\xi^\top X=\mathbb{E}\sin-\xi^\top X$. But as $\cos$ is symmetric, $\mathbb{E}\cos\xi^\top X=\mathbb{E}\cos-\xi^\top X$ and so $\phi_X(\xi)=\phi_X(-\xi)$ resulting in $\phi_{X}(\xi)=\phi_{-X}(\xi)$, see for example proposition 13.2 (iii) in these lecture notes. Consequently, as the characteristic function uniquely determines the probability distribution of $X$, we conclude that $\mu_X(A)=\mu_X(-A)$ for any $A\in\mathcal{B}(\mathbb{R}^d)$.

Indeed, depending on how one defines the characteristic function, you could generalise by defining $\varphi_X(\xi)=\mathbb{E}\mathrm{e}^{\xi^\top(X-\mathbb{E}X)}$, which would then have the conclusion that $\mu_X$ is symmetric around $\mathbb{E}X$ if and only if $\varphi_X(\xi)\in\mathbb{R}$ for all $\xi\in\mathbb{R}^d$.

For a good reference, although extremely old, I found the book probability and measure by Patrick Billingsley to be very insightful, but of course any modern textbook on measure theoretic probability will likely cover a similar case.