Connection between distribution and characteristic function.

Question

Let $X$ be a $\mathbb{R}^d$-valued random variable. Denote by $\mathbb{P}_{X}$ its distribution and $\phi_{X}$ its characteristic function. I want to prove that the following are equivalent

1. $\mathbb{P}_{-X} = \mathbb{P}_{X}$
2. $\phi_{X} = \phi_{-X}$
3. $\phi_{X}(t) = \mathbb{E}[\cos(\langle t, X \rangle]$ for all $t \in \mathbb{R}$
4. $\phi_X \in \mathbb{R}$.

I am having trouble with finding a connection between $\phi$ and $\mathbb{P}$ and therefore can't prove any of the directions $(x) \implies (1)$, where $x \in \{2,3,4\}$.

I have proven the equivalence of the last three statements like this and would appreciate any feedback:

(3) $\implies$ (4). By definition we have \begin{equation*} \phi_{X}(t) \overset{\textrm{Def.}}{=} \mathbb{E}[\exp(i \langle t ,X \rangle)] \overset{\textrm{L}}{=} \mathbb{E}[\cos(\langle t ,X \rangle)] + i \mathbb{E}[\sin(\langle t ,X \rangle)] \overset{!}{=} \mathbb{E}[\cos(\langle t, X \rangle)]. \end{equation*} Therefore we have $i \mathbb{E}[\sin(\langle t ,X \rangle)] = 0$.

(4) $\implies$ (2). For all $a,b \in \mathbb{R}$ we have \begin{equation*} \phi_{a X + b}(t) = e^{i t b} \phi_{X}(at) \qquad \text{and} \qquad \phi_X(-t) = \overline{\phi_X(t)}. \end{equation*} This implies \begin{equation*} \phi_{-X}(t) = \phi_{X}(-t) = \overline{\phi_{X}(t)} \end{equation*}

(2) $\implies$ (3): By definition we shall have \begin{equation*} \mathbb{E}[\cos(\langle t, X \rangle)] + i \mathbb{E}[\sin(\langle t, X \rangle)] \mathbb{E}[\sin(\langle t, -X \rangle)], \end{equation*} which is equivalent to \begin{equation*} \mathbb{E}[\cos(\langle t, X \rangle)] + i \mathbb{E}[\sin(\langle t, X \rangle)] = \mathbb{E}[\cos(\langle t, X \rangle)] - i \mathbb{E}[\sin(\langle t, X \>)]. \end{equation*} This implies \begin{equation*} i \mathbb{E}[\sin(\langle t, X \rangle)] = - i \mathbb{E}[\sin(\langle t, X \rangle)] \implies i \mathbb{E}[\sin(\langle t, X \rangle)] = 0 \end{equation*}

Answer 1

This is a comment, transformed into an answer:

The equivalence (1) <-> (2) is one of the basic and important properties of the characteristic function, that two random variables (vectors) $X$ and $Y$ have the same probability distribution if and only if their characteristic functions are the same. The proof can be found in every basic probability book. Then plug in $X=X$ and $Y=-X$. Otherwise, your proofs look correct to me, aside from some cosmetic problems.

Regarding the cosmetic problems:

1. In your property 3.) you state "for all $t\in \mathbb{R}$". It should rather be $t\in \mathbb{R}^d$ to make sense of the product $\langle t,X\rangle$.
2. In your implication 2.) $\Rightarrow$ 3.) your first statement $\begin{equation*} \mathbb{E}[\cos(\langle t, X \rangle)] + i \mathbb{E}[\sin(\langle t, X \rangle)] \mathbb{E}[\sin(\langle t, -X \rangle)], \end{equation*}$ is not really meaningful, because it is first of all not a logical statement.
3. The equality sign $\stackrel{L}{=}$ in your implication 3.) to 4.) is not really meaningful in that context (assuming it means equality in distribution. Otherwise, you should define it, please), because left and right-hand side of the equation are non-random real numbers (for fixed $t$).

Answer 2

In general, the reason you are having trouble deducing (1) from any of (2), (3), or (4) is that it requires a key fact about characteristics functions whose proof is not so trivial: you need to be able to "go backwards" from a characteristics function to a random variable, and there is no simple inversion formula that applies in the general case of random variables that do not possess a density. Assuming this property as a given, there is a simple argument as follows:

Proof that (2) implies (1), assuming the uniqueness theorem:

Translating into standard probabilistic notations, condition (1) is that $X$ and $-X$ have the same distribution, whereas condition (2) is that $X$ and $-X$ have the same characteristic function. Thus, the implication is a direct consequence of the uniqueness theorem for characteristic function, which states that two random variables $X$ and $Y$ have the same characteristic function if and only if they have the same distribution.

---

Now a few remarks about your proofs, as you requested.

(3) implies (4).

This follows directly from the observation that $\mathbb E\cos(\langle t,X\rangle)\in\mathbb R$ since the expectation of a real-valued random variable is necessarily real-valued. (Apply this observation to the random variable $\cos(\langle t,X\rangle)$.)

(4) implies (2).

It is unnecessary to introduce $a,b$ as you did. Instead, simply compute $$ \overline{\mathbb Ee^{i\langle t,X\rangle}}=\mathbb Ee^{-i\langle t,X\rangle}=\mathbb Ee^{i\langle t,-X\rangle}, $$ and thus the condition $\mathbb Ee^{i\langle t,X\rangle}\in\mathbb R$ implies that $$ \mathbb Ee^{i\langle t,X\rangle}=\overline{\mathbb Ee^{i\langle t,X\rangle}}=\mathbb Ee^{i\langle t,-X\rangle}, $$ which yields (2).

(2) implies (3).

By hypothesis, $\mathbb Ee^{i\langle t,X\rangle}=\mathbb Ee^{i\langle t,-X\rangle}$. Averaging the two quantities (that are equal) yields $$ \mathbb Ee^{i\langle t,X\rangle}=\frac{\mathbb Ee^{i\langle t,X\rangle}+\mathbb Ee^{i\langle t,-X\rangle}}{2}=\mathbb E\left[\frac{e^{i\langle t,X\rangle}+e^{-i\langle t,X\rangle}}{2}\right]=\mathbb E\cos(\langle t,X\rangle). $$

---

How to prove the uniqueness theorem.

This result is typically proven using more advanced tools from analysis, specifically the Stone-Weierstrass approximation theorem (or equivalent results). The idea is to show that, for all measurable sets $A\subseteq \mathbb R^n$, the quantity $$ \mathbb P(X\in A)=\int_{\mathbb R^d}1_{x\in A}\ d\mathbb P(x) $$ can be approximated by quantities of the form $$ \mathbb Ee^{i\langle t,X\rangle}=\int_{\mathbb R^d}e^{i\langle t,x\rangle}\ d\mathbb P(x). $$ In fact, it amounts to showing that the characteristic function $1_{x\in A}$ is well-approximated by linear combinations of the functions $e^{i\langle t,x\rangle}$. First, one can start by approximating $A$ by a sequence of bounded subsets $A_n=A\cap B(0,n)$. Thus it suffices to show the approximation when $A$ is bounded, in which case it can be done by observing that $1_{x\in A}$ is an element of $L^2(\mathbb P)$ and the functions $e^{i\langle t,x\rangle}$ are an orthonormal basis for $L^2(\mathbb P)$. (The latter state can be shown using the Stone-Weierstrass theorem...) Anyway, my point is that this result is of a different caliber than the other equivalences above, and your understanding of the proof depends on your familiarity with classical analytical results.