Connection between distribution and characteristic function.

characteristic-functionsprobability distributionsprobability theory

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*}

Best Answer

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 $t$).
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