> 
> 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\*}