Is the function $ (\cos t)^2$ the characteristic function of some distribution

Question

Is the function $(\cos t)^2$ the characteristic function of some distribution? Why or why not.

I already checked hat it is uniformly continuous and it values $1$ when $t = 0$.

Answer 1

Step 1: Show that $t \mapsto \cos(t)$ is a characteristic function. Hint: Consider a random variable $X$ which satisfies $$\mathbb{P}(X=c) = p \qquad \qquad \mathbb{P}(X=-c) = 1-p$$ for suitable suitable constants $p \in (0,1)$ and $c \in \mathbb{R}$.

Step 2: Show that if $t \mapsto \varphi(t)$ is a characteristic function, then $$t \mapsto \varphi(t)^2$$ is a characteristic function. Hint: Let $X$ be a random variable with characteristic function $\varphi$, and let $X'$ be an independent copy of $X$. Consider the characteristic function of $Y:=X+X'$.

Answer 2

Since $\cos^2 t=\frac{1}{4}(e^{-2it}+2+e^{2it})$, it's the characteristic function of the distribution satisfying $P(X=-2)=P(X=2)=\frac{1}{4},\,P(X=0)=\frac{1}{2}$.

Answer 3

Recall the property of characteristic functions that for $X \perp \!\!\! \perp Y$ we have $$\varphi_{X+Y}(t)=\varphi_X(t)\varphi_Y(t).$$

A discrete random variable $X_1$ where $P(X_1=1)=P(X_1=-1)=\frac{1}{2}$ has $$\varphi_{X_1}(t)=\cos(t).$$

Let $X_2$ have the same distribution as $X_1$, and $X_1 \perp \!\!\! \perp X_2$ therefore $$\varphi_{X_1+X_2}(t)=\varphi_{X_1}(t)\varphi_{X_2}(t)=\cos^2(t).$$

This result also shows that $\cos^n(t)$ is a characteristic function for any finite $n$.