Show that : $1 - |\phi(t)| \ge \frac{1-|\phi(2t)|}{4}$

Question

Show that : $1 - |\phi(t)| \ge \frac{1-|\phi(2t)|}{4}$ where $\phi(t)$ is a characteristic function .

I am able to prove another part of the question $\Re(1-\phi(t)) \ge \Re(\frac{1-\phi(2t)}{4})$ . How to derive the first part using this information ?

$\textbf{Note}$ : $\phi$ is a characteristic function and $\Re F$ is real part of $F$ .

I am unable to show the first part . Any kind of help is appreciated .

Answer 1

If $X$ is a random variable with characteristic function $\phi$ then for a sutable $s$ we have $1-|\phi (t)|=1-\phi (t)e^{is}=E(1-e^{i(tX+s)})\geq E(1-\cos (tX+s)$ . Now use the following inequality: $1-\cos x \geq \frac 1 4 (1-\cos (2x)$. [This can be proved easily using the fact that $1-\cos y=2\sin^{2}(\frac y 2)$ and $\sin (2u)=2\sin u \cos u$]. We now have $1-|\phi (t)|\geq E\frac 1 4 (1-e^{it(X+s)})$. Finish the proof by noting that $|\phi (2t)|=|Ee^{2it X}e^{i2s}|\geq E(\cos (2t X+2s))$

Proof using the inequality you have proved already: Fix $t \neq 0$. Let $\psi (u)=\phi (u)e^{ius}$ where $s$ is chosen such that $\phi (t)e^{its}=|\phi (t)|$. If $X$ has characterisitic function $\phi$ then $X+s$ has characterisitic function $\psi$. Apply the inequality you already know to the characterisitic function $\psi$. You get $$1-|\phi (t)|$$ $$ =1-\psi (t)$$ $$\geq \frac 1 4 (1-\Re \psi (2t))$$ $$ \geq \frac 1 4(1-|\psi (2t)|$$ $$=\frac 1 4(1-|\phi (2t)|).$$

Answer 2

Maybe you can consider $1-\cos tx\geq4(1-\cos 2tx)$ (relevant to estimation of the real part)