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

Solution 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 $$1-|\phi (t)|$$ $$ =1-\psi (t)$$ $$\geq \frac 1 4 (1-\Re \psi (2t))$$ $$ \geq \frac 1 4(1-|\psi (2t)|$$ $$=\frac 1 4(1-|\phi (2t)|).$$