Suppose Random variable $X$ ~ Bernoulli $( p )$ .
How can we prove that
$E[(X-p)^4]$ $\leq$ $p^4 + ( 1- p)^4$. ?
I know that $E[(X-p)^2]$ = $Var[X]$ and $E[X^2]= Var[X] + E[X]^2$
Cross Validated is a question and answer site for people interested in statistics, machine learning, data analysis, data mining, and data visualization. It only takes a minute to sign up.
Sign up to join this communitySuppose Random variable $X$ ~ Bernoulli $( p )$ .
How can we prove that
$E[(X-p)^4]$ $\leq$ $p^4 + ( 1- p)^4$. ?
I know that $E[(X-p)^2]$ = $Var[X]$ and $E[X^2]= Var[X] + E[X]^2$
Let $Y = X - p$ then the characteristic function of $Y$ define as $\phi_Y(t) = \mathbb{E}[e^{tY}]$ is $e^{-tp} \phi_X(t)$ where $\phi_X(t)$ is th characteristic function of a Bernoulli and is $1-p + pe^t$.
Thus, $\phi_Y(t) = (1-p)e^{-tp} + pe^{t(1-p)}$.
It's easy to see that $\phi^{(n)}_Y(t) = p(1-p)^ne^{t(1-p)} + (-1)^np^n(1-p)e^{-tp}$.
It follows from $\mathbb{E}[(X-p)^n] = \phi_Y^{(n)}(0)$ that $$ \mathbb{E}[(X-p)^4] = p(1-p)^4 + (1-p)p^4 \leq (1-p)^4 + p^4 $$