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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 $E[X^2]= Var[X] + E[X]^2$

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    $\begingroup$ The expectation is the sum of two terms (which depend only on $p$), so why not write down this sum and maximize it? The sum, when compared to the right hand side and simplified, becomes very simple indeed. $\endgroup$ – whuber Jul 25 at 19:20
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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 $$ \mathbb{E}[(X-p)^4] = p(1-p)^4 + (1-p)p^4 \leq (1-p)^4 + p^4 $$

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    $\begingroup$ Why not just write the last line immediately from the definition of expectation? :-) $\endgroup$ – whuber Jul 25 at 19:29
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    $\begingroup$ I feel dumb right now.... I guess I saw a moment of order n and directly thought of the characteristic function without even thinking about just writing the expectation $\endgroup$ – winperikle Jul 25 at 19:38
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    $\begingroup$ ... but your approach is perhaps more readily generalized! $\endgroup$ – jbowman Jul 25 at 19:38
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    $\begingroup$ @jbowman E.g., it would be interesting to show the right hand side is an upper bound for any distribution of expectation $p$ supported on $[0,1]$ :-). $\endgroup$ – whuber Jul 25 at 19:41
  • $\begingroup$ Thank you both ! It is always good to know all approaches! Because I might use this moment generating function in other cases ( Eg. other distributions ) $\endgroup$ – GAGA Jul 25 at 22:44

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