# Distribution Bounded

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

• 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. – whuber Jul 25 at 19:20

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 $$\mathbb{E}[(X-p)^4] = p(1-p)^4 + (1-p)p^4 \leq (1-p)^4 + p^4$$

• Why not just write the last line immediately from the definition of expectation? :-) – whuber Jul 25 at 19:29
• 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 – winperikle Jul 25 at 19:38
• ... but your approach is perhaps more readily generalized! – jbowman Jul 25 at 19:38
• @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 $[0,1]$ :-). – whuber Jul 25 at 19:41
• 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 ) – GAGA Jul 25 at 22:44