Show that $E(Y^4) \geq (E[Y^2])^{2}$ using Jensen's Inequality

by Boy Wonder   Last Updated September 12, 2019 04:20 AM

I'm a bit stuck I need to show that $E[Y^{4}] \geq (E[Y^2])^2$ where $Y$ has finite second moment using Jensen's Inequality.

I know that $Y^{4}$ has finite second moment so using jensen's inequality I have:

$E[Y^4] \geq (E[Y])^4 =E([Y])^{2} \cdot E([Y])^{2}$

But then I get stuck since I need $(E[Y^2])^2$ but I have things in terms of $E[Y]$. Thanks for the help.

Or is this right: $E[Y^4]=E[(Y^2)^2)] \geq (E(Y^2))^2$ since $y^2$ is convex? Thanks

Answers 2

Do you see that $E(X^2)\geq (E(X))^2$ by Jensen's inequality? Let $X=Y^2$.

September 11, 2019 15:36 PM

Since it seems you are confused, let me include a self-contained proof that does not involve Jensen's inequality, to make it crystal clear why this is true. Consider the quantity which is obviously non-negative: $$ \mathbb E\bigl(Y^2-\mathbb E(Y^2)\bigr)^2\geq 0. $$ Now expand the square and collect like terms to learn that $$ \mathbb E\bigl(Y^2-\mathbb E(Y^2)\bigr)^2=\mathbb E(Y^4)-\bigl(\mathbb E(Y^2)\bigr)^2. $$ Therefore $\mathbb E(Y^4)\geq \bigl(\mathbb E(Y^2)\bigr)^2.$

September 12, 2019 04:12 AM

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