# 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

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

user658409
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 $$\mathbb E(Y^4)\geq \bigl(\mathbb E(Y^2)\bigr)^2.$$

pre-kidney
September 12, 2019 04:12 AM