# cauchy schwarz inequality

• Jan 30th 2010, 12:45 AM
kkjs358
cauchy schwarz inequality
For any X € L = L^1 prove that |EX| ≤ E|X|

For any X € L = L^2 prove that |EX|^2 ≤ E|X|^2

I was trying to prove those facts but I was not sure how to...
I figured out that I need to use the cauchy schwarz inequality but i did not clearly understand how to apply it (Crying)
I started the questions like this for the first one,

a b ≤ |a b| ≤ |a||b|
I started the questions like this for the second one,
var (X) = E(X^2)-(EX)^2

but this is how far I came, can someone plz help me?(Crying)

• Jan 30th 2010, 08:27 AM
matheagle
You can use Jensen's inequality here.
• Jan 30th 2010, 01:03 PM
kkjs358
how? im not too sure about Jensen's inequality... I don't really know about
• Jan 30th 2010, 02:42 PM
Laurent
Quote:

Originally Posted by kkjs358
how? im not too sure about Jensen's inequality... I don't really know about

Do you mean you don't know Jensen's inequality? You can prove these elementary inequalities without it.

For the first one, you have to separate the positive and negative parts of $X$: by definition, $E[X]=E[X_+]-E[X_-]$, and both terms are nonnegative so that $|E[X]|\leq E[X_+]+E[X_-]=E[X_++X_-]$; finally, $X_++X_-=|X|$.

For the second one, on one hand the quantity $E[(X-E[X])^2]$ is clearly nonnegative (and it is called the variance of $X$), and on the other hand if you expand the square inside the expectation and use linearity of expectation you get that $E[(X-E[X])^2]=E[X^2]-E[X]^2$. The nonnegativity thus gives you the inequality you need.

Alternatively, you can use Cauchy-Schwarz inequality: $E[X]=E[X\cdot 1]\leq E[X^2]^{1/2}E[1]^{1/2}=E[X^2]^{1/2}$. This is ok as well.

And $E[X]=E[{\rm sign}(X)\sqrt{|X|}\sqrt{|X|}]\leq E[|X|]^{1/2}E[|X|]^{1/2}$ $=E[|X|]$ where ${\rm sign}(X)=\pm 1$ according to the sign of X, and subsequently $-E[X]=E[-X]\leq E[|X|]$ as well so that $|E[X]|\leq E[|X|]$, but this is not a very direct way of proving it at all (I mean, the inequality is way more elementary than it could seem from this argument), I don't recommand it.
• Jan 31st 2010, 12:09 AM
kkjs358
wow... it looks very simple. I never thought i could use elementary inequalities
I was trying to use famous theories like cauchy-schwarz or Jensen's inequality (this one i don't really know)

thank you so much for your help!!