# Thread: Jensen inequality

1. ## Jensen inequality

Hello!

This is a quick one, if you guys can help...

I'd like to make sure that the Jensen inequality comes out with the equal sign for the following problem (i is just an index):

Given n independent random varables Y(i) i=1,2,...,n each with E[Y(i)] = 0,

obviusly E[Y(1)+Y(2)+...+Y(n)] = |E[Y(1)+Y(2)+...+Y(n)]| = 0

Since the function "absolute value" is convex, for Jensen inequality we know that:

|E[Y(1)+Y(2)+...+Y(n)]| leq E[|Y(1)+Y(2)+...+Y(n)|]

But I'm pretty sure that this is the case for the two member being equal.

Do you agree?

All random variables have expectation = 0, therefore the expectation of their sum is = 0 and so is the expectation of the absolute value of their sum.

Am I correct?

Thanks!

2. Originally Posted by paolopiace
Hello!

This is a quick one, if you guys can help...

I'd like to make sure that the Jensen inequality comes out with the equal sign for the following problem (i is just an index):

Given n independent random varables Y(i) i=1,2,...,n each with E[Y(i)] = 0,

obviusly E[Y(1)+Y(2)+...+Y(n)] = |E[Y(1)+Y(2)+...+Y(n)]| = 0

Since the function "absolute value" is convex, for Jensen inequality we know that:

|E[Y(1)+Y(2)+...+Y(n)]| leq E[|Y(1)+Y(2)+...+Y(n)|]

But I'm pretty sure that this is the case for the two member being equal.

Do you agree?

All random variables have expectation = 0, therefore the expectation of their sum is = 0 and so is the expectation of the absolute value of their sum.

Am I correct?

Thanks!
|Y(1)+Y(2)+...+Y(n)|>=0

So

E[|Y(1)+Y(2)+...+Y(n)|]>0

unless |Y(1)+Y(2)+...+Y(n)|=0 with probability 1

RonL.

3. ## Ok, Let's put it differently...

...given n independent random varables Y(i) i=1,2,...,n each with E[Y(i)] = 0,

I'm required to prove that E[|Y(1)+Y(2)+...+Y(n)|] < infinity

Any hint?

4. Originally Posted by paolopiace
...given n independent random varables Y(i) i=1,2,...,n each with E[Y(i)] = 0,

I'm required to prove that E[|Y(1)+Y(2)+...+Y(n)|] < infinity

Any hint?

This would follow immediately from the triangle inequality if $\displaystyle E(X)=0$ implies that $\displaystyle E(|X|)<\infty$

Also with $\displaystyle n=1$ the problem is equivalent to $\displaystyle E(X)=0$ implies that $\displaystyle E(|X|)<\infty$

RonL

5. ## RonL, Thanks...

... I browsed through the Web looking for a formal proof of the implication.

No success.

If you know any, I would appreciate if you can post a link to some proof.

6. Originally Posted by paolopiace
... I browsed through the Web looking for a formal proof of the implication.

No success.

If you know any, I would appreciate if you can post a link to some proof.
As all the integrals are Lesbegue (in the continuous case) the existance of the expectation $\displaystyle E(X)$ implies the Lesbegue integrability of $\displaystyle x ~p(x)$, but that
implies the absolute integrability of $\displaystyle x ~p(x)$ , but $\displaystyle p(x)$ is positive so the integral $\displaystyle |x| ~p(x)$ exists and is finite. But this last is $\displaystyle E(|X|)<\infty$.

Does not work is we allow improper integrals.

RonL

7. ## Simply marvelous!

Ron I'd like to thank you better than this...