0
$\begingroup$

I wish to prove $$ \left(\sum_{i=1}^n a_ib_i\right)^2 \leq \left(\sum_{i=1}^n a_i^2\right)\left(\sum_{i=1}^n b_i^2\right) $$ for really numbers $a_1,\dots,a_n,b_1,\dots,b_n$, by using Hölder inequality.

My plan was to define random variables $X = a_i$ with probability $\frac1n$ and $Y=b_i$ with probability $\frac1n$. Then, using Hölder's, I can get

$$ E\vert X\cdot Y\vert \leq \left[\left(\frac{1}{n}\sum_{i=1}^n a_i^2\right)\left(\frac{1}{n}\sum_{i=1}^n b_i^2\right)\right]^{1/2} $$ and by squaring that I can get the RHS correct.

However, the LHS is not correct with this definition of $X,Y$. Specifically, $X\cdot Y = a_ib_j$ with probability $\frac{1}{n^2}$ so $$ E(X\cdot Y) =\frac{1}{n^2}\sum_{i=1}^na_i\sum_{j=1}^n b_j $$ which is where I'm stuck. Also, I don't quite see what $E\vert XY\vert$ would be since $a_i b_j$ can be negative, so the absolute value matters.

If $XY$ was uniformly distributed over $\{a_ib_i\}$ then I would be golden.

Hints are much appreciated.

$\endgroup$
  • 1
    $\begingroup$ To prove C-S using probability theory you have check that any theorems from probability that are used were not derived using C-S at any stage, which means thoroughly checking their proofs. Are you familiar with the elementary algebraic proof of C-S? $\endgroup$ – DanielWainfleet Jun 21 '17 at 5:52
0
$\begingroup$

In the standard formulation of Holder's inequality, the $p=q=2$ case is just the Cauchy-Schwarz inequality.

For your probabilistic version, you need to align your variables $X$ and $Y$; throw a fair $n$-sided die, and let $X=a_i$ and $Y=b_i$ when it comes up $i$.

$\endgroup$
  • $\begingroup$ I see. I was thinking about having $X,Y$ being dependent, but could not visualize how to make it happen. With the roll of the dice I understand. Thank you. I will accept once the required time passes. $\endgroup$ – user106860 Jun 21 '17 at 5:05
2
$\begingroup$

Holder it's the following thing.

Let $a_1$, $a_2$,..., $a_n$, $b_1$, $b_2$,..., $b_n$, $\alpha$ and $\beta$ be positive numbers. Prove that: $$(a_1+a_2+...+a_n)^{\alpha}(b_1+b_2+...+b_n)^{\beta}\geq$$$$\geq\left(\left(a_1^{\alpha}b_1^{\beta}\right)^{\frac{1}{\alpha+\beta}}+\left(a_2^{\alpha}b_2^{\beta}\right)^{\frac{1}{\alpha+\beta}}+...+\left(a_n^{\alpha}b_n^{\beta}\right)^{\frac{1}{\alpha+\beta}}\right)^{\alpha+\beta}$$

Thus, by Holder for $\alpha=\beta=1$ we obtain: $$\sum_{i=1}^na_i^2\sum_{i=1}^nb_i^2\geq\left(\sum_{i=1}^n\sqrt{a_i^2b_i^2}\right)^2=\left(\sum_{i=1}^n|a_ib_i|\right)^2\geq\left(\sum_{i=1}^na_ib_i\right)^2.$$ Done!

$\endgroup$
  • $\begingroup$ I found the algebra manipulation going from the double sum to the right interesting, thanks. $\endgroup$ – user106860 Jun 21 '17 at 5:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.