Return to Answer

Not that I have ever seen this used anywhere (so this might be the wrong direction), but since the second moment method is essentially using Cauchy-Schwarz on the variable $X=X 1_{X>0}$, can't we use Holder's inequality on this expression to obtain higher order inequalities? edit: and obtain something like $P(X>0)\geq (\frac{E[X]^p}{E[X^p]})^{1/(p-1)}$ (assuming $X\geq 0$ and $p>1$).

Not that I have ever seen this used anywhere (so this might be the wrong direction), but since the second moment method is essentially using Cauchy–Schwarz on the variable $X=X 1_{X>0}$, can't we use Hölder's inequality on this expression to obtain higher order inequalities? edit: and obtain something like $P(X>0)\geq \left(\frac{E[X]^p}{E[X^p]}\right)^{1/(p-1)}$ (assuming $X\geq 0$ and $p>1$).

Not that I have ever seen this used anywhere (so this might be the wrong direction), but since the second moment method is essentially using Cauchy Schwartz on the variable $X=X 1_{X>0}$, can't we use Holder's inequality on this expression to obtain higher order inequalities? edit: and obtain something like $P(X>0)\geq (\frac{E[X]^p}{E[X^p]})^{1/(p-1)}$ (assuming $X\geq 0$ and $p>1$).

Not that I have ever seen this used anywhere (so this might be the wrong direction), but since the second moment method is essentially using Cauchy-Schwarz on the variable $X=X 1_{X>0}$, can't we use Holder's inequality on this expression to obtain higher order inequalities? edit: and obtain something like $P(X>0)\geq (\frac{E[X]^p}{E[X^p]})^{1/(p-1)}$ (assuming $X\geq 0$ and $p>1$).

Not that I have ever seen that used anywhere (so this might be the wrong direction), but since the second moment method is essentially using Cauchy Schwartz on the variable $X=X 1_{X>0}$, can't we use Holder's inequality on this expression to obtain higher order inequalities?

Not that I have ever seen this used anywhere (so this might be the wrong direction), but since the second moment method is essentially using Cauchy Schwartz on the variable $X=X 1_{X>0}$, can't we use Holder's inequality on this expression to obtain higher order inequalities? edit: and obtain something like $P(X>0)\geq (\frac{E[X]^p}{E[X^p]})^{1/(p-1)}$ (assuming $X\geq 0$ and $p>1$).