Fourth(?) moment method for minimum value

I would like to lower bound the quantity $\Pr[X\ge t, Y\ge t]=\Pr[\min(X,Y)\ge t]$ using the small moments of $X$, $Y$ and $XY$. In particular I am interested in the case where $E[X]=E[Y]=0$, but $E[X Y]>0$.

In "The Fourth Moment Method" Berger shows that if $\frac{E[X^4]}{E[X^2]^2}\le b$, then $\Pr[X\ge \sqrt{E[X^2]}/(4\sqrt{t})]\ge1/(4^{4/3}t)$. Alternative formulations of this principle include the Paley–Zygmund inequality.

Using the fact that $\min(X,Y)=\frac12(X+Y-|X-Y|)$, we get that that $$E[\min(X,Y)^2] = E[\min(X^2,Y^2)] = \tfrac12\left(E[X^2+Y^2]-E[|X^2-Y^2|]\right),\\ E[\min(X,Y)^4] = E[\min(X^4,Y^4)] = \tfrac12\left(E[X^4+Y^4]-E[|X^4-Y^4|]\right).$$ Now using $\frac{E[X^2]^{3/2}}{E[X^4]^{1/2}} \le E[|X|] \le \sqrt{E[X^2]}$ we can get a bound: $$\frac{E[\min(X,Y)^4]}{E[\min(X,Y)^2]^2} = 2\frac{E[X^4+Y^4]-E[|X^4-Y^4|]}{(E[X^2+Y^2]-E[|X^2-Y^2|])^2} \le 2\frac{E[X^4+Y^4]-\sqrt{E[(X^4-Y^4)^2]}}{\left(E[X^2+Y^2]-\frac{E[(X^2-Y^2)^2]^{3/2}}{E[(X^2-Y^2)^4]^{1/2}}\right)^2}.$$

Besides being ugly, this has ended up using the eighth moment of $X$ and $Y$, rather than just the fourth. Is this necessary? Or is there a nicer way taking better advantage of the $\min$ function?

Update: We may wonder what the ideal result would be. If we let $\hat{X}=[X,Y]^T$, $\Sigma=\text{Cov}(\hat X)$ and $Z=\hat X ^T\Sigma^{-1}\hat X$, then Markov's (or the multivariate Chebyshev) inequality tells us $\Pr[Z \ge \epsilon] \le E[Z]/\epsilon=2/\epsilon$. If we assume $\text{Var}[X]=\text{Var}[Y]=1$ then $\min(X,Y)\ge t$ implies $Z\ge t^T\Sigma^{-1}t=\frac{2t^2}{1+\text{Cov}(X,Y)}$; and so $\Pr[X,Y\ge t]\le\frac{1+\text{Cov}(X,Y)}{t^2}$.

With Paley–Zygmund we then get $\Pr[Z\ge 2t] \ge 4(1-t)^2/E[Z^2]$, which is a fourth moment bound. Of course this is a lower bound on the entire space outside the ellipse, however we might(?) hope that it is also correct up to a constant factor for any convex region at distance t, if $X$ and $Y$ are nice enough?

• I would suggest asking this on math.stackexchange.com or mathoverflow.net. – Ryan O'Donnell Jul 27 '17 at 1:13