Talagrand's convex distance inequality ($\mathbb{P}(A)\mathbb{P}(d_T(X,A)\ge s)\le e^{-s^2/4}$)

On a finite product of standard probability spaces, for any measurable event $A$ of positive probability,Consequently $\mathbb{P}(A)\mathbb{P}(d_T(X,A)\ge s)\le e^{-s^2/4}$. Here $d_T$ is the Talagrand convex distance. The distance-event conclusion can be read with outer probability when necessary.

Let the largest last-coordinate section have probability $a$, and another section probability $ar$. Mix mismatch vectors from these two sections with weights $\lambda ,1-\lambda$. Their squared norm is at most $\lambda d_1^2+(1-\lambda )d_2^2+(1-\lambda {)}^2$. Hölder's inequality and induction bound that section's exponential integral by $a^{-1}{\inf }_{\lambda }{r}^{-\lambda }{e}^{(1-\lambda {)}^2/4}$. The scalar estimate for Talagrand product induction makes it at most $(2-r)/a$. Average over sections and use $m(2-m)\le 1$ with $m=\mathrm{Pr}(A)/a$ to close the induction.

For $a\in [0,1]$,where the endpoint $t=1$ is used directly if $a=0$. Put $L=-\log a$. For $L\le 1/2$, choose $t=2L$ and set $h(L)=\log (2-e^{-L})-L+L^2$. Then $h(0)=h^{\prime }(0)=0$ and $h^{\prime \prime }(L)=2-2e^L/(2e^L-1{)}^2\ge 0$, proving the bound. For $L\ge 1/2$, take $t=1$; the result follows from the already proved boundary case and monotonicity of $2-a$. This estimate closes the section-by-section induction for the exponential Talagrand convex distance moment.