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Use the independent product measure on the finite coordinate spaces. For $A\ne \varnothing$, let $V_A(x)$ be the convex hull of its mismatch vectors $(1_{x_i\ne y_i}{)}_{i=1}^n$, $y\in A$. The distance in the question equalsTo verify this, let $v_{\ast }$ be the nearest point. Any unit vector $\alpha$ satisfies ${\min }_{v\in V_A}\alpha \cdot v\le \alpha \cdot v_{\ast }\le \Vert v_{\ast }{\Vert }_2$. Conversely the projection property $v_{\ast }\cdot (v-v_{\ast })\ge 0$ gives equality for $\alpha =v_{\ast }/\Vert v_{\ast }{\Vert }_2$ when $v_{\ast }\ne 0$. If $v_{\ast }=0$, nonnegative weights give supremum zero. Linear functionals have the same minimum on the hull as on its original vectors, so this is exactly the quantified definition of the Talagrand convex distance.

We prove the stronger section induction for the Talagrand exponential moment,Zero-probability atoms may be discarded, replacing $A$ by its supported part: this preserves its probability and can only increase its distance. Thus assume all coordinate atoms have positive mass. The induction starts with zero coordinates, when a positive-probability event is the singleton space and its distance is zero.

Split the last coordinate into values $j$, with probabilities $q_j$, and write $A_j$ for the corresponding section in the first $n-1$ coordinates. Put $a_j=\mathrm{Pr}(A_j)$, choose a largest section $A_s$ of probability $a>0$, and set $r_j=a_j/a\in [0,1]$. For a point $(x^{\prime },j)$, mix nearest mismatch vectors from $A_j$ and $A_s$ with weights $\lambda ,1-\lambda$. Their final coordinates are zero and, if $j\ne s$, one. Convexity of the squared norm givesFor an empty $A_j$, use $\lambda =0$ directly. If $j=s$, the final-coordinate term is an unnecessary nonnegative allowance, so the bound still holds.

Exponentiate, integrate in $x^{\prime }$, and use Hölder's inequality and the induction hypothesis. For $r_j>0$ this givesWe supply the scalar estimate for Talagrand product induction used in the last step. Set $L=-\log r$. For $0\le L\le 1/2$, choose $1-\lambda =2L$, making the expression $e^{L-L^2}$. The functionhas $h(0)=h^{\prime }(0)=0$ andThus $e^{L-L^2}\le 2-e^{-L}$. For $L\ge 1/2$, choose $\lambda =0$, giving $e^{1/4}$; the already proved boundary case and monotonicity of $2-e^{-L}$ give the same bound. For $r=0$, the direct empty-section bound is $e^{1/4}/a\le 2/a$.

Now average over $j$. With $m={\sum }_j{q}_j{r}_j=\mathrm{Pr}(A)/a\in (0,1]$, the exponential integral is at most $(2-m)/a$. Since $m(2-m)\le 1$, this is at most $1/(am)=1/\mathrm{Pr}(A)$, closing the induction.

Finally apply the Markov inequality to the nonnegative exponential. For the distance-tail event ${\overline{A}}_t$,This proves Talagrand's convex distance inequality for the finite product. If $\mathrm{Pr}(A)=0$, the required probability-product inequality is immediate, including the empty-event case with distance defined as infinity; no finite exponential-moment assertion is needed there.

For $x\in \Omega$ and nonempty $A$, letThe Talagrand convex distance isEquivalently,For completeness, let $v_{\ast }$ be a closest point of the finite convex hull to zero. If $v_{\ast }\ne 0$, minimality gives $v_{\ast }\cdot (v-v_{\ast })\ge 0$ for every $v\in V$. Taking $\alpha =v_{\ast }/\Vert v_{\ast }\Vert$ proves that the supremum is at least $\Vert v_{\ast }\Vert$. Conversely each permitted $\alpha$ has ${\inf }_{v\in V}\alpha \cdot v\le \alpha \cdot v_{\ast }\le \Vert v_{\ast }\Vert$ by Cauchy-Schwarz. The weights are nonnegative because the hull has nonnegative coordinates. If $v_{\ast }=0$, both sides are zero. This proves the dual definition used below.

We prove Talagrand's convex distance inequality by induction on the number of factors. Write $\Omega ={\Omega }^{\prime }\times X_n$, with measure $\mu \otimes \nu$, and $F_A=d_T(\cdot ,A{)}^2$. For a last-coordinate value $z$, let $A_z=\{y^{\prime }:(y^{\prime },z)\in A\}$. Choose a section $B=A_b$ of largest probability $m=\mu (B)>0$, and put $a_z=\mu (A_z)/m\in [0,1]$. The total event probability is $p=\mathrm{Pr}(A)=m{\sum }_z\nu (z)a_z$.

If $A_z$ has positive probability, combine mismatch vectors from $A_z$ and from $B$. A weight $t$ on the second section contributes at most $t$ in the last coordinate. Convexity of the squared Euclidean norm gives, for every $0\le t\le 1$,The last-coordinate contribution is actually zero if $z=b$, so the displayed upper bound remains valid. Hölder's inequality and the induction hypothesis yieldFor a zero-probability section use $t=1$ directly, so no division by zero or exponent of an infinite quantity is needed.

The scalar estimate for Talagrand product induction isHere is its proof. For $a>0$ set $L=-\log a$. If $L\le 1/2$, take $t=2L$; the logarithm of the left expression is $L-L^2$. Put $h(L)=\log (2-e^{-L})-L+L^2$. Then $h(0)=h^{\prime }(0)=0$ andsince $(2u-1{)}^2-u=(4u-1)(u-1)\ge 0$ for $u=e^L\ge 1$. Thus $h\ge 0$. If $L\ge 1/2$, choose $t=1$. The already proved boundary case gives $e^{1/4}\le 2-e^{-1/2}\le 2-a$. The same direct choice covers $a=0$.

Average the resulting section bounds over $z$:where the last inequality is equivalent to $(m-p{)}^2\ge 0$. The induction starts on the zero-coordinate singleton, where every positive-probability event is the whole space and the moment is one. HenceWe assume $\mathrm{Pr}(A)>0$; a zero-probability event has no finite right-hand bound to prove. Markov's inequality also gives

For the subsequence application, let $L(x)$ be the longest increasing subsequence length for independent sequence coordinates. A witness of length $\ell$ uses only $\ell$ positions. If $L(x)=\ell \ge b$ and $y$ satisfies $L(y)\le a<b$, at least $\ell -a$ witness positions must change: all unchanged witness positions remain in increasing order. Put weights $1/\sqrt{\ell }$ on the witness and zero elsewhere. The dual distance definition giveswhere the latter expression increases with $\ell$ for $a\ge 0$. Therefore the increasing-subsequence certificate concentration bound isIf the lower event has probability zero the conclusion is immediate.

Let $m$ be an integer median, so $\mathrm{Pr}(L\le m),\mathrm{Pr}(L\ge m)\ge 1/2$. Taking the two thresholds in the bound gives, for $t>0$,For $t>m$ the lower event is empty. In a nonempty sequence $m\ge 1$. Integration of these tails gives an explicit mean-median estimate: use $m+t\le 2m$ for $t\le m$ and $m+t\le 2t$ for $t\ge m$, obtainingWriting $\mu =\mathbb{E}L$, the median property also gives $m\le 2\mu$. Consequently, for a universal constant $C$, all $u\ge 0$ satisfyThus the subsequence length is concentrated near its mean on the scale $\sqrt{\mu }$, rather than the generic coordinate-exposure scale $\sqrt{n}$. This argument applies directly to finite-valued independent sequences. Independent continuous samples follow by refining finite coordinate discretizations; the strict-order pattern stabilizes almost surely when ties have probability zero. A uniform random permutation is the rank sequence of independent continuous samples, so it has the same subsequence-length concentration. Arbitrarily dependent random sequences are not covered by the product-space assumption.

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.