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Assume that the sequence $\{X_n\}$ are i.i.d with uniform distribution on $[0,1]$. Now, define the length longest increasing subsequent as $$L_n=\max\{k:X_{i_1}<X_{i_2}<...<X_{i_k}, 1\leq i_1<i_2<...<i_k\leq n\}.$$

(1) Show that $\mathbb P[L_n\geq 2e\sqrt{n}]<\exp\{-2e\sqrt{n}\}$

(2) Show that for every $a>\frac{1}{3}$ there exists a constant b such that for all large $n$ we have: $$\mathbb P[|L_n-\mathbb E[L_n]|\geq n^a]\leq\exp\{-n^b\}.$$

My attempt:

(1) It is not difficult to prove that the upper bound is $\mathbb P[L_n\geq k]\leq \begin{pmatrix}n\\ k\end{pmatrix}\cdot\frac{1}{k!}$. However, I cannot connect this bound with the bound given in the question.

(2) Using Azuma's inequality, I can prove that $$\mathbb P[|L_n-\mathbb E[L_n]|\geq t]\leq2\exp\bigg\{-\frac{t^2}{2n}\bigg\}.$$

Again, I don't know how to get rid of the coefficient 2 outside the exponential and also the coefficient 1/2 inside the exponential.

If you have any suggestions, it would be greatly appreciated. Thank you!

Cyno Benette
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  • You can also check some related questions: https://math.stackexchange.com/questions/1622315 And other materials: Romik, Dan. The surprising mathematics of longest increasing subsequences. No. 4. Cambridge University Press, 2015. – Vezen BU Apr 18 '24 at 08:22

1 Answers1

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For (1), using some classical bounds of binomial coefficients, e.g., Proof for the upper bound and lower bound for binomial coefficients., we get $$ \binom{n}{k}/k! \leq \frac{n^k}{(\frac{k}{e})^{2k}}. $$ Hence, $$ \Pr[L_n \geq 2e\sqrt{n}] \leq \frac{n^k}{(\frac{k}{e})^{2k}}\bigg\rvert_{k = 2e\sqrt{n}} \leq 4^{-k} < \exp(-k). $$

For (2), can you compute the expectation and use tail bounds?

Vezen BU
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  • [+1] Can we exactly determine the distribution of $L_n$? – Amir Apr 18 '24 at 08:10
  • @Vezen BU Thank you, can you explain a little bit more how you get the second inequality (I mean how to get $4^{-k}$. Also, for (2), for tail bounds, you refer to what inequality? – Cyno Benette Apr 18 '24 at 23:41
  • For the second inequality, you just put $k = 2e\sqrt{n}$ into the term. For (2), I believe you can start with Markov and Chebyshev. – Vezen BU Apr 19 '24 at 07:45