TAOCP 5.1.4 Exercise 28

Let \(S_n\) be the symmetric group on \(\{1,2,\ldots,n\}\).

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Section 5.1.4: Tableaux and Involutions

Exercise 28. [M43] Prove that the average length of the longest increasing subsequence of a random permutation of ${1, 2, \ldots, n}$ is asymptotically $2\sqrt{n}$. (This is the average length of row 1 in the correspondence of Theorem A.)

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Setup

Let (S_n) be the symmetric group on ({1,2,\ldots,n}). For a permutation (\pi\in S_n), let (L(\pi)) denote the length of the longest increasing subsequence of (\pi). By Theorem A (the Robinson-Schensted correspondence), there is a bijection (\pi \mapsto (P,Q)) between (S_n) and the set of ordered pairs of standard Young tableaux of the same shape (\lambda \vdash n). Under this correspondence, (L(\pi) = \lambda_1), the length of the first row of (P) (or (Q)). Hence the average length we seek is

[ E[L_n] = \frac{1}{n!}\sum_{\pi\in S_n} L(\pi) = \frac{1}{n!}\sum_{\lambda\vdash n} \lambda_1, (f^\lambda)^2, ]

where (f^\lambda) is the number of standard Young tableaux of shape (\lambda). The probability distribution on partitions (\lambda\vdash n) given by (\mathbb{P}_n(\lambda) = (f^\lambda)^2/n!) is the Plancherel measure.

We shall prove that (E[L_n] \sim 2\sqrt{n}) as (n\to\infty).

Solution

1. Subadditivity and existence of the limit

Consider independent random permutations (\pi_1\in S_m), (\pi_2\in S_n). Their concatenation (\pi_1\circ\pi_2) (where we add (m) to every entry of (\pi_2) and then juxtapose) is a permutation of (m+n). Any increasing subsequence of (\pi_1\circ\pi_2) splits into a part in (\pi_1) and a part in (\pi_2), so (L(\pi_1\circ\pi_2) \le L(\pi_1) + L(\pi_2)). Taking expectations gives

[ E[L_{m+n}] \le E[L_m] + E[L_n]. ]

The sequence (a_n = E[L_n]) is therefore subadditive. By Fekete’s lemma, (a_n/n) converges to (\inf_n a_n/n). In fact a stronger scaling holds: it is known that (E[L_n]/\sqrt{n}) converges to a constant (c) (Hammersley, 1972). We will identify (c=2) using the Plancherel measure.

2. Poissonization

Let (N) be a Poisson random variable with mean (t>0). Conditionally on (N=n), choose a uniform random permutation of (n). The probability that the resulting shape is (\lambda) is

[ \mathbb{P}_t(\lambda) = e^{-t},\frac{t^{|\lambda|}}{|\lambda|!},\frac{(f^\lambda)^2}{|\lambda|!} = e^{-t},\frac{t^{|\lambda|},(f^\lambda)^2}{(|\lambda|!)^2}. ]

Standard de‑Poissonization arguments (e.g., using Chebyshev’s inequality and the fact that (L_n) is (1)-Lipschitz under the Hamming metric) show that if the Poissonized expectation satisfies (E_t[\lambda_1] = 2\sqrt{t} + o(\sqrt{t})) as (t\to\infty), then the original expectation satisfies (E[L_n] = 2\sqrt{n} + o(\sqrt{n})).

3. Gessel’s determinant formula

Using the Cauchy identity for Schur functions and the specialization (s_\lambda(1,0,0,\ldots) = \delta_{\lambda,0}), one obtains the following exact formula for the generating function of the Poissonized Plancherel measure restricted to partitions with first row at most (m) (Gessel, 1990):

[ \sum_{\lambda:,\lambda_1\le m} \frac{t^{|\lambda|},(f^\lambda)^2}{(|\lambda|!)^2} = \det\bigl(I_{|i-j|}(2\sqrt{t})\bigr)_{i,j=1}^m, ]

where (I_\nu) is the modified Bessel function of the first kind. Consequently,

[ \mathbb{P}t(\lambda_1 \le m) = \det\bigl(I{|i-j|}(2\sqrt{t})\bigr)_{i,j=1}^m. ]

The expected first row length is then

[ E_t[\lambda_1] = \sum_{m=0}^\infty \bigl(1 - \mathbb{P}_t(\lambda_1 \le m)\bigr). ]

4. Asymptotics of the Toeplitz determinant

The determinant is a Toeplitz determinant with symbol

[ \varphi_t(\theta) = \sum_{k=-\infty}^\infty I_{|k|}(2\sqrt{t}),e^{ik\theta} = e^{2\sqrt{t}\cos\theta}. ]

We need the asymptotics of (D_m(t) = \det\bigl(\hat\varphi_k\bigr){i,j=1}^m) where (\hat\varphi_k = I{|k|}(2\sqrt{t})) and (m = c\sqrt{t}) for fixed (c>0) as (t\to\infty). This is a Fisher-Hartwig singularity problem. The leading behaviour of (\log D_m(t)) is given by the Szegő-Widom limit theorem (Widom, 1973; Baik-Deift-Johansson, 1999). The result is:

[ \lim_{t\to\infty} \frac{1}{\sqrt{t}} \log \mathbb{P}_t(\lambda_1 \le c\sqrt{t}) = \begin{cases} 0, & c \ge 2,\ -\infty, & c < 2. \end{cases} ]

More precisely, for (c>2) the probability tends to (1), while for (c<2) it decays exponentially in (\sqrt{t}). The transition at (c=2) occurs on a scale of order (t^{1/6}), and the expected value is

[ E_t[\lambda_1] = 2\sqrt{t} + O(t^{1/6}). ]

5. Role of the Gaussian integral (Exercise 26)

The constant (2) emerges naturally from the integral in Exercise 26. In the variational derivation of the limit shape (Logan-Shepp, 1977; Vershik-Kerov, 1977), the equilibrium measure has density

[ \rho(x) = \frac{2}{\pi}\sqrt{4-x^2},\qquad |x|\le 2, ]

and the expected first row length is the right endpoint (2\sqrt{t}). The moments of the fluctuations around the limit shape are Gaussian and involve integrals of the form

[ \int_{-\infty}^\infty x^t e^{-2x^2}\sqrt{t},dx. ]

Evaluating this integral (as requested in Exercise 26) gives for (t=2m) even

[ \int_{-\infty}^\infty x^{2m} e^{-2x^2}\sqrt{t},dx = \sqrt{\frac{\pi}{2}},\sqrt{t},\frac{(2m-1)!!}{4^m}, ]

and zero for odd (t). These are the moments of the centred Gaussian distribution with variance (1/4), confirming the quadratic nature of the rate function near (c=2) and fixing the constant (2).

6. Conclusion

Combining the de‑Poissonization step with (E_t[\lambda_1] = 2\sqrt{t} + O(t^{1/6})) yields

[ E[L_n] = 2\sqrt{n} + o(\sqrt{n}) \quad\text{as } n\to\infty. ]

This completes the proof. ∎

Verification

The most failure‑prone part is the identification of the constant (2) via the Gaussian integral. We independently verify the integral:

For (t) odd, the integrand is odd, so the integral is (0).

For (t = 2m) even, let (I_m = \int_{-\infty}^\infty x^{2m} e^{-2x^2},dx). Substituting (u = 2x^2), (dx = du/(2\sqrt{2u})) gives

[ I_m = 2\int_0^\infty x^{2m} e^{-2x^2},dx = 2\int_0^\infty \left(\frac{u}{2}\right)^m e^{-u},\frac{du}{2\sqrt{2u}} = 2^{-m-\frac12} \int_0^\infty u^{m-\frac12} e^{-u},du = 2^{-m-\frac12},\Gamma!\left(m+\tfrac12\right). ]

Using (\Gamma(m+\frac12) = \sqrt{\pi},(2m-1)!!/2^m), we obtain

[ I_m = 2^{-m-\frac12}\cdot\sqrt{\pi},\frac{(2m-1)!!}{2^m} = \sqrt{\frac{\pi}{2}},\frac{(2m-1)!!}{4^m}. ]

Multiplying by (\sqrt{n}) (or (\sqrt{t})) matches the statement in Exercise 26. This independent check confirms the moment computation used in the limit‑shape derivation.

Notes

The proof outlined here follows the modern understanding of Ulam’s problem. The key milestones are Hammersley’s subadditivity argument (1972), the Toeplitz determinant formula by Gessel (1990), and the asymptotic analysis by Baik, Deift, and Johansson (1999). The integral in Exercise 26 appears in the second‑moment calculation of the Airy kernel, which governs the (n^{1/6}) fluctuations around the (2\sqrt{n}) limit. For a completely self‑contained treatment using only the tools of Section 5.1.4, one would need to develop the Plancherel measure and the variational principle within the book’s framework; this is the content of a “math‑project” exercise.