Moments of Maximum: Segment of AR(1)

1 Partial Correlations

Set $ℓ = 4$ for simplicity. Fix $1 \leq i < j \leq 4$

. There exists a unique ordered pair

m < n

such that

(i, j, m, n} = {1, 2, 3, 4}

. Define a

3 \times 3

matrix

P = ⎛ ⎜ ⎜ ⎝ \begin{matrix} 1 & r_{i, m n} & r_{i, j m} r_{i, m n} & 1 & r_{i, j n} r_{i, j m} & r_{i, j n} & 1 \end{matrix} ⎞ ⎟ ⎟ ⎠

which captures all correlations among $X_{i} - X_{m}$ , $X_{i} - X_{n}$ , $X_{i} - X_{j}$ . Let $P_{a b}$ be the cofactor of the element $p_{a b}$ in the expansion of the determinant of $P$ . The partial correlation $r_{i, m n . j}$ between $X_{i} - X_{m}$ and $X_{i} - X_{n}$ , given $X_{i} - X_{j}$ , is prescribed by

r_{i, m n . j} = - \frac{P_{12}}{\sqrt{P_{11} P_{22}}} = \frac{r_{i, m n} - r_{i, j m} r_{i, j n}}{\sqrt{(1 - r_{i, j m}^{2}) (1 - r_{i, j n}^{2})}} .

The notation here [1] differs from that used elsewhere [2, 3, 4]. In words, $r_{i, m n . j}$ measures the linear dependence of $X_{i} - X_{m}$ and $X_{i} - X_{n}$ in which the influence of $X_{i} - X_{j}$ is removed. Define finally a $2 \times 2$ matrix

R_{i, j} = (\begin{matrix} 1 & r_{i, m n . j} r_{i, m n . j} & 1 \end{matrix})

as was to be done.

Set now $ℓ = 5$ . Fix $1 \leq i < j \leq 5$ . There exists a unique ordered triple $m < n < o$ such that $(i, j, m, n, o} = {1, 2, 3, 4, 5}$ . The preceding discussion extends naturally, supplementing the case $(m, n)$ by additional possible cases $(m, o)$ and $(n, o)$ . Define finally a $3 \times 3$ matrix

R_{i, j} = ⎛ ⎜ ⎜ ⎝ \begin{matrix} 1 & r_{i, m n . j} & r_{i, m o . j} r_{i, m n . j} & 1 & r_{i, n o . j} r_{i, m o . j} & r_{i, n o . j} & 1 \end{matrix} ⎞ ⎟ ⎟ ⎠ .

We could go on for larger $ℓ$ , but this is all that is needed for our purposes.

Again, set $ℓ = 5$ . Fix $1 \leq i < j < k \leq 5$ . There exists a unique ordered pair $m < n$ such that $(i, j, k, m, n} = {1, 2, 3, 4, 5}$ . Define a $4 \times 4$ matrix

Q = ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ \begin{matrix} 1 & r_{i, m n} & r_{i, j m} & r_{i, k m} r_{i, m n} & 1 & r_{i, j n} & r_{i, k n} r_{i, j m} & r_{i, j n} & 1 & r_{i, j k} r_{i, k m} & r_{i, k n} & r_{i, j k} & 1 \end{matrix} ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠

which captures all correlations among $X_{i} - X_{m}$ , $X_{i} - X_{n}$ , $X_{i} - X_{j}$ , $X_{i} - X_{k}$ . Let $Q_{a b}$ be the cofactor of the element $q_{a b}$ in the expansion of the determinant of $Q$ . The partial correlation $r_{i, m n . j k}$ between $X_{i} - X_{m}$ and $X_{i} - X_{n}$ , given $X_{i} - X_{j}$ and $X_{i} - X_{k}$ , is prescribed by

	$r_{i, m n . j k}$	$= - \frac{Q_{12}}{\sqrt{Q_{11} Q_{22}}}$
		$= \frac{r_{i, m n} - r_{i, j m} r_{i, j n} - r_{i, k m} r_{i, k n} + r_{i, k m} r_{i, j n} r_{i, j k} + r_{i, j m} r_{i, k n} r_{i, j k} - r_{i, m n} r_{i, j k}^{2}}{\sqrt{(1 - r_{i, j m}^{2} - r_{i, k m}^{2} + 2 r_{i, j m} r_{i, k m} r_{i, j k} - r_{i, j k}^{2}) (1 - r_{i, j n}^{2} - r_{i, k n}^{2} + 2 r_{i, j n} r_{i, k n} r_{i, j k} - r_{i, j k}^{2})}} .$

In words, $r_{i, m n . j k}$ measures the linear dependence of $X_{i} - X_{m}$ and $X_{i} - X_{n}$ in which the influence of $X_{i} - X_{j}$ and $X_{i} - X_{k}$ is removed. Define finally a $2 \times 2$ matrix

R_{i, j k} = (\begin{matrix} 1 & r_{i, m n . j k} r_{i, m n . j k} & 1 \end{matrix})

as was to be done.

Set now $ℓ = 6$ . Fix $1 \leq i < j < k \leq 6$ . There exists a unique ordered triple $m < n < o$ such that $(i, j, k, m, n, o} = {1, 2, 3, 4, 5, 6}$ . The preceding discussion extends naturally, supplementing the case $(m, n)$ by additional possible cases $(m, o)$ and $(n, o)$ . Define finally a $3 \times 3$ matrix

R_{i, j k} = ⎛ ⎜ ⎜ ⎝ \begin{matrix} 1 & r_{i, m n . j k} & r_{i, m o . j k} r_{i, m n . j k} & 1 & r_{i, n o . j k} r_{i, m o . j k} & r_{i, n o . j k} & 1 \end{matrix} ⎞ ⎟ ⎟ ⎠ .

We could go on for larger $ℓ$ , but this is all that is needed for our purposes.

2 Small Segments

For convenience, define

	$h (x, y, z)$	$= \frac{1 - x}{4 π} \cdot \frac{1 + x - y - z}{\sqrt{4 (1 - x) (1 - y) - (1 - x - y + z)^{2}}}$
		$= \frac{1 - x}{4 π} \cdot \frac{1 + x - y - z}{\sqrt{4 (1 - x) (1 - z) - (1 - x - z + y)^{2}}}$
		$= \frac{1 - x}{4 π} \cdot \frac{1 + x - y - z}{\sqrt{(1 - x) (3 + x) - y (2 + y) - z (2 + z) + 2 (x y + x z + y z)}} .$

The latter expression, while more cumbersome, exhibits symmetry in $y$ , $z$ .

If $ℓ = 2$ , then $Φ_{ℓ - 2} (R_{i, j}) = 1$ and $Φ_{ℓ - 3} (R_{i,, j k}) = 0$ . We have

\begin{matrix} E (M) = 2 \sqrt{\frac{1 - ρ_{12}}{4 π}} = \sqrt{\frac{1 - ρ_{12}}{π}}, & E (M^{2}) = 1. \end{matrix}

If $ℓ = 3$ , then $Φ_{ℓ - 2} (R_{i, j}) = \frac{1}{2}$ and $Φ_{ℓ - 3} (R_{i,, j k}) = 1$ . We have

E (M) = \frac{1}{\sqrt{4 π}} (\sqrt{1 - ρ_{12}} + \sqrt{1 - ρ_{13}} + \sqrt{1 - ρ_{23}}),

E (M^{2}) = 1 + 2 h (ρ_{12}, ρ_{13}, ρ_{23}) + 2 h (ρ_{13}, ρ_{12}, ρ_{23}) + 2 h (ρ_{23}, ρ_{12}, ρ_{13}) .

In formula (3.6) for $E (M)$ in [1], $\sqrt{(π)}$ should be replaced by $\sqrt{(2 π)}$ .

If $ℓ = 4$ , then

\begin{matrix} Φ_{ℓ - 2} (R_{i, j}) = \frac{1}{4} + \frac{1}{2 π} arcsin (r_{i, m n . j}), & Φ_{ℓ - 3} (R_{i,, j k}) = \frac{1}{2} . \end{matrix}

In general, $r_{i, m n . j} \neq r_{j, m n . i}$ and thus symmetry fails for $E (M)$ . We have

	$E (M)$	$= \frac{1}{\sqrt{4 π}} [\sqrt{1 - ρ_{12}} \cdot (\frac{1}{4} + \frac{1}{2 π} arcsin (r_{1, 34.2})) + \sqrt{1 - ρ_{12}} \cdot (\frac{1}{4} + \frac{1}{2 π} arcsin (r_{2, 34.1}))$
		$+ \sqrt{1 - ρ_{13}} \cdot (\frac{1}{4} + \frac{1}{2 π} arcsin (r_{1, 24.3})) + \sqrt{1 - ρ_{13}} \cdot (\frac{1}{4} + \frac{1}{2 π} arcsin (r_{3, 24.1}))$
		$+ \sqrt{1 - ρ_{14}} \cdot (\frac{1}{4} + \frac{1}{2 π} arcsin (r_{1, 23.4})) + \sqrt{1 - ρ_{14}} \cdot (\frac{1}{4} + \frac{1}{2 π} arcsin (r_{4, 23.1}))$
		$+ \sqrt{1 - ρ_{23}} \cdot (\frac{1}{4} + \frac{1}{2 π} arcsin (r_{2, 14.3})) + \sqrt{1 - ρ_{23}} \cdot (\frac{1}{4} + \frac{1}{2 π} arcsin (r_{3, 14.2}))$
		$+ \sqrt{1 - ρ_{24}} \cdot (\frac{1}{4} + \frac{1}{2 π} arcsin (r_{2, 13.4})) + \sqrt{1 - ρ_{24}} \cdot (\frac{1}{4} + \frac{1}{2 π} arcsin (r_{4, 13.2}))$
		$+ \sqrt{1 - ρ_{34}} \cdot (\frac{1}{4} + \frac{1}{2 π} arcsin (r_{3, 12.4})) + \sqrt{1 - ρ_{34}} \cdot (\frac{1}{4} + \frac{1}{} 2 π arcsin (r_{4, 12.3}))],$

	$E (M^{2})$	$= 1 + h (ρ_{12}, ρ_{13}, ρ_{23}) + h (ρ_{12}, ρ_{14}, ρ_{24}) + h (ρ_{13}, ρ_{12}, ρ_{23}) + h (ρ_{13}, ρ_{14}, ρ_{34})$
		$+ h (ρ_{14}, ρ_{12}, ρ_{24}) + h (ρ_{14}, ρ_{13}, ρ_{34}) + h (ρ_{23}, ρ_{12}, ρ_{13}) + h (ρ_{23}, ρ_{24}, ρ_{34})$
		$+ h (ρ_{24}, ρ_{12}, ρ_{14}) + h (ρ_{24}, ρ_{23}, ρ_{34}) + h (ρ_{34}, ρ_{13}, ρ_{14}) + h (ρ_{34}, ρ_{23}, ρ_{24}) .$

In formula (3.7) for $E (M)$ in [1], a factor $1 / \sqrt{2 π}$ should be inserted in front of the summation.

If $ℓ = 5$ , then

Φ_{ℓ - 2} (R_{i, j}) = \frac{1}{2} - \frac{1}{4 π} arccos (r_{i, m n . j}) - \frac{1}{4 π} arccos (r_{i, m o . j}) - \frac{1}{4 π} arccos (r_{i, n o . j}),

Φ_{ℓ - 3} (R_{i,, j k}) = \frac{1}{4} + \frac{1}{2 π} arcsin (r_{i, m n . j k}) .

Symmetry now fails for both $E (M)$ and $E (M^{2})$ . We have

	$E (M)$	$= \frac{1}{\sqrt{4 π}} [\sqrt{1 - ρ_{12}} \cdot (\frac{1}{2} - \frac{1}{4 π} arccos (r_{1, 34.2}) - \frac{1}{4 π} arccos (r_{1, 35.2}) - \frac{1}{4 π} arccos (r_{1, 45.2}))$
		$+ \sqrt{1 - ρ_{12}} \cdot (\frac{1}{2} - \frac{1}{4 π} arccos (r_{2, 34.1}) - \frac{1}{4 π} arccos (r_{2, 35.1}) - \frac{1}{4 π} arccos (r_{2, 45 .1}))$
		$+ \sqrt{1 - ρ_{13}} \cdot (\frac{1}{2} - \frac{1}{4 π} arccos (r_{1, 24.3}) - \frac{1}{4 π} arccos (r_{1, 25.3}) - \frac{1}{4 π} arccos (r_{1, 45 .3}))$
		$+ \sqrt{1 - ρ_{13}} \cdot (\frac{1}{2} - \frac{1}{4 π} arccos (r_{3, 24.1}) - \frac{1}{4 π} arccos (r_{3, 25.1}) - \frac{1}{4 π} arccos (r_{3, 45.1})) + \dots],$

	$E (M^{2})$	$= 1 + h (ρ_{12}, ρ_{13}, ρ_{23}) \cdot (\frac{1}{4} + \frac{}{1} 2 π arcsin (r_{1, 45.23})) + h (ρ_{12}, ρ_{23}, ρ_{13}) \cdot (\frac{1}{4} + \frac{1}{2 π} arcsin (r_{2, 45.13}))$
		$+ h (ρ_{12}, ρ_{14}, ρ_{24}) \cdot (\frac{1}{4} + \frac{1}{2 π} arcsin (r_{1, 35.24})) + h (ρ_{12}, ρ_{24}, ρ_{14}) \cdot (\frac{1}{4} + \frac{1}{2 π} arcsin (r_{2, 35.14}))$
		$+ h (ρ_{12}, ρ_{15}, ρ_{25}) \cdot (\frac{1}{4} + \frac{1}{2 π} arcsin (r_{1, 34.25})) + h (ρ_{12}, ρ_{25}, ρ_{15}) \cdot (\frac{1}{4} + \frac{1}{2 π} arcsin (r_{2, 34.15})) + \dots,$

a total of $20$ terms and $61$ terms, respectively.

If $ℓ = 6$ , then $E (M)$ contains non-elementary functions which require numerical integration (beyond our present scope). In contrast,

	$E (M^{2})$	$= 1 + h (ρ_{12}, ρ_{13}, ρ_{23}) \cdot (\frac{1}{2} - \frac{}{1} 4 π arccos (r_{1, 45.23}) - \frac{1}{4 π} arccos (r_{1, 46.23}) - \frac{1}{4 π} arccos (r_{1, 56.23}))$
		$+ h (ρ_{12}, ρ_{23}, ρ_{13}) \cdot (\frac{1}{2} - \frac{1}{4 π} arccos (r_{2, 45.13}) - \frac{1}{4 π} arccos (r_{2, 46.13}) - \frac{1}{4 π} arccos (r_{2, 56.13}))$
		$+ h (ρ_{12}, ρ_{14}, ρ_{24}) \cdot (\frac{1}{2} - \frac{1}{4 π} arccos (r_{1, 35.24}) - \frac{1}{4 π} arccos (r_{1, 36.24}) - \frac{1}{4 π} arccos (r_{1, 56.24}))$
		$+ h (ρ_{12}, ρ_{24}, ρ_{14}) \cdot (\frac{1}{2} - \frac{1}{4 π} arccos (r_{2, 35.14}) - \frac{1}{4 π} arccos (r_{2, 36.14}) - \frac{1}{4 π} arccos (r_{2, 56.14}))$
		$+ h (ρ_{12}, ρ_{15}, ρ_{25}) \cdot (\frac{1}{2} - \frac{1}{4 π} arccos (r_{1, 34.25}) - \frac{1}{4 π} arccos (r_{1, 36.25}) - \frac{1}{4 π} arccos (r_{1, 46.25}))$
		$+ h (ρ_{12}, ρ_{25}, ρ_{15}) \cdot (\frac{1}{2} - \frac{1}{4 π} arccos (r_{2, 34.15}) - \frac{1}{4 π} arccos (r_{2, 36.15}) - \frac{1}{4 π} arccos (r_{2, 46.15}))$
		$+ h (ρ_{12}, ρ_{16}, ρ_{26}) \cdot (\frac{1}{2} - \frac{1}{4 π} arccos (r_{1, 34.26}) - \frac{1}{4 π} arccos (r_{1, 35.26}) - \frac{1}{4 π} arccos (r_{1, 45.26}))$
		$+ h (ρ_{12}, ρ_{26}, ρ_{16}) \cdot (\frac{1}{2} - \frac{1}{4 π} arccos (r_{2, 34.16}) - \frac{1}{4 π} arccos (r_{2, 35.16}) - \frac{1}{4 π} arccos (r_{2, 45.16})) + \dots,$

a total of $121$ terms. In formula (3.9) for $E (M^{2})$ in [1], a constant term $1$ should be inserted in front of the first summation; further, the last summation should be taken over both $k \neq i$ and $k \neq j$ (not merely $k \neq i$ ).

3 Time Series

Consider a discrete-time stationary first-order autoregressive process

\begin{matrix} X_{t} = ρ X_{t - 1} + \sqrt{1 - ρ^{2}} \cdot ε_{t}, & - \infty < t < \infty, & | ρ | < 1 \end{matrix}

where $ε_{t}$ is $N (0, 1)$ white noise. The $ℓ \times ℓ$ covariance matrix $R$ has $i j^{th}$ element

ρ_{i j} = ρ^{| j - i |}

which leads to certain simplifications. Let us make the reliance of $M$ on $ℓ$ explicit. We have

E (M_{2}) = \sqrt{\frac{1 - ρ}{π}},

E (M_{3}) = \sqrt{\frac{1 - ρ}{π}} + \sqrt{\frac{1 - ρ^{2}}{4 π}},

	$E (M_{4}) = \frac{1}{4 π} \sqrt{\frac{1 - ρ}{π}} [π + 2 arcsin (1 - \frac{2}{3 - ρ})] + \frac{1}{2 π} \sqrt{\frac{1 - ρ}{π}} [π + 2 arcsin (\frac{1 + 2 ρ - ρ^{2}}{\sqrt{(3 - ρ) (3 + ρ + ρ^{2} - ρ^{3})}})]$
	$+ \frac{1}{2 π} \sqrt{\frac{1 - ρ^{2}}{π}} [π + 2 arcsin (\frac{(1 - ρ)^{2}}{\sqrt{(3 - ρ) (3 + ρ + ρ^{2} - ρ^{3})}})] + \frac{1}{4 π} \sqrt{\frac{1 - ρ^{3}}{π}} [π + 2 arcsin (1 - \frac{2}{3 + ρ + ρ^{2} - ρ^{3}})]$

but $E (M_{5})$ is too lengthy to record here. In the limit as $ρ \to 0$ , we obtain

\frac{1}{\sqrt{π}}, \frac{3}{2 \sqrt{π}}, \frac{3}{\sqrt{π}} [1 - \frac{1}{π} arcsec (3)], \frac{5}{\sqrt{π}} [1 - \frac{3}{2 π} arcsec (3)]

for $ℓ = 2, 3, 4, 5$ and these are consistent with well-known values [5] corresponding to independent $X_{t}$ . Figure 1 displays $E (M_{ℓ})$ as functions of $ρ$ . The left-hand endpoint is at $(- 1, \sqrt{2 / π})$ and $\sqrt{2 / π}$ is unsurprisingly the mean of a standard half-normal distribution. The right-hand endpoint is at $(1, 0)$ . Associated with $ℓ = 3, 4, 5$ are maximum points with $ρ$ equal to

\frac{1 - \sqrt{5}}{2} = - 0.6180339887498948482045868...,

\begin{matrix} - 0.4973597615161907364022217..., & - 0.43364768431626561412 75672... \end{matrix}

respectively. Closed-form expressions for the latter two quantities remain open.

We also have

E (M_{3}^{2}) = 1 + \frac{(1 - ρ) \sqrt{(3 - ρ) (1 + ρ)}}{2 π},

E (M_{4}^{2}) = 1 + \frac{3 + \sqrt{(3 - ρ) (3 + ρ + ρ^{2} - ρ^{3})} + ρ [1 - 2 ρ - 2 ρ^{2} - ρ^{3} + ρ^{4} - ρ \sqrt{(3 - ρ) (3 + ρ + ρ^{2} - ρ^{3})}]}{2 π \sqrt{(1 + ρ) (3 + ρ + ρ^{2} - ρ^{3})}}

but $E (M_{5}^{2})$ and $E (M_{6}^{2})$ are too lengthy to record here. In the limit as $ρ \to 0$ , we obtain

1 + \frac{\sqrt{3}}{2 π}, 1 + \frac{\sqrt{3}}{π}, 1 + \frac{5 \sqrt{3}}{2 π} [1 - \frac{1}{π} arcsec (4)], 1 + \frac{5 \sqrt{3}}{π} [1 - \frac{3}{2 π} arcsec (4)]

for $ℓ = 3, 4, 5, 6$ and these again are consistent with well-known values [5]. Associated with $ℓ = 3, 4, 5, 6$ are maximum points with $ρ$ equal to

1 - \sqrt{2} = - 0.4142135623730950488016887,

\begin{matrix} - 0.3879232988398265768779440..., & - 0.35992671048296895553 67968... \end{matrix}

- 0.3406053067160525788737944...

respectively. Closed-form expressions for the latter three quantities remain open. Figure 2 displays $V (M_{ℓ}) = E (M_{ℓ}^{2}) - E {(M_{ℓ})}_{ℓ}^{2}$ as functions of $ρ$ . The left-hand endpoint is at $(- 1, 1 - 2 / π)$ ; the right-hand endpoint is at $(1, 1)$ . Unlike $E (M_{ℓ})$ or $E (M_{ℓ}^{2})$ , the variance is strictly increasing throughout the interval. An intuitive reason for such behavior would be good to establish someday.

4 Proof of Revision

Our general formula for $E (M^{2})$ looks somewhat different from that presented by Afonja [1]. To demonstrate the equivalence of the two formulas, it suffices to prove that if $j \neq i$ , $k \neq i$ and $k \neq j$ , then

\frac{1}{2 π} r_{i, j i} \cdot \frac{r_{i, k i} - r_{i, j k} r_{i, j i}}{\sqrt{1 - r_{i, j k}^{2}}} = h (ρ_{i j}, ρ_{i k}, ρ_{j k}) .

The left-hand side is equal to

	$\frac{1}{2 π} \sqrt{\frac{1 - ρ_{i j}}{2}} \cdot \frac{\sqrt{\frac{1 - ρ_{i k}}{2}} - \frac{1 - ρ_{i j} - ρ_{i k} + ρ_{j k}}{\sqrt{4 (1 - ρ_{i j}) (1 - ρ_{i k})}} \sqrt{\frac{1 - ρ_{i j}}{2}}}{\sqrt{1 - \frac{(1 - ρ_{i j} - ρ_{i k} + ρ_{j k})^{2}}{4 (1 - ρ_{i j}) (1 - ρ_{i k})}}}$
	$= \frac{1}{4 π} \sqrt{1 - ρ_{i j}} \cdot \frac{\sqrt{1 - ρ_{i k}} - \frac{1 - ρ_{i j} - ρ_{i k} + ρ_{j k}}{\sqrt{4 (1 - ρ_{i k})}}}{\sqrt{1 - \frac{}{(1 - ρ_{i j} - ρ_{i k} + ρ_{j k})^{2}} 4 (1 - ρ_{i j}) (1 - ρ_{i k})}}$
	$= \frac{1}{4 π} \sqrt{1 - ρ_{i j}} \cdot \frac{2 (1 - ρ_{i k}) - (1 - ρ_{i j} - ρ_{i k} + ρ_{j k})}{\sqrt{4 (1 - ρ_{i k}) - \frac{(1 - ρ_{i j} - ρ_{i k} + ρ_{j k})^{2}}{1 - ρ_{i j}}}}$
	$= \frac{1}{4 π} (1 - ρ_{i j}) \cdot \frac{1 + ρ_{i j} - ρ_{i k} - ρ_{j k}}{\sqrt{4 (1 - ρ_{i j}) (1 - ρ_{i k}) - (1 - ρ_{i j} - ρ_{i k} + ρ_{j k})^{2}}}$

which is the right-hand side, as was to be shown.

5 Proof from First Principles

An exercise in [6] suggests that formulas for $E (M_{2})$ and $V (M_{2})$ should be derived from

max {X_{1}, X_{2}} = \frac{1}{2} (X_{1} + X_{2}) + \frac{1}{2} | X_{1} - X_{2} | .

It is instructive to similarly prove our formula for $E (M_{3})$ , using instead

	$max {X_{1}, X_{2}, X_{2}}$	$= max {max {X_{1}, X_{2}}, max {X_{2}, X_{3}}}$
		$= \frac{1}{2} (max {X_{1}, X_{2}} + max {X_{2}, X_{3}}) + \frac{1}{2} \| max {X_{1}, X_{2}} - max {X_{2}, X_{3}} \|$
		$= \frac{1}{4} ((X_{1} + X_{2}) + (X_{2} + X_{3}) + \| X_{1} - X_{2} \| + \| X_{2} - X_{3} \|)$
		$+ \frac{1}{4} \| (X_{1} + X_{2}) - (X_{2} + X_{3}) + \| X_{1} - X_{2} \| - \| X_{2} - X_{3} \| \| .$

Define $Y = X_{1} - X_{2}$ and $Z = X_{3} - X_{2}$ . Clearly $(Y, Z)$ is bivariate normally distributed with vector mean zero and covariance matrix

Also

\frac{1}{4} E | Y | = \frac{σ_{y}}{4} \sqrt{\frac{2}{π}} = \sqrt{\frac{1 - ρ_{12}}{4 π}},

\frac{1}{4} E | Z | = \frac{σ_{z}}{4} \sqrt{\frac{2}{π}} = \sqrt{\frac{1 - ρ_{23}}{4 π}} .

The four integrals (depending on signs of $Y$ and $Z$ ) underlying

\frac{1}{4} E | (Y + | Y |) - (Z + | Z |) | = \sqrt{\frac{1 - ρ_{13}}{4 π}}

can all be evaluated (however tediously). Because

X_{1} - X_{3} = (X_{1} - X_{2}) - (X_{3} - X_{2}) = Y - Z

we suspect that a more elegant proof ought to be available. Ideas on bridging this gap would be welcome.

In more detail, letting

f (y, z) = \frac{1}{2 π \sqrt{1 - ξ^{2}} σ_{y} σ_{z}} exp [- \frac{1}{} 2 (1 - ξ^{2}) (\frac{y^{2}}{σ_{y}^{2}} - \frac{2 ξ y z}{σ_{y} σ_{z}} + \frac{z^{2}}{σ_{z}^{2}})]

denote the bivariate normal density, we obtain

\infty \int 0 \infty \int 0 2 | y - z | f (y, z) d y d z = \frac{1}{\sqrt{2 π}} [- (1 - ξ) (σ_{y} + σ_{z}) + 2 \sqrt{σ_{y}^{2} - 2 ξ σ_{y} σ_{z} + σ_{z}^{2}}]

when $Y > 0$ and $Z > 0$ ;

\infty \int 0 0 \int - \infty 2 z f (y, z) d y d z = \frac{(1 - ξ) σ_{z}}{\sqrt{2 π}}

when $Y < 0$ and $Z > 0$ ;

0 \int - \infty \infty \int 0 2 y f (y, z) d y d z = \frac{(1 - ξ) σ_{y}}{\sqrt{2 π}}

when $Y > 0$ and $Z < 0$ ; and $0$ when $Y < 0$ and $Z < 0$ . Adding these contributions and dividing by $4$ , we verify

	$\frac{1}{2 \sqrt{2 π}} \sqrt{σ_{y}^{2} - 2 ξ σ_{y} σ_{z} + σ_{z}^{2}}$	$= \frac{1}{2 \sqrt{π}} \sqrt{(1 - ρ_{12}) - (ρ_{13} - ρ_{12} - ρ_{23} + 1) + (1 - ρ_{23})}$
		$= \sqrt{\frac{1 - ρ_{13}}{4 π}}$

as was desired.

Calculating the variance of $M_{3}$ from first principles has not been attempted. The variance of the median ( $50 %$ -tile) is also of interest, appearing explicitly in [7] for $ℓ = 3$ but under the assumption of independence.

An alternative probability density-based derivation of

E (M_{2})

and

V (M_{2})

can be found in [8, 9]. See also [10] for the expected range of a normal sample, [11] for the expected absolute maximum, and [12] for other aspects of AR(1).

6 Large Segments

Assuming $ρ_{i j}$ depends only on $| j - i | = d$ , Berman [13, 14, 15] proved that if either

\begin{matrix} lim d \to \infty ρ (d) \cdot ln (d) = 0 & or & \infty \sum d = 1 ρ (d)^{2} < \infty, \end{matrix}

then

lim ℓ \to \infty P {\sqrt{2 ln (ℓ)} (M_{ℓ} - a_{ℓ}) \leq x} = exp (- e^{- x})

where

a_{ℓ} = \sqrt{2 ln (ℓ)} - \frac{1}{2} \frac{ln (ln (ℓ)) + ln (4 π)}{\sqrt{2 ln (ℓ)}} .

Further, the two hypotheses on $ρ (d)$ cannot be significantly weakened. This theorem clearly applies for a first-order autoregressive process, although we note that $a_{ℓ}$ does not incorporate lag-one correlation $ρ$ at all. A more precise asymptotic result might do so.

Other relevant works in the literature include [16, 17, 18, 19, 20, 21, 22, 23, 24]. In particular, Figure 2 of [19] depicts the density of AR(1) maximum for $ℓ = 5$ and $ρ = - 9 / 10$ , $- 8 / 10$ , …, $8 / 10$ , $9 / 10$ .

7 Acknowledgements

Raymond Kan [25] symbolically evaluated the integrals in Section 5, at my request, for the special case

	$max {X_{1}, X_{2}, X_{2}}$	$= max {max {X_{1}, X_{2}}, max {X_{2}, X_{3}}}$
		$= \frac{1}{2} (max {X_{1}, X_{2}} + max {X_{2}, X_{3}}) + \frac{1}{2} \| max {X_{1}, X_{2}} - max {X_{2}, X_{3}} \|$
		$= \frac{1}{4} ((X_{1} + X_{2}) + (X_{2} + X_{3}) + \| X_{1} - X_{2} \| + \| X_{2} - X_{3} \|)$
		$+ \frac{1}{4} \| (X_{1} + X_{2}) - (X_{2} + X_{3}) + \| X_{1} - X_{2} \| - \| X_{2} - X_{3} \| \| .$

Moments of Maximum: Segment of AR(1)

Number of Sign Changes: Segment of AR(1)

Piercing Diametral Disks Induced by Edges of Maximum Spanning Tree

Learning the Number of Autoregressive Mixtures in Time Series Using the Gap Statistics

P not equal to NP

Towards Automated Let's Play Commentary

Maximum Number of Steps of Topswops on 18 and 19 Cards

Enforcing stationarity through the prior in vector autoregressions

1 Partial Correlations

2 Small Segments

3 Time Series

4 Proof of Revision

5 Proof from First Principles

6 Large Segments

7 Acknowledgements

Moments of Maximum: Segment of AR(1)

Related Research

Number of Sign Changes: Segment of AR(1)

Piercing Diametral Disks Induced by Edges of Maximum Spanning Tree

Learning the Number of Autoregressive Mixtures in Time Series Using the Gap Statistics

P not equal to NP

Towards Automated Let's Play Commentary

Maximum Number of Steps of Topswops on 18 and 19 Cards

Enforcing stationarity through the prior in vector autoregressions

1 Partial Correlations

2 Small Segments

3 Time Series

4 Proof of Revision

5 Proof from First Principles

6 Large Segments

7 Acknowledgements