12F (Probability)

For

x \geq m

X_{n} = x

implies

M_{n} \geq m

. For

x < m

, reflect every step after the first visit to

m

x

with unrestricted paths ending at

2 m - x

. Therefore

P (M_{n} \geq m, X_{n} = x) = {P (X_{n} = x), P (X_{n} = 2 m - x), x \geq m, x < m .

ii

Summing part i and reindexing gives

P (M_{n} \geq m) = P (X_{n} = m) + 2 x > m \sum P (X_{n} = x) .

Subtracting the formula for

m + 1

yields

P (M_{n} = m) = P (X_{n} = m) + P (X_{n} = m + 1) .

Using part ii and symmetry of the simple random walk,

E X_{n}^{2} - E M_{n}^{2} = \sum_{k \geq 1} (2 k - 1) P (X_{n} = k) > 0

for

n \geq 1

. Thus $E M_{n}^{2} < E X_{n}^{2} = n$ .

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