31.6. HITTING THIS BEFORE THAT 855

Consider this last term. By the definition, [τa = τb] corresponds to M (t) never hittingeither a or b. Since M (0) = 0, this can only happen if M (t) has values in [a,b] . Therefore,this last term satisfies

aP([τa = τb]∩ [M∗ > 0]) ≤∫[τa=τb]∩[M∗>0]

M (∞)dP

≤ bP([τa = τb]∩ [M∗ > 0]) (31.12)

Obviously the following inequality holds because on the left you have

aP([τa = τb]∩ [M∗ > 0])

and on the right you have the larger bP([τa = τb]∩ [M∗ > 0]) . That 0 is in the middlefollows from 31.11.

aP([τa = τb]∩ [M∗ > 0])+aP([τa < τb])+bP([τb < τa])≤

0≤ bP([τa = τb]∩ [M∗ > 0])+aP([τa < τb])+bP([τb < τa]) (31.13)

Note that [τb < τa] , [τa < τb]⊆ [M∗ > 0] and so

[τb < τa]∪ [τa < τb]∪ ([τa = τb]∩ [M∗ > 0]) = [M∗ > 0] (31.14)

The following diagram may help in keeping track of the various substitutions.

[τa < τb] [τb < τa] [τb = τa]∩ [M∗ > 0]

Left side of 31.13

From 31.14, this yields on substituting for P([τa < τb])

0 ≥ aP([τa = τb]∩ [M∗ > 0])+a [P([M∗ > 0])−P([τa ≥ τb]∩ [M∗ > 0])]+bP([τb < τa])

and so since [τa ̸= τb]⊆ [M∗ > 0] ,

0≥ a [P([M∗ > 0])−P([τa > τb])]+bP([τb < τa])

−aP([M∗ > 0])≥ (b−a)P([τb < τa]) (31.15)

Next use 31.14 to substitute for P([τb < τa])

0≥ aP([τa = τb]∩ [M∗ > 0])+aP([τa < τb])+bP([τb < τa])

= aP([τa = τb]∩ [M∗ > 0])+aP([τa < τb])

+b [P([M∗ > 0])−P([τa ≤ τb]∩ [M∗ > 0])]

= aP([τa ≤ τb]∩ [M∗ > 0])+b [P([M∗ > 0])−P([τa ≤ τb]∩ [M∗ > 0])]

and so(b−a)P([τa ≤ τb])≥ bP([M∗ > 0]) (31.16)