[jason1990]

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Rereading jason's solution, this is making a bit more sense: jason has started his frog at 0, and asked how many points in (-L,L) get visited as L gets large. The vast majority of points left of the frog's starting point never get visited, since the frog drifts to the right.

That is, the fraction of points in (-L,0) that get visited approaches 0 as L becomes large, trivially; while the fraction of points in (0,L) approaches ~0.854 as L becomes large (and this is the one that takes all the work.)

Yes, the original post was a bit ill-defined.

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I just want to clarify something I said earlier. If the frog starts at any fixed point x, then the fraction of points in [-L,0] that get visited will also go to 0, since L will eventually be very far to the left of x. So this "principal value" limit is 1 - M, no matter where he starts. It then follows that it is also 1 - M if his starting point has any probability distribution on the integers. So starting the frog at 0 is not necessary for this result.

Moreover, the limit, as (L_1,L_2) -> (-infinity, infinity), of the fraction of missed sites in [L_1,L_2] does not exist.

The only way I can see to get around this is to somehow start the frog "at -infinity." I am not claiming that it is impossible to rigorously make sense of this. I just cannot presently see a sensible way to do it.