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pmsrw3

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You have a tube, open at the top. Raindrops fall into the mouth of the tube at a mean rate i per second, 0 <= i < 1, in a Poisson process. There's a hole in the bottom of the tube. When there's water in the tube, it flows out at a constant rate of 1 drop/sec. (Ignore that this is not physically plausible. You can imagine that there's a peristaltic pump hooked up if it makes you feel better.) What is the mean amount of water in the tube at steady-state?

I came up with a complicated solution to this (which I will describe, if anyone wishes) in the form of a sum. But when I evaluated my solution numerically, I found that, within round-off error, computed values equaled the simple result <v> = i/(2(1 - i)). This is also plausible and has correct limiting behavior at 0 and 1. Since then I've been trying hard without success to come up with a proof or even fairly good rationale for that formula.

Thanks in advance for any ideas.