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Consider the specific example where a row of light bulbs (initially off), arbitrarily turn on at different rates. For a light bulb at position $x$, call this rate $f(x)$. This is an irreversible process, so if a light is on, it stays on.

Now consider $p(x,t)$ the probability of a light bulb turning on at position $x$ and time $t$ (provided it has not yet been turned on). How are $f$ and $p$ related?

sam wolfe
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  • What do you mean by "rate"? Probably not a rate of change, since the change is discrete. Also probably not a frequency, since it's not periodic. – Vercassivelaunos Nov 10 '23 at 13:04
  • I think this is a problem with multiple Poisson processes, one for each light, looking for first occurrence. – Ethan Bolker Nov 10 '23 at 13:04
  • @Vercassivelaunos I should have clarified, but Ethan Bolker is right, this turning on should be interpreted as a Poisson process with rate $f$. The time of turning on is exponentially distributed with rate $1/f$. – sam wolfe Nov 10 '23 at 14:33

2 Answers2

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$f(x) = K$ looks like i.e. light bulb $x$, is off, turns on at, and stays on after, $K$ units of time.

$P(x_{on}, t < K) = 0$ and $P(x_{on}, t \geq K) = 1$

Hudjefa
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If there is an exponential distribution with rate $f(x)$,

then the probability that light $x$ is on by time $t$ is $p(x,t)=1-e^{-t \, f(x)}$

and is off with probability $1-p(x,t) = e^{-t \, f(x)}$

with a density for the time of switching on $\frac{dp}{dt} = f(x)\, e^{-t \, f(x)}$.

The probability all the lights are on by time $t$ is $\prod\limits_x P(x,t)=\prod\limits_x (1-e^{-t \, f(x)})$.

Henry
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