Originally posted by DeepThought
Can you apply this to a single radioactive atom?
No. h has got to be a specific observed value of a continuous random variable (such as, say, 2.234 ) , not merely a single event/non-event.
Say you have a load of live electric wires, each with a different voltage and it is this voltage that is the continuous random variable x we are considering. Lets say you are given info from a trustworthy source that the range of voltages in the wires is of, say, a continuous uniform distribution from 0 volt as its lower limit to some maximum upper limit of voltage, but you are not told what that upper limit is, only that it exists.
You then randomly pick one of these live wires and make a measurement with a volt meter. Say the reading you get just happens to be exactly 1.5 volts. Here in this context that will mean:
h = 1.5
hence the equation:
density( cumulative(h) = p ) = 1 where 0≤p≤1
density( cumulative(1.5) = p ) = 1 where 0≤p≤1
and now what that equation says about the cumulative of h, i.e. cumulative(1.5), is that, from your limited information, you should assume the probability of cumulative(1.5) being between, say, 0.1 and 0.2 of the actual distribution of x (so your unknown upper limit of the uniform distribution is somewhere between 1.5/0.1 = 15 volts, corresponding to that 0.1, and 1.5/0.2 = 7.5 volts, corresponding to that 0.2 ), is the same as cumulative(1.5) being between, say, 0.2 and 0.3 of the actual distribution of x. (and that probability in each case being 0.1 ). And the probability of cumulative(1.5) being less that half ( <0.5 ) of the actual distribution is the same as cumulative(1.5) being more than half ( >0.5 ), of the actual distribution (and that probability in each case being 0.5 so the probability of 1.5 being more than the median of the actual distribution is the same as it being less than the median of the actual distribution ). In other words, for any given magnitude of range of what proportion cumulative(1.5) is of the actual distribution, the probability is the same regardless of i.e independent of the values that range starts and ends with hence that is why my equation says the probability density
of "cumulative(h) = p" is "=1".
I hope I haven't made any part of that confusing; I confess I struggle somewhat to explain that concisely.
P.S. still not sure but I now think it probably is correct to call it "posterior probability" even though it cannot be derived from the standard equation for posterior probability. If that is right, I think I ought to invent a special name for this new kind of posterior probability. It is "new" because, 1, it cannot come form the standard equation for posterior probability, 2, unusually for legitimate probability, there exists no prior probabilities for the evidence thus; 3, it doesn't come from updating priors.