So again, I wasn’t referring to the expected value of the number of steps, but instead how we should update after learning about the time – that is, I wasn’t talking about E[k|t], but instead P(k|t)/P(k) for various k.
Let’s dig into this. From Bayes, we have: P(k|t)/P(k)=P(t|k)/P(t). As you say, P(t|k) ~ kt^(k-1). We have the pesky P(t) term, but we can note that for any value of t, this will yield a constant, so we can discard it and recognize that now we don’t get a value for the update, but instead just a relative value (we can’t say how large the update is at any individual k, but we can compare the updates for different k). We are now left with P(k|t)/P(k) ~ kt^(k-1), holding t constant. Using the empirical value on Earth of t=(4.5/5.5)=0.82, we get P(k|t=0.82)/P(k) ~ k*0.82^(k-1).
If we graph this, we get:
which apparently has its maximum at 5. That is, whatever the expected value for the number of steps is after considering the time, if we do update on the time, the largest update is in favor of there having been 5 steps. Compared to other plausible numbers for k, the update is weak, though – this partiuclar piece of evidence is a <2x update on there having been 5 steps compared to there having been 2 steps or 10 steps; the relative update for 5 steps is only even ~5x the size of the update for 20 steps.
Considering the general case (where we don’t know t), we can find the maximum of the update by setting the derivative of kt^(k-1) equal to zero. This derivative is (k ln(t) + 1)t^(k-1), and so we need k∗ln(t)=−1, or k=−1/ln(t). If we replace t with x/(x+1), such that x corresponds to the naive number of steps as I was calculating before, then that’s k=−1/ln(x/(x+1)). Here’s what we get if we graph that:
This is almost exactly my original guess (though weirdly, ~all values for k are ~0.5 higher than the corresponding values of x).
So again, I wasn’t referring to the expected value of the number of steps, but instead how we should update after learning about the time – that is, I wasn’t talking about E[k|t], but instead P(k|t)/P(k) for various k.
Let’s dig into this. From Bayes, we have: P(k|t)/P(k)=P(t|k)/P(t). As you say, P(t|k) ~ kt^(k-1). We have the pesky P(t) term, but we can note that for any value of t, this will yield a constant, so we can discard it and recognize that now we don’t get a value for the update, but instead just a relative value (we can’t say how large the update is at any individual k, but we can compare the updates for different k). We are now left with P(k|t)/P(k) ~ kt^(k-1), holding t constant. Using the empirical value on Earth of t=(4.5/5.5)=0.82, we get P(k|t=0.82)/P(k) ~ k*0.82^(k-1).
If we graph this, we get:
which apparently has its maximum at 5. That is, whatever the expected value for the number of steps is after considering the time, if we do update on the time, the largest update is in favor of there having been 5 steps. Compared to other plausible numbers for k, the update is weak, though – this partiuclar piece of evidence is a <2x update on there having been 5 steps compared to there having been 2 steps or 10 steps; the relative update for 5 steps is only even ~5x the size of the update for 20 steps.
Considering the general case (where we don’t know t), we can find the maximum of the update by setting the derivative of kt^(k-1) equal to zero. This derivative is (k ln(t) + 1)t^(k-1), and so we need k∗ln(t)=−1, or k=−1/ln(t). If we replace t with x/(x+1), such that x corresponds to the naive number of steps as I was calculating before, then that’s k=−1/ln(x/(x+1)). Here’s what we get if we graph that:
This is almost exactly my original guess (though weirdly, ~all values for k are ~0.5 higher than the corresponding values of x).
I agree with those computations/results. Thank you