I could have said that the beauty was simulated floor(5^x) times where x is a random real between 0 and n
Ah, I see now what you mean. Disregarding this new problem for the moment, you can still formulate my original expression on a per-interview basis, and it will still have the same Cesàro sum because it still diverges in the same manner; it just does so more continuously. If you envision a graph of an isomorphic series of my original expression, it will have “saw teeth” where it alternates between even and odd coin flips, and if you formulate that series on a per-interview basis, those saw teeth just get increasingly longer, which has no impact on the Cesàro sum (because the series alternates between those saw teeth).
Concerning your new problem, it can still be expressed as a series with a Cesàro sum, it’s just a lot more complicated. If I were you, I’d first try to find the simplest variant of that problem with the same properties. Still, the fact that this is solvable in an analogous way should be clear, because you can essentially solve the “floor(5^x) times where x is a random real between 0 and n” part with a series for x (similar to the one for the original problem) and then have a series of those series for n. Basically you’re adding another dimension (or recursion level), but not doing anything fundamentally different.
I haven’t actually done the math yet, but I don’t believe you. I think that if your terms are “per interview” then the more recent chunk of 100% even will overpower all the older stuff because there are so many of them, and the series of averages will oscillate.
I take it that my approach was not discussed in the heated debate you had? Because it seems a good exercise for grad students.
Also, I don’t understand why you think a per interview series would net fundamentally different results than a per coin toss series. I’d be interested in your reports after you (or your colleagues) have done the math.
I have no need for the new continuous question, if you are happy with saying that a per day analysis is no less arbitrary than a coin flip analysis.
The math is proving to be too much work to write up, so ill just tell you why I think there is a difference between per day and per coin flip. In the per coin flip, you take each of the possible coin flip sequences with equal weight when you are taking the averages of the partial sums in the Cesàro sums. In the per day analysis, you are putting much much more weight on the coin flip sequences which have more flips, because there are many more days which include them.
The “many more days that include them” is the 3^n part in my expression that is missing from any per day series. This 3^n is the sum of all interviews in that coin flip sequence (“coin flip sequence” = “all the interviews that are done because one coin flip showed up tails”, right?) and in the per day (aka per interview) series the exact same sum exists, just as 3^n summands.
In both cases, the weight of the later coin flip sequences increases, because the number of interviews (3^n) increases faster than the probabilistic weight of the coin flip (1/2^n) decreases.
However, this doesn’t mean that there exists no Cesàro sum. In fact the existence of such a sum can be proven for my original expression because the quotient of the last two numerators (if we include both odd and even coin flips) of the isomorphic series is always 3:1, regardless of wether the last coin flip was even or odd. (The same thing can be said for the quotient of the last 3^n and 3^(n-1) summands of your series. Basically, the per day series is just a dragged out per coin flip series.)
The reason why my estimation for the Cesàro sum is 0.5 is that if we express that quotient in a way that one coin state is written first, then it alternates between 3:1 and 1:3, which results in 1:1 which is 0.5. Obviously this is not exact maths, but it’s a good way for a quick estimation. (Alternatively, you could intuitively infer that if there exists a Cesàro sum it must be 0.5, because whether you look for even or odd coin flips gets increasingly irrelevant as the series approaches infinity.)
Also, since I haven’t previously touched upon the subject of the isomorphic series: If we call my original expression f, then we can construct a function g where g(n) = f(n)-f(n-1) with f(-1) = 0, and a series a = g(0) + g(1) + g(2) + …
That is wrong. Sorry. floor(3^x) doesn’t work because sqrt(3)<2. Try floor(5^x).
Ah, I see now what you mean. Disregarding this new problem for the moment, you can still formulate my original expression on a per-interview basis, and it will still have the same Cesàro sum because it still diverges in the same manner; it just does so more continuously. If you envision a graph of an isomorphic series of my original expression, it will have “saw teeth” where it alternates between even and odd coin flips, and if you formulate that series on a per-interview basis, those saw teeth just get increasingly longer, which has no impact on the Cesàro sum (because the series alternates between those saw teeth).
Concerning your new problem, it can still be expressed as a series with a Cesàro sum, it’s just a lot more complicated. If I were you, I’d first try to find the simplest variant of that problem with the same properties. Still, the fact that this is solvable in an analogous way should be clear, because you can essentially solve the “floor(5^x) times where x is a random real between 0 and n” part with a series for x (similar to the one for the original problem) and then have a series of those series for n. Basically you’re adding another dimension (or recursion level), but not doing anything fundamentally different.
I haven’t actually done the math yet, but I don’t believe you. I think that if your terms are “per interview” then the more recent chunk of 100% even will overpower all the older stuff because there are so many of them, and the series of averages will oscillate.
I take it that my approach was not discussed in the heated debate you had? Because it seems a good exercise for grad students.
Also, I don’t understand why you think a per interview series would net fundamentally different results than a per coin toss series. I’d be interested in your reports after you (or your colleagues) have done the math.
We did not discuss Cesàro sums.
I have no need for the new continuous question, if you are happy with saying that a per day analysis is no less arbitrary than a coin flip analysis.
The math is proving to be too much work to write up, so ill just tell you why I think there is a difference between per day and per coin flip. In the per coin flip, you take each of the possible coin flip sequences with equal weight when you are taking the averages of the partial sums in the Cesàro sums. In the per day analysis, you are putting much much more weight on the coin flip sequences which have more flips, because there are many more days which include them.
The “many more days that include them” is the 3^n part in my expression that is missing from any per day series. This 3^n is the sum of all interviews in that coin flip sequence (“coin flip sequence” = “all the interviews that are done because one coin flip showed up tails”, right?) and in the per day (aka per interview) series the exact same sum exists, just as 3^n summands.
In both cases, the weight of the later coin flip sequences increases, because the number of interviews (3^n) increases faster than the probabilistic weight of the coin flip (1/2^n) decreases.
However, this doesn’t mean that there exists no Cesàro sum. In fact the existence of such a sum can be proven for my original expression because the quotient of the last two numerators (if we include both odd and even coin flips) of the isomorphic series is always 3:1, regardless of wether the last coin flip was even or odd. (The same thing can be said for the quotient of the last 3^n and 3^(n-1) summands of your series. Basically, the per day series is just a dragged out per coin flip series.)
The reason why my estimation for the Cesàro sum is 0.5 is that if we express that quotient in a way that one coin state is written first, then it alternates between 3:1 and 1:3, which results in 1:1 which is 0.5. Obviously this is not exact maths, but it’s a good way for a quick estimation. (Alternatively, you could intuitively infer that if there exists a Cesàro sum it must be 0.5, because whether you look for even or odd coin flips gets increasingly irrelevant as the series approaches infinity.)
Also, since I haven’t previously touched upon the subject of the isomorphic series: If we call my original expression f, then we can construct a function g where g(n) = f(n)-f(n-1) with f(-1) = 0, and a series a = g(0) + g(1) + g(2) + …
Does that all make sense?