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Author | Topic: On the proportion of Nucleotides in the Genome and what it can tell us about Evolutio | |||||||||||||||||||||||
Dr Adequate Member (Idle past 284 days) Posts: 16113 Joined: |
I think it does Dr A - the expectation value of the excess of heads or tails is the square root of n, where n is the number of tosses My emphasis. Do we not expect any particular random walk to meet the heads = tails axis infinitely many times as n tends to infinity? Especially bearing in mind that this particular random walk has a floor and a ceiling. But thanks for telling me that, 'cos I'd been wondering what the function was. Do you have a reference? But I'm not sure that it's relevant to my point. Edited by Dr Adequate, : No reason given. Edited by Dr Adequate, : No reason given.
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Dr Jack Member Posts: 3514 From: Immigrant in the land of Deutsch Joined: Member Rating: 8.7 |
The
The reason for this is quite obvious if you think about it: because all future flips occur independently of those flips that have already happened, the chances are that there will not be a trend reversing any initial discrepency from zero. Edited by Mr Jack, : Added explaination Edited by Mr Jack, : Meant mode not median
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Dr Adequate Member (Idle past 284 days) Posts: 16113 Joined: |
The median number of times for a sequence of heads and tails to cross the zero line is 0, with all higher numbers occur with decreasing probibility. How can zero be the median?
The reason for this is quite obvious if you think about it: because all future flips occur independently of those flips that have already happened, the chances are that there will not be a trend reversing any initial discrepency from zero. No trend, certainly. What of it? --- I think you're still confusing the average difference from zero as n tends to infinity with the path of a particular random walk. --- In any case, we are here discussing a random walk with a floor and a ceiling.
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jacortina Member (Idle past 5083 days) Posts: 64 Joined: |
How can zero be the median? Talking about the median in this context actually means (the way I see it) comparing multiple sequences of coin flipping. If, in each of five sequences, the number of times the zero-difference point is crossed are (0,0,0,2,5), the median is zero. With six sequences (0,0,0,0,3,4), the median is also zero. Zero can certainly be the middle number of an ordered set with an odd number of members or the average of the two middle numbers of an ordered set with an even number of members. Basically, claiming a zero median is claiming that more than half the time, a sequence will NOT cross the zero-difference line.
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Dr Jack Member Posts: 3514 From: Immigrant in the land of Deutsch Joined: Member Rating: 8.7 |
How can zero be the median? Sorry, I meant mode. My bad.
I think you're still confusing the average difference from zero as n tends to infinity with the path of a particular random walk. No, I'm not. More paths will not ever cross the zero line than any other particular number of crosses; what is more as the number of crosses increases, the number of paths that have that number of crosses decreases.
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Dr Adequate Member (Idle past 284 days) Posts: 16113 Joined: |
Talking about the median in this context actually means (the way I see it) comparing multiple sequences of coin flipping. One might call it the "mean median".
If, in each of five sequences, the number of times the zero-difference point is crossed are (0,0,0,2,5), the median is zero. With six sequences (0,0,0,0,3,4), the median is also zero. Zero can certainly be the middle number of an ordered set with an odd number of members or the average of the two middle numbers of an ordered set with an even number of members. Basically, claiming a zero median is claiming that more than half the time, a sequence will NOT cross the zero-difference line.
How can zero be the median? Talking about the median in this context actually means (the way I see it) comparing multiple sequences of coin flipping. If, in each of five sequences, the number of times the zero-difference point is crossed are (0,0,0,2,5), the median is zero. With six sequences (0,0,0,0,3,4), the median is also zero. Zero can certainly be the middle number of an ordered set with an odd number of members or the average of the two middle numbers of an ordered set with an even number of members. Basically, claiming a zero median is claiming that more than half the time, a sequence will NOT cross the zero-difference line. Right, that's exactly what is being claimed. Now let's think this through. In the first two tosses of the coin, we will either get HH or TT, in which case we will not meet the h = t axis --- or we will get HT or TH, in which case we will meet the h = t axis. So after just the first two flips of the coin, it's fifty-fifty that we shall have met the h = t axis once. Now, further flips of the coin can only increase, not reduce, the probability that we meet the h = t axis at least once. Therefore, for n > 2, the "mean median" cannot be zero.
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NosyNed Member Posts: 8996 From: Canada Joined: |
So after just the first two flips of the coin, it's fifty-fifty that we shall have met the h = t axis once. Meeting is not crossing. In these examples the crossings are zero. (presuming I understand what is being put forward)
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Dr Jack Member Posts: 3514 From: Immigrant in the land of Deutsch Joined: Member Rating: 8.7 |
Nosy, it correct, I was discussing crossings, not meetings.
And, as I said, I meant mode not median. Again, my bad.
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Dr Adequate Member (Idle past 284 days) Posts: 16113 Joined: |
Sorry, I meant mode. My bad. Oh, now you're going to bring the mode into this? Apart from wondering why, I'd like to see your reasoning.
No, I'm not. More paths will not ever cross the zero line than any other particular number of crosses; what is more as the number of crosses increases, the number of paths that have that number of crosses decreases. I didn't follow that. Let me talk about the mean. You remember that, the thing we use when doing statistics and talking about expected values? Well, rather than figure it out from first principles, I just got my computer to toss lots of coins for me. Each number in the right hand column is the mean of a thousand trials.
number of coin tosses number of times heads - tails = 0 2: 0.500 4: 0.853 8: 1.438 16: 2.360 32: 3.644 64: 5.165 128: 8.226 256: 11.866 512: 17.422 1024: 23.681 2048: 34.958 4096: 52.290 8192: 72.220 16384: 101.483 32768: 139.231 65536: 201.394 131072: 275.738 262144: 420.173 524288: 560.344 1048576: 826.211 I therefore stand by my claim that the number of times we'll meet the h = t axis will indeed tend to infinity with the number of coin tosses. If you would like to dispute that, perhaps you could try thinking for a few moments about where, in that case, the asymptote would be. Edited by Dr Adequate, : No reason given. Edited by Dr Adequate, : No reason given. Edited by Dr Adequate, : No reason given.
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Dr Adequate Member (Idle past 284 days) Posts: 16113 Joined: |
Meeting is not crossing. In these examples the crossings are zero. (presuming I understand what is being put forward) Apparently not. If the number of heads minus the number of tails is zero, then we have indeed met the h = t axis.
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Dr Jack Member Posts: 3514 From: Immigrant in the land of Deutsch Joined: Member Rating: 8.7 |
Mean is an extremely poor choice for a situation such as this - especially when considering the infinite limit - because of the large value distortion effect, and the non-normal distribution of the probabilities.
The most probable number of crosses is 0, followed by 1, followed by 2, followed by 3, etc.
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Dr Jack Member Posts: 3514 From: Immigrant in the land of Deutsch Joined: Member Rating: 8.7 |
But you've not crossed it unless the next toss is the same.
HTH meets the zero line, it does not cross itHTT meets and then crosses the zero line. Of the possible 3 toss sequences, then, HTT and THH cross the line once; the other six do not cross it at all.
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Dr Jack Member Posts: 3514 From: Immigrant in the land of Deutsch Joined: Member Rating: 8.7 |
After looking it up, I realise I have erred. In fact, the result I stated holds for a finite random walk, but it does not hold for an infinite random walk.
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Dr Adequate Member (Idle past 284 days) Posts: 16113 Joined: |
But you've not crossed it unless the next toss is the same. So, you are not merely wrong, you're also calculating the wrong thing. Wrongly. * facepalm * Edited by Dr Adequate, : No reason given.
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Dr Adequate Member (Idle past 284 days) Posts: 16113 Joined: |
Mean is an extremely poor choice for a situation such as this - especially when considering the infinite limit - because of the large value distortion effect, and the non-normal distribution of the probabilities. The most probable number of crosses is 0, followed by 1, followed by 2, followed by 3, etc. * jaw drops * I'm going to guess that you're not a mathematician.
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