isov
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April 15, 2014, 10:22:04 PM |
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Quick TA update: - 6H candle color/volume: BREAKING - huge green candle beats the previous reds AND action afterwards is very promising, conclusion: probability of recent reversal further increased - Bid/ask strengh at market (Bitstamp): slippage to sell 5k: $84, slippage to buy: $143, conclusion: potential for extreme volatility - Trendline comparison: we are now at -0.341 log units. The trendline is at $1,015 and rising $7 per day, conclusion: rock bottom (note: it is not necessary that 'rock bottom' will change until the parabolic uptrend starts, because the trendline is itself rising) - Sentiment: short covering/panic buys are starting - Prognosis: getting better; probability for going <400 gets smaller by the day, long-term buy zone And only 1 day after... Quick TA update (at $500): - 6H candle color/volume: after a row of tall reds we have turned exclusively tall green, conclusion: reversal seems confirmed- Bid/ask strengh at market (Bitstamp): slippage to sell 5k: $77, slippage to buy: $83, conclusion: potential for high volatility - Trendline comparison: we are now at -0.310 log units. The trendline is at $1,022 and rising $7 per day, conclusion: rock bottom (note: it is not necessary that 'rock bottom' will change until the parabolic uptrend starts, because the trendline is itself rising) - Sentiment: awaiting a breakout or a pullback - Prognosis: reversal seems confirmed and <400 fades, long-term buy zone Good times ahead it appears! Hey, my fortune cookie from today at Panda Express said, "YOU WILL BE COMING INTO A FORTUNE" Does that count as confirmation!? Missing Panda express, maybe the best fast food joint there is. I guess it U.S. only. One week macd has also turned to declining red, that should be a nice sign.
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EuroTrash
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April 15, 2014, 10:32:30 PM |
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Here is the Blockchain adjusted transaction quantity chart, for 180 days using the 7-day moving average. The rightmost 4 days appear to confirm the reversal of the price trend. Thanks. I noticed we also had something similar hapoening between May and June 2013, and confirmed a reversal in price trend, albeit it was not the bottom.
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<=== INSERT SMART SIGNATURE HERE ===>
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frienemy
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Activity: 235
Merit: 100
I was promised da moon
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April 15, 2014, 10:34:59 PM |
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Quick TA update: - 6H candle color/volume: BREAKING - huge green candle beats the previous reds AND action afterwards is very promising, conclusion: probability of recent reversal further increased - Bid/ask strengh at market (Bitstamp): slippage to sell 5k: $84, slippage to buy: $143, conclusion: potential for extreme volatility - Trendline comparison: we are now at -0.341 log units. The trendline is at $1,015 and rising $7 per day, conclusion: rock bottom (note: it is not necessary that 'rock bottom' will change until the parabolic uptrend starts, because the trendline is itself rising) - Sentiment: short covering/panic buys are starting - Prognosis: getting better; probability for going <400 gets smaller by the day, long-term buy zone And only 1 day after... Quick TA update (at $500): - 6H candle color/volume: after a row of tall reds we have turned exclusively tall green, conclusion: reversal seems confirmed- Bid/ask strengh at market (Bitstamp): slippage to sell 5k: $77, slippage to buy: $83, conclusion: potential for high volatility - Trendline comparison: we are now at -0.310 log units. The trendline is at $1,022 and rising $7 per day, conclusion: rock bottom (note: it is not necessary that 'rock bottom' will change until the parabolic uptrend starts, because the trendline is itself rising) - Sentiment: awaiting a breakout or a pullback - Prognosis: reversal seems confirmed and <400 fades, long-term buy zone Good times ahead it appears! Hey, my fortune cookie from today at Panda Express said, "YOU WILL BE COMING INTO A FORTUNE" Does that count as confirmation!? Missing Panda express, maybe the best fast food joint there is. I guess it U.S. only. One week macd has also turned to declining red, that should be a nice sign. The name Panda Express says it all. That's the express train the panda bears are taking in order to catch upt to our beloved choo choo train so that they don't miss it. I was expecting a strong dent today after hitting 500 before we go further up again, but that hasn't come true. This short term recovery has quite a bite to it, leaving quite nice prospects for the long term. I'm tired of the bears saying we have to break thousands of downward trendlines before we can go up again. Once every single one of those lines is broken, we ARE already up again.
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MCTRL_751 > END OF LINE
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BitChick
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Activity: 1148
Merit: 1001
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April 15, 2014, 11:11:27 PM |
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The name Panda Express says it all. That's the express train the panda bears are taking in order to catch upt to our beloved choo choo train so that they don't miss it.
Ha Ha Ha. The irony that it was "Panda Express" was lost on me until you said that. That makes it an even greater sign that we are on the "Panda Express" train to the moon now.
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1BitcHiCK1iRa6YVY6qDqC6M594RBYLNPo
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SlipperySlope
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April 15, 2014, 11:33:06 PM |
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At the moment, China is just waking up and volume is smoothly climbing on Huobi's 30 minute chart as displayed at Bitcoin Wisdom. On the daily Bitstamp chart, $550 appears to be the trend line of resistance.
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rpietila (OP)
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April 16, 2014, 07:17:17 AM |
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I have not thought it through, wanted to have feedback! From my background the intuitive thing is to use F1 measure as a score. Or area under ROC curve. There are good wikipedia articles about these. These seemed to be concerned with binary outcomes. I want a metric that is continuous. I am developing one currently
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HIM TVA Dragon, AOK-GM, Emperor of the Earth, Creator of the World, King of Crypto Kingdom, Lord of Malla, AOD-GEN, SA-GEN5, Ministry of Plenty (Join NOW!), Professor of Economics and Theology, Ph.D, AM, Chairman, Treasurer, Founder, CEO, 3*MG-2, 82*OHK, NKP, WTF, FFF, etc(x3)
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rpietila (OP)
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April 16, 2014, 07:30:05 AM |
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I have not thought it through, wanted to have feedback! From my background the intuitive thing is to use F1 measure as a score. Or area under ROC curve. There are good wikipedia articles about these. These seemed to be concerned with binary outcomes. I want a metric that is continuous. I am developing one currently Eureka! It is this simple: - Every predictor gives two prices in log scale eg. "In 2014-5-16 the price is between 2.7 and 2.85 (roughly 500 and 700)" - When the actual price is known, you take min [ abs ( actual - upper_limit); abs ( actual - lower_limit) ]- Whoever has the lowest average error after a reasonable number of predictions (predictions can be renewed as often as you wish regardless of their maturity) is the best! - Proof omitted
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HIM TVA Dragon, AOK-GM, Emperor of the Earth, Creator of the World, King of Crypto Kingdom, Lord of Malla, AOD-GEN, SA-GEN5, Ministry of Plenty (Join NOW!), Professor of Economics and Theology, Ph.D, AM, Chairman, Treasurer, Founder, CEO, 3*MG-2, 82*OHK, NKP, WTF, FFF, etc(x3)
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yenom
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April 16, 2014, 08:19:33 AM |
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Eureka! It is this simple: - Every predictor gives two prices in log scale eg. "In 2014-5-16 the price is between 2.7 and 2.85 (roughly 500 and 700)" - When the actual price is known, you take min [ abs ( actual - upper_limit); abs ( actual - lower_limit) ]- Whoever has the lowest average error after a reasonable number of predictions (predictions can be renewed as often as you wish regardless of their maturity) is the best! - Proof omitted I would be very grateful if you could explain this to a simpleton like myself.
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phatsphere
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April 16, 2014, 09:16:11 AM |
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Those shorters may have a very hard time if Houbi breaks 2800CNY.
for the record: broken at around 10:00 UTC on 10k volume. There's nothing significant about 2800. Breaking 3200 would be significant. for the record: 3200 has been broken yesterday at 2300 UTC (on low volume). volume increased until 3450 and ended with a sharp 10k decline to retest 3200. this could be the first wave of pullbacks, but basically i had to reply to your earlier posting ;-)
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MahaRamana
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April 16, 2014, 11:17:48 AM |
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At the moment, China is just waking up and volume is smoothly climbing on Huobi's 30 minute chart as displayed at Bitcoin Wisdom. On the daily Bitstamp chart, $550 appears to be the trend line of resistance. Looks like we hit that ceiling. Next attempt will probably go through
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phatsphere
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April 16, 2014, 11:54:47 AM |
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My attempt at drawing random lines: and a 1h zoom:
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BitchicksHusband
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April 16, 2014, 12:19:45 PM |
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Eureka! It is this simple: - Every predictor gives two prices in log scale eg. "In 2014-5-16 the price is between 2.7 and 2.85 (roughly 500 and 700)" - When the actual price is known, you take min [ abs ( actual - upper_limit); abs ( actual - lower_limit) ]- Whoever has the lowest average error after a reasonable number of predictions (predictions can be renewed as often as you wish regardless of their maturity) is the best! - Proof omitted I would be very grateful if you could explain this to a simpleton like myself. Whoever was the closest to the actual price with the narrowest range was the best. I think he's being a little facetious here, because this is, of course, obvious.
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1BitcHiCK1iRa6YVY6qDqC6M594RBYLNPo
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bitfair
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April 16, 2014, 01:06:34 PM Last edit: April 16, 2014, 01:19:09 PM by bitfair |
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Eureka! It is this simple: - Every predictor gives two prices in log scale eg. "In 2014-5-16 the price is between 2.7 and 2.85 (roughly 500 and 700)" - When the actual price is known, you take min [ abs ( actual - upper_limit); abs ( actual - lower_limit) ]- Whoever has the lowest average error after a reasonable number of predictions (predictions can be renewed as often as you wish regardless of their maturity) is the best! - Proof omitted I would be very grateful if you could explain this to a simpleton like myself. Whoever was the closest to the actual price with the narrowest range was the best. I think he's being a little facetious here, because this is, of course, obvious. Except that it doesn't actually work. Proof by counterexample: Imagine a forecast range of 50-100. If the outcome if 95, i.e. within the range, the formula produces a score of 5. However, if the outcome is 105, i.e. outside the range, the formula produces a score of 5. But clearly, the first situation should score better, but with this formula it does not! QED? Edit: I can think of more examples where it doesn't work too, can I leave those as an exercise to the reader? Edit 2: For those wondering how to do it properly, I suggest searching the meteorology literature - it's much more comprehensive on this issue than the financial/economic/econometric literature.
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SlipperySlope
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April 16, 2014, 01:45:45 PM Last edit: April 16, 2014, 02:05:21 PM by SlipperySlope |
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Eureka! It is this simple: - Every predictor gives two prices in log scale eg. "In 2014-5-16 the price is between 2.7 and 2.85 (roughly 500 and 700)" - When the actual price is known, you take min [ abs ( actual - upper_limit); abs ( actual - lower_limit) ]- Whoever has the lowest average error after a reasonable number of predictions (predictions can be renewed as often as you wish regardless of their maturity) is the best! - Proof omitted I would be very grateful if you could explain this to a simpleton like myself. Whoever was the closest to the actual price with the narrowest range was the best. I think he's being a little facetious here, because this is, of course, obvious. Except that it doesn't actually work. Proof by counterexample: Imagine a forecast range of 50-100. If the outcome if 95, i.e. within the range, the formula produces a score of 5. However, if the outcome is 105, i.e. outside the range, the formula produces a score of 5. But clearly, the first situation should score better, but with this formula it does not! QED? Edit: I can think of more examples where it doesn't work too, can I leave those as an exercise to the reader? Edit 2: For those wondering how to do it properly, I suggest searching the meteorology literature - it's much more comprehensive on this issue than the financial/economic/econometric literature. Suppose rpietila's formula was amended to yield zero in the case that the prediction is within the range. Then your offered counterexample fails. Did you have others that would prove the amended scoring formula invalid? Note, to anyone interested, that the reduction of the price to log10 form allows the predictions to be compared on widely differing timescales, in which price values might be 10x larger or smaller. rpietila has been talking about deviation from the log10 trendline in terms of these log10 deltas. Accordingly, I added a column to my own logistic model of bitcoin prices which produces this log10 delta for each day's data that I record. https://docs.google.com/spreadsheet/ccc?key=0ArD8rjI3DD1WdFIzNDFMeEhVSzhwcEVXZDVzdVpGU2c
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bitfair
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April 16, 2014, 01:55:10 PM |
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Suppose rpietila's formula was amended to yield zero in the case that the prediction is within the range. Then your offered counterexample fails. Did you have others that would prove the amended scoring formula invalid?
Note, to anyone interested, that the reduction of the price to log10 form allows the predictions to be compared on widely differing timescales, in which price values might be 10x larger or smaller. rpietila has been talking about deviation from the log10 trendline in terms of these log10 deltas.
Fair enough, let's do another one: Imagine the actual outcome is 95. A prediction of 50-100 would score 5. A prediction of 89-100 would also score 5, although it is obviously much better than the first one. So the amended formula does not correctly rank the predictions either (at least not intuitively). Let me see if I can think of another one... Log scale or not doesn't really matter for comparing predictions to each other (by which I mean ranking/ordering them). Edit: Came to think of another counterexample for the amended formula: if the prediction is 50-100 and the outcome is 101, the amended formula gives score 0. If the outcome is 150, the amended formula also gives score 0, although the first case is obviously better than the second. (You might argue that this is "fair", though, but that is a subjective argument: the amended formula still clearly fails to correctly rank/order the two predictions).
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bb113
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April 16, 2014, 02:03:23 PM |
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Eureka! It is this simple: - Every predictor gives two prices in log scale eg. "In 2014-5-16 the price is between 2.7 and 2.85 (roughly 500 and 700)" - When the actual price is known, you take min [ abs ( actual - upper_limit); abs ( actual - lower_limit) ]- Whoever has the lowest average error after a reasonable number of predictions (predictions can be renewed as often as you wish regardless of their maturity) is the best! - Proof omitted I would be very grateful if you could explain this to a simpleton like myself. Whoever was the closest to the actual price with the narrowest range was the best. I think he's being a little facetious here, because this is, of course, obvious. Except that it doesn't actually work. Proof by counterexample: Imagine a forecast range of 50-100. If the outcome if 95, i.e. within the range, the formula produces a score of 5. However, if the outcome is 105, i.e. outside the range, the formula produces a score of 5. But clearly, the first situation should score better, but with this formula it does not! QED? Edit: I can think of more examples where it doesn't work too, can I leave those as an exercise to the reader? Edit 2: For those wondering how to do it properly, I suggest searching the meteorology literature - it's much more comprehensive on this issue than the financial/economic/econometric literature. Suppose rpietila's formula was amended to yield zero in the case that the prediction is within the range. Then your offered counterexample fails. Did you have others that would prove the amended scoring formula invalid? Note, to anyone interested, that the reduction of the price to log10 form allows the predictions to be compared on widely differing timescales, in which price values might be 10x larger or smaller. rpietila has been talking about deviation from the log10 trendline in terms of these log10 deltas. What you want to do is reward for the prediction being within the interval and penalize for distance from the interval as well as for the width of the interval. For example: I= upper_limit -lower_limit W=Weight of Interval width penalty P= Actual
if( P>lower_limit AND P<upper_limit){ D=0 }else{ D=min [ abs ( P - upper_limit); abs ( P - lower_limit) ] }
Score= D+W*I
W could even be 1, IDK. Lowest score wins.
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envy2010
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April 16, 2014, 02:05:32 PM |
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Predicting a range is clearly useful, but having the value fall into that range or hit the mean value of the range does not always mean the prediction is better.
Example: Prediction 1: 90-100; Prediction 2: 50-150. Actual: 101 The first is clearly the better prediction, although the actual is outside the range and further from the mean than the second.
What you really want to do is 1) reward a predicted mean value (i.e. max+min/2) that is close to the actual; 2) reward a smaller range preferentially to a larger range.
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SlipperySlope
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April 16, 2014, 02:10:11 PM |
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Here is a chart illustrating the Log10 delta of actual bitcoin price from my manually-fitted logistic model trendline. Rpietila has been discussing Log10 deviation from the Log10 trend as an indicator for buy, sell or hold. It is clear from the chart how such limits could be derived from the April 2013 and November 2103 bubbles.
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bitfair
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April 16, 2014, 02:12:31 PM |
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What you want to do is reward for the prediction being within the interval and penalize for distance from the interval as well as for the width of the interval. For example: I= upper_limit -lower_limit W=Weight of Interval width penalty P= Actual
if( P>lower_limit AND P<upper_limit){ D=0 }else{ D=min [ abs ( P - upper_limit); abs ( P - lower_limit) ] }
Score= D+W*I
W could even be 1, IDK. Lowest score wins. Sorry for coming off as very negative today, I really don't mean to smack down everything, but this wouldn't work either. My counterexample: I would choose upper_limit = lower_limit = 0 and win every time (assuming P is not zero)! I would get D=0 and I=0, which would give me Score=0, regardless of W. Furthermore 0 is the lowest possible score since D>=0 and I>=0, so I would win every time!
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bb113
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April 16, 2014, 02:17:18 PM |
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What you want to do is reward for the prediction being within the interval and penalize for distance from the interval as well as for the width of the interval. For example: I= upper_limit -lower_limit W=Weight of Interval width penalty P= Actual
if( P>lower_limit AND P<upper_limit){ D=0 }else{ D=min [ abs ( P - upper_limit); abs ( P - lower_limit) ] }
Score= D+W*I
W could even be 1, IDK. Lowest score wins. Sorry for coming off as very negative today, I really don't mean to smack down everything, but this wouldn't work either. My counterexample: I would choose upper_limit = lower_limit = 0 and win every time (assuming P is not zero)! I would get D=0 and I=0, which would give me Score=0, regardless of W. Furthermore 0 is the lowest possible score since D>=0 and I>=0, so I would win every time! No. D is the deviation from the interval and is added to the weighted interval width. I think you misread it.
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