notme
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September 11, 2013, 03:59:44 AM |
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Max was $103.03. I don't think you know how medians work.
Not everyone is a math genius like yourself, so please forgive me oh Lord of the Numbers for utilizing the median as the literal center point of 2 numbers rather than the midpoint in a range data plots. I was taught the difference between mean, median, and mode in the 3rd grade. Forgive me for assuming you received a similar education.
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notme
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September 11, 2013, 04:02:48 AM |
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Max was $103.03. I don't think you know how medians work.
I am not going to argue, can you ELI5, I am not wanting to do the number crunching just understand the principal behind the method of prediction. Okay, you have a list of numbers. You put them in order. Median is the number middle (or average of the two middle numbers if there is an even count). You can't say anything about the max given the minimum and median other than max >= median. Those are just the outliers of the random sample I posted, a larger sampling would produce outliers even farther out. The thing that is really hard to know though is how well the probability of those outliers occurring matches with the probability of predicting them—the future is tricksy like that. Good to know. The future is indeed tricksy.
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rampantparanoia
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September 11, 2013, 04:05:10 AM |
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Max was $103.03. I don't think you know how medians work.
Not everyone is a math genius like yourself, so please forgive me oh Lord of the Numbers for utilizing the median as the literal center point of 2 numbers rather than the midpoint in a range data plots. I was taught the difference between mean, median, and mode in the 3rd grade. Forgive me for assuming you got a similar education. In an effort not to throw this entire thread off topic with mere slander back and forth, I choose to withhold my response.
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derpinheimer
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September 11, 2013, 04:36:14 AM |
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I have moved away from using convergence/divergence calibrations and have instead worked the indicators I have been using into a set of statistically based oscillators, some of which I actually use for the projection, and others I just use for diagnostics. As well, I have moved more toward a Monte Carlo approach to doing the projection. Here is the median case result of a projection under a given set of assumptions plus a sampling of the raw outputs. This is using calibrations presuming $148.91 as an intermediate peak. The time and price targets are actually treated independently. I expect to post updates as dictated by new data, updates to my assumptions, or updates to methodology. $75.20 median Wednesday, October 23, 2013 $52.64 Wednesday, October 09, 2013 $55.67 Thursday, November 21, 2013 $57.65 Saturday, November 16, 2013 $59.95 Tuesday, December 17, 2013 $61.92 Monday, October 28, 2013 $62.03 Wednesday, October 16, 2013 $63.80 Wednesday, October 16, 2013 $64.31 Saturday, October 12, 2013 $64.98 Thursday, October 17, 2013 $66.61 Tuesday, October 22, 2013 $68.22 Monday, November 04, 2013 $70.25 Tuesday, October 15, 2013 $70.32 Sunday, October 20, 2013 $71.22 Tuesday, December 03, 2013 $71.78 Sunday, October 20, 2013 $72.69 Monday, November 04, 2013 $73.74 Sunday, October 20, 2013 $74.48 Monday, October 21, 2013 $74.96 Sunday, October 27, 2013 $74.98 Sunday, October 06, 2013 $75.42 Friday, October 18, 2013 $76.17 Monday, October 14, 2013 $77.37 Saturday, November 02, 2013 $78.76 Thursday, October 24, 2013 $78.85 Friday, November 01, 2013 $79.09 Tuesday, November 05, 2013 $79.11 Sunday, November 03, 2013 $79.30 Sunday, October 06, 2013 $79.36 Tuesday, October 29, 2013 $80.87 Saturday, October 19, 2013 $81.09 Sunday, October 27, 2013 $82.69 Wednesday, October 16, 2013 $83.17 Friday, October 18, 2013 $83.69 Friday, October 11, 2013 $85.57 Tuesday, November 05, 2013 $87.84 Monday, October 28, 2013 $90.03 Saturday, October 12, 2013 $90.91 Monday, November 11, 2013 $95.85 Thursday, October 24, 2013 $103.03 Sunday, November 03, 2013 I cant understand what these numbers are at all. Median price on a given day? I know thats not right but its the only thing I can see it representing.
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adpinbr
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September 11, 2013, 07:36:38 AM |
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I have moved away from using convergence/divergence calibrations and have instead worked the indicators I have been using into a set of statistically based oscillators, some of which I actually use for the projection, and others I just use for diagnostics. As well, I have moved more toward a Monte Carlo approach to doing the projection. Here is the median case result of a projection under a given set of assumptions plus a sampling of the raw outputs. This is using calibrations presuming $148.91 as an intermediate peak. The time and price targets are actually treated independently. I expect to post updates as dictated by new data, updates to my assumptions, or updates to methodology. $75.20 median Wednesday, October 23, 2013
This is as clear as mud for me, if you are projecting a $75.20 median on X date then the price will likely be somewhere between $150 and $0 ![Huh](https://bitcointalk.org/Smileys/default/huh.gif) I hate to burden you but I think I need an ELI5. Before I was attempting a crude projection based on the use of median values, for price movement, time progression, and the components of the reversal indicator. With the addition of new data points these median values change slightly resulting in a variously changing forecast. This is neither principled, or rigorous, but it was easy so that's what I went with while working on improvements to the methodology that are more principled and rigorous, but are more work and are more calculation intensive. Before, I was using indicators calibrated to either converge or diverge with price movement. But the calibration was not very robust. So, I reworked them so that they behave more like oscillators, and the calibration is based on a statistical measure instead of a fixed value, or a value determined by a calibrated convergence. The points to which the indicators are calibrated is an intermediate price peak, this is one of many assumptions that go into the projection. The purpose of the reversal indicator is to show where a putative trend reversal has occurred. When this happens, that is the price and time target returned by the projection. What drives the projection is a simulation of price development based on a random variable that has a distribution similar to price development that occurred over a period of time in the price history. Because it is a random variable it will produce different values for the projection for each trial. What I am interested in is the distribution of these outcomes. This tells me a couple of things: What is most likely to happen given the distributions driving price development in the past, and how current price development statistically differs from past price development. I figured that most people would be most interested in the median case, so I broke that out. But I also figured that some people might be interested in exactly what kind of variability can be expected from the output, so I included that. It is slightly more precise that saying we are going to "zero"—or "to da moon!!1!" I don't follow....are we still supposed to be going to around 180 towards the end of the month as you previously predicted, or is this new data declaring previous statements invalid, placing us in a down trend?
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Tzupy
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September 11, 2013, 08:15:05 PM |
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Is it a possibility that the price will fluctuate between say 110$ and 150$, instead of following the natural course of a bear market? The reason I'm asking is because today the long term parabolic SAR flipped sides again, now showing an upward trend.
My recognition of the EW says that we should experience a large drop no later than 2-3 days from now ( 13-14 September ), and then enter a fake bull market. If this does not happen ( and right now it seems very unlikely ), then I think EW will be useless for long term predictions.
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Sometimes, if it looks too bullish, it's actually bearish
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MoreFun
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WePower.red
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September 11, 2013, 08:42:33 PM |
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Is it a possibility that the price will fluctuate between say 110$ and 150$, instead of following the natural course of a bear market? The reason I'm asking is because today the long term parabolic SAR flipped sides again, now showing an upward trend.
My recognition of the EW says that we should experience a large drop no later than 2-3 days from now ( 13-14 September ), and then enter a fake bull market. If this does not happen ( and right now it seems very unlikely ), then I think EW will be useless for long term predictions.
Similar thoughts there, weekend could be dumping. Expecting $115 +- $2.
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Miz4r
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September 11, 2013, 11:23:01 PM |
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Some rich whale is using your peak/reversal predictions as a contra-indicator I think. ![Tongue](https://bitcointalk.org/Smileys/default/tongue.gif)
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Bitcoin = Gold on steroids
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DrJoeGrine
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September 12, 2013, 12:13:53 AM |
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lolz
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DrJoeGrine
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September 12, 2013, 01:00:38 AM |
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Yes great oracle. Some whale with millions to invest is using your predictions. LOLZ
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DrJoeGrine
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September 12, 2013, 01:27:33 AM |
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Yes great oracle. Some whale with millions to invest is using your predictions. LOLZ
Oh, THAT's what you are laughing about. I thought you were laughing at me. I wouldn't make fun of you. You are sharing your methodology to provide another perspective of how to interpret Bitcoin fluctuations. Its been a good read and I have been following for some time now. I personally use strictly charts to provide my predictions and appreciate getting another take of the whale psychology.
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Wekkel
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yes
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September 12, 2013, 05:26:14 AM |
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Am I correct in understanding that the median price according to the projection has decreased the last week (now more around $100 than $160+)?
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Tzupy
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September 12, 2013, 08:31:57 AM |
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I may have an explanation for the whale buys of yesterday: in a normal bear market the $ funds would have been declining sharply. This is not the case, at least not yet. That's why the genuine rally succeeded 3 days ago, and some smart whales decides to take a gamble, buying now and hoping ( calculating? ) that the market will rise to 150$ or above. On the 31st of August after the whale buys there was a rally that drove the price up to 148, and such a rally ( by panic buyers ) is still possible today 12th September.
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Sometimes, if it looks too bullish, it's actually bearish
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RationalSpeculator
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This bull will try to shake you off. Hold tight!
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September 12, 2013, 01:16:58 PM |
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All Y'all seemed to be confused by the raw output list, so here's a scatter plot. Yeah—scatter plots are fun. ![](https://ip.bitcointalk.org/?u=https%3A%2F%2Fi.imgur.com%2Fj0t5rDO.png&t=665&c=jYUBMyRwKy4BJQ) Sorry, I would normally do a lot more trials but I am still in the middle of testing some updates and that takes too long. Do I understand correctly that you expect the price to go to around $60 before the end of the year? How much percent chance do you give this? 50%? 90%? Also, I'm confused on what you base this?
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RationalSpeculator
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This bull will try to shake you off. Hold tight!
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September 12, 2013, 03:38:58 PM |
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Do I understand correctly that you expect the price to go to around $60 before the end of the year?
How much percent chance do you give this? 50%? 90%?
Also, I'm confused on what you base this?
That is the median case, but this is just a toy. What I am looking at is an indicator which shows the likelihood that an intermediate reversal has occurred. It works by comparing how BTCOBV and $OBV track compared to price development in order to detect what regime of feedback effects is dominant. Then I go back to look at the distribution of the change in price from bin to bin in the last leg before the last reversal and take the compliment of it to form a random variable going forward, there are actually 5 random variables involved which have some dependency re. the development of the reversal indicator, to provide a random trial of where it might turn again in the future, this is the output of the trial. I am not actually certain that it is meaningful to assign a point probability to a particular outcome. It would be more meaningful to say, for instance, that there is a percentile case for price being above or below a target, or before or after a target. The median case as a rule of thumb would be the 50th percentile presuming a well-behaved Gaussian distribution. But I don't think that is a safe assumption. For one thing, I know that there are dependencies that I don't have worked into the (Markov chain) Monte Carlo yet. That is chinese for me. I have no idea what your answer to my question is. I'm willing to pay to get your opinion if that is what it takes. In such case, I would ask: explain like I'm a 5 year old. Are you able/willing to?
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adpinbr
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September 12, 2013, 04:39:23 PM |
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check out his signature, i have donated to him and others who give good/honest advice, we are all friends here but still the man deserves some compensation
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Odalv
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September 12, 2013, 07:37:21 PM |
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The latest. $78.47 Friday, October 25, 2013 <—Median
you forgot prepend 1 => $178,47 :-)
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Tzupy
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September 12, 2013, 07:47:25 PM |
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chodpaba, thank you for the detailed explanation, I saved it for later reading. Do you think you'll have to adjust your prediction if we'll have very soon a peak higher than 148.7$?
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Sometimes, if it looks too bullish, it's actually bearish
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Hfleer
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Changing avatars is currently not possible.
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September 12, 2013, 08:11:31 PM |
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It's nice to see BTC reaching for higher prices again recently. Lets see what can happen with ATM possibilities and whatever other news is out.
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derpinheimer
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September 13, 2013, 03:40:12 PM |
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Here is a little something I have been experimenting with. The one week call.
I basically use the same assumptions that I am using for the reversal projection, except that in the reporting I am picking off samples from a specific time target. I am specifically targeting the VWA one week out. And with that, here's a test call:
Friday, September 20, 2013 $130.27
I approve of this prediction(projection)!
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