Topic: Predict the price of Bitcoin on 1 April 2016 - win CBX (Crypto Bullion) - page 5. (Read 11350 times)
April 28, 2015, 10:44:22 PM
837$
April 28, 2015, 08:11:37 PM
$7340.58
April 28, 2015, 06:29:44 PM
$525
April 28, 2015, 03:23:50 PM
315 USD
April 28, 2015, 03:22:47 PM
$30,000.00 USD
because
a) This would still be less than the change in 2013, and
b) What would be the point of winning if the current trend hasn't changed by then?
because
a) This would still be less than the change in 2013, and
b) What would be the point of winning if the current trend hasn't changed by then?
April 09, 2015, 01:39:10 AM
As for me I would say ~$400.
April 08, 2015, 09:15:46 PM
The median is simply the middle number of the sample here. The data isn't distributed normally, the mean is over $5000 and the SD over $26000.
Should have used log scale (seriously!) i.e. use the geometric mean instead of arithmetic mean,
and compute the SD by exp(sqrt(sum((log(P(i)) - log(Pmean))^2))).
Hey, thanks. I have ~$480 for the geometric mean, and ~10.5 for the geometric SD. If I understand this right, the geometric standard deviation is multiplicative, so 1sigma would be 480*10.5, 2sigma 480*10.5^2, while -1sigma would be 480/10.5?
Yes, that's correct, thanks!
April 08, 2015, 06:09:39 PM
$444
April 08, 2015, 05:14:42 PM
$423
April 08, 2015, 04:32:53 PM
The median is simply the middle number of the sample here. The data isn't distributed normally, the mean is over $5000 and the SD over $26000.
Should have used log scale (seriously!) i.e. use the geometric mean instead of arithmetic mean,
and compute the SD by exp(sqrt(sum((log(P(i)) - log(Pmean))^2))).
Hey, thanks. I have ~$480 for the geometric mean, and ~10.5 for the geometric SD. If I understand this right, the geometric standard deviation is multiplicative, so 1sigma would be 480*10.5, 2sigma 480*10.5^2, while -1sigma would be 480/10.5?
April 08, 2015, 07:33:03 AM
2849$
That's my guess.
That's my guess.
April 08, 2015, 07:20:21 AM
The median is simply the middle number of the sample here. The data isn't distributed normally, the mean is over $5000 and the SD over $26000.
Should have used log scale (seriously!) i.e. use the geometric mean instead of arithmetic mean,
and compute the SD by exp(sqrt(sum((log(P(i)) - log(Pmean))^2))).
Can't we just go ahead and determine the median values with the 5 most accepted ways there are? Hell, we could even introduce the super-meta-mean-index
April 08, 2015, 05:22:46 AM
where do you all take these numbers from?
Real numbers, unlike 'real money', are inflationary and infinitely divisible. Their Central Bank can issue an infinite amount of them in an arbitrarily short time. In fact, more than you can count.
April 08, 2015, 02:34:48 AM
where do you all take these numbers from?
April 08, 2015, 12:44:43 AM
$1512
April 08, 2015, 12:00:53 AM
$843
April 07, 2015, 07:04:53 PM
$62
April 07, 2015, 06:38:49 PM
The median is simply the middle number of the sample here. The data isn't distributed normally, the mean is over $5000 and the SD over $26000.
Should have used log scale (seriously!) i.e. use the geometric mean instead of arithmetic mean,
and compute the SD by exp(sqrt(sum((log(P(i)) - log(Pmean))^2))).
April 07, 2015, 05:32:15 PM
^ Thank you for combining the responses! Nicely done.
April 07, 2015, 05:24:00 PM
Did you use a Gaussian function to have it? A normal distribution is very good to understand probability...
The median is simply the middle number of the sample here. The data isn't distributed normally, the mean is over $5000 and the SD over $26000.
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