Go back to


Related calculators


Newly added calculators




.....The site is being constantly updated, so come back to check new updates.....

If you find any bug or need any improvements in solution report it here

Crypto fear and greed index

Detailed derivation of exponential regression of Bitcoin



The detailed derivation below is based on above data. If you want you can change and calculate again


Regression is not rocket science. You can also do that with the help of pen and paper. Below it is derived in simple steps


$$ \displaylines{---} $$
$$ \displaylines{Let\;P\;is\;price\;of\;bitcoin,\;T\;is\;year \\ \\ For\;simplicity\;we\;create\;another\;variable\;x \\ \\ Where\;x\;=\;T-2010 \\ \\ Now\;we\;have\;to\;find\;x\;values\;for\;corresponding\;T\;values \\ \\ Using\;x\;=\;T-2010 \\ \\ } $$
$$ No $$
$$ T $$
$$ x $$
$$ 1 $$
$$ 2011 $$
$$ 1 $$
$$ 2 $$
$$ 2012 $$
$$ 2 $$
$$ 3 $$
$$ 2013 $$
$$ 3 $$
$$ 4 $$
$$ 2014 $$
$$ 4 $$
$$ 5 $$
$$ 2015 $$
$$ 5 $$
$$ 6 $$
$$ 2016 $$
$$ 6 $$
$$ 7 $$
$$ 2017 $$
$$ 7 $$
$$ 8 $$
$$ 2018 $$
$$ 8 $$
$$ 9 $$
$$ 2019 $$
$$ 9 $$
$$ 10 $$
$$ 2020 $$
$$ 10 $$
$$ 11 $$
$$ 2021 $$
$$ 11 $$
$$ \displaylines{} $$
$$ \displaylines{\\ \\ The\;curve\;to\;be\;fitted\;is\;P=ae^{bx}\; \\ \\ Take\;log\;on\;both\;side\;to\;the\;base\;e \\ \\ \log\;_{e}\;P\;=\;\log\;_{e}\;a\;+\;bx \\ \\ It\;is\;in\;the\;form\;y=A+Bx.\;Where\;y\;=\;\log\;_{e}\;P,\;A\;=\;\log\;_{e}\;a,\;B\;=\;b \\ \\ Now\;we\;have\;to\;apply\;Linear\;Regression \\ \\ Before\;applying\;we\;have\;to\;find\;y\;values\;for\;corresponding\;P\;values \\ \\ Using\;y\;=\;\log\;_{e}\;P \\ \\ } $$
$$ No $$
$$ P $$
$$ y $$
$$ 1 $$
$$ 11 $$
$$ 2.4 $$
$$ 2 $$
$$ 12 $$
$$ 2.48 $$
$$ 3 $$
$$ 98 $$
$$ 4.58 $$
$$ 4 $$
$$ 520 $$
$$ 6.25 $$
$$ 5 $$
$$ 266 $$
$$ 5.58 $$
$$ 6 $$
$$ 567 $$
$$ 6.34 $$
$$ 7 $$
$$ 4149 $$
$$ 8.33 $$
$$ 8 $$
$$ 6212 $$
$$ 8.73 $$
$$ 9 $$
$$ 9973 $$
$$ 9.21 $$
$$ 10 $$
$$ 11894 $$
$$ 9.38 $$
$$ 11 $$
$$ 45000 $$
$$ 10.71 $$
$$ \displaylines{} $$
$$ \displaylines{\\ \\ Now\;we\;have\;to\;find\;mean\;of\;x\;values \\ \\ Mean = \frac{\sum_{i=1}^{n}x_{i}}{n} \\ \\ \,=\frac{1+2+3+4+5+6+7+8+9+10+11}{11} \\ \\ \,=\frac{66}{11} \\ \\ \Rightarrow \bar{x}= 6.0 \\ \\ Then\;we\;have\;to\;find\;mean\;of\;y\;values \\ \\ Mean = \frac{\sum_{i=1}^{n}y_{i}}{n} \\ \\ \,=\frac{2.4+2.48+4.58+6.25+5.58+6.34+8.33+8.73+9.21+9.38+10.71}{11} \\ \\ \,=\frac{73.99000000000001}{11} \\ \\ \Rightarrow \bar{y}= 6.73 \\ \\ Now,\;make\;below\;table \\ \\ } $$
$$ No $$
$$ x_{i} $$
$$ y_{i} $$
$$ (x_{i} - \bar{x}) $$
$$ (y_{i} - \bar{y}) $$
$$ (x_{i} - \bar{x})*(y_{i} - \bar{y}) $$
$$ (x_{i} - \bar{x})^2 $$
$$ 1 $$
$$ 1 $$
$$ 2.4 $$
$$ -5.0 $$
$$ -4.33 $$
$$ 21.65 $$
$$ 25.0 $$
$$ 2 $$
$$ 2 $$
$$ 2.48 $$
$$ -4.0 $$
$$ -4.25 $$
$$ 17.0 $$
$$ 16.0 $$
$$ 3 $$
$$ 3 $$
$$ 4.58 $$
$$ -3.0 $$
$$ -2.15 $$
$$ 6.45 $$
$$ 9.0 $$
$$ 4 $$
$$ 4 $$
$$ 6.25 $$
$$ -2.0 $$
$$ -0.48 $$
$$ 0.96 $$
$$ 4.0 $$
$$ 5 $$
$$ 5 $$
$$ 5.58 $$
$$ -1.0 $$
$$ -1.15 $$
$$ 1.15 $$
$$ 1.0 $$
$$ 6 $$
$$ 6 $$
$$ 6.34 $$
$$ 0.0 $$
$$ -0.39 $$
$$ -0.0 $$
$$ 0.0 $$
$$ 7 $$
$$ 7 $$
$$ 8.33 $$
$$ 1.0 $$
$$ 1.6 $$
$$ 1.6 $$
$$ 1.0 $$
$$ 8 $$
$$ 8 $$
$$ 8.73 $$
$$ 2.0 $$
$$ 2.0 $$
$$ 4.0 $$
$$ 4.0 $$
$$ 9 $$
$$ 9 $$
$$ 9.21 $$
$$ 3.0 $$
$$ 2.48 $$
$$ 7.44 $$
$$ 9.0 $$
$$ 10 $$
$$ 10 $$
$$ 9.38 $$
$$ 4.0 $$
$$ 2.65 $$
$$ 10.6 $$
$$ 16.0 $$
$$ 11 $$
$$ 11 $$
$$ 10.71 $$
$$ 5.0 $$
$$ 3.98 $$
$$ 19.9 $$
$$ 25.0 $$
$$ Total $$
$$ $$
$$ $$
$$ $$
$$ $$
$$ 90.75 $$
$$ 110.0 $$
$$ \displaylines{} $$
$$ \displaylines{\\ \\ \\ \\ B= \frac{\sum_{i=1}^{n}(x_{i} - \bar{x})(y_{i} - \bar{y})}{\sum_{i=1}^{n}(x_{i} - \bar{x})^{2}} \\ \\ From\;the\;table\;total\;we\;will\;get\;numerator\;and\;denominator \\ \\ \Rightarrow \frac{90.75}{110.0} \\ \\ \Rightarrow \mathbf{\color{Red}{0.82}} \\ \\ A = \bar{y}-B* \bar{x} \\ \\ \Rightarrow 6.73-0.82*6.0 \\ \\ \Rightarrow \mathbf{\color{Red}{1.81}} \\ \\ Equation\;of\;line\;\Rightarrow\;y\;=\;A+B*x \\ \\ \Rightarrow y = 1.81+0.82*x \\ \\ Using\;A\;=\;\log\;_{e}\;a \\ \\ \Rightarrow a\;=\;e^{A} \\ \\ \Rightarrow a\;=\;e^{1.81} \\ \\ \Rightarrow a\;=\;6.11 \\ \\ b\;=\;B\;=\;0.82 \\ \\ After\;putting\;back\;these\;in\;original\;equation\;P=ae^{bx} \\ \\ We\;get\;P\;=\;6.11e^{0.82x} \\ \\ Now\;replace\;x\;with\;T-\;2010 \\ \\ \Rightarrow We\;get\;P\;=\;6.11e^{0.82(T-\;2010)} } $$

Below is the graph of Regression we made