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# Detailed derivation of exponential regression of Bitcoin

Enter years Use ',' or new line to seperate between values: 2011, 2012, 2013, 2014, 2015, 2016, 2017, 2018, 2019, 2020, 2021 11, 12, 98, 520, 266, 567, 4149, 6212, 9973, 11894, 45000

## 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

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




$$90.75$$
$$110.0$$
$$\displaylines{}$$
$$\displaylines{\\ \\ \\ \\ B= \frac{\sum_{i=1}^{n}(x_{i} - \bar{x})(y_{i} - \bar{y})}{\sum_{i=1}^{n}(x_{i} - \bar{x})^{2}} \\ \\ From\;the\;table\;total\;we\;will\;get\;numerator\;and\;denominator \\ \\ \Rightarrow \frac{90.75}{110.0} \\ \\ \Rightarrow \mathbf{\color{Red}{0.82}} \\ \\ A = \bar{y}-B* \bar{x} \\ \\ \Rightarrow 6.73-0.82*6.0 \\ \\ \Rightarrow \mathbf{\color{Red}{1.81}} \\ \\ Equation\;of\;line\;\Rightarrow\;y\;=\;A+B*x \\ \\ \Rightarrow y = 1.81+0.82*x \\ \\ Using\;A\;=\;\log\;_{e}\;a \\ \\ \Rightarrow a\;=\;e^{A} \\ \\ \Rightarrow a\;=\;e^{1.81} \\ \\ \Rightarrow a\;=\;6.11 \\ \\ b\;=\;B\;=\;0.82 \\ \\ After\;putting\;back\;these\;in\;original\;equation\;P=ae^{bx} \\ \\ We\;get\;P\;=\;6.11e^{0.82x} \\ \\ Now\;replace\;x\;with\;T-\;2010 \\ \\ \Rightarrow We\;get\;P\;=\;6.11e^{0.82(T-\;2010)} }$$