Linear Regression Forecasting Exchange Rates

This project obtains data from the Central Bank of Armenia's open API, covering the data from a specified start date for any of the 30 supported currencies.

ISO	Currency	ISO	Currency
USD	US Dollar	GBP	British Pound Sterling
AUD	Australian Dollar	EUR	Euro
XDR	IMF Special Drawing Rights	IRR	Iranian Rial
PLN	Polish Złoty	CAD	Canadian Dollar
INR	Indian Rupee	NOK	Norwegian Krone
JPY	Japanese Yen	SEK	Swedish Krona
CHF	Swiss Franc	CZK	Czech Republic Koruna
CNY	Chinese Yuan (Renminbi)	SGD	Singapore Dollar
BRL	Brazilian Real	AED	UAE Dirham
KGS	Kyrgyzstani Som	KZT	Kazakhstani Tenge
RUB	Russian Ruble	UAH	Ukrainian Hryvnia
UZS	Uzbekistani Som	BYN	Belarusian Ruble
TJS	Tajikistani Somoni	GEL	Georgian Lari
HKD	Hong Kong Dollar	XAU	Gold (troy ounce)
XAG	Silver (troy ounce)	NZD	New Zealand Dollar

The data is then processed using linear regression to forecast exchange rate, which is compared with the actual rate.

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Usage/Examples

To install the necessary dependencies, once you've cloned the repo, cd to the directory in which 'requirements.txt' is located, and run

pip3 install -r requirements.txt

Before proceeding, make sure to configure the exchange rate data fetching. In the exchange_rate_data/data_fetcher.py file, specify the ISO code of the desired currency and the start date in the format YYYY-MM-DD:

iso_codes = 'USD'
start_date = '2023-09-01'

Once you've configured the data fetcher, you can run the application as follows:

python3 main.py

After executing the application, it will produce a graphical representation illustrating exchange rates, linear regression analysis, and a forecast.

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Linear regression

Linear regression represents a regression model that describes the relationship between one (dependent) variable and one or more other variables (factors, regressors, independent variables) with a linear functional dependency. Let's consider a linear regression model in which the dependent variable depends on only one factor. In this case, the function describing the relationship of $y$ with $x$ will have the following form:

$f(x) = w_{0} + w_{1} \ast x$

The task is to find the weight coefficients $w_{0}$ and $w_{1}$, such that this line best fits the original data. To achieve this, we define an error function, the minimization of which will determine the values of $w_{0}$ and $w_{1}$ using the method of least squares:

$\begin{aligned} MSE = \frac{1}{n} \ast \sum\limits_{i=0}^{n}(y_{i} - f(x_{i}))^{2} \end{aligned}$

Or, by substituting the model equation:

$\begin{aligned} MSE = \frac{1}{n} \ast \sum\limits_{i=0}^{n}(y_{i} - w_{0} - w_{1} \ast x_{i} )^{2} \end{aligned}$

Minimizing the $MSE$ error function involves finding the partial derivatives with respect to $w_{0}$ and $w_{1}$:

$\begin{aligned} \frac{\delta MSE(w_{0}, w_{1})}{\delta w_{0}} = -\frac{2}{n} \ast \sum\limits_{i=0}^{n}(y_{i} - w_{0} - w_{1} \ast x_{i} )^{2} \end{aligned}$

$\begin{aligned} \frac{\delta MSE(w_{0}, w_{1})}{\delta w_{1}} = -\frac{2}{n} \ast \sum\limits_{i=0}^{n}((y_{i} - w_{0} - w_{1} \ast x_{i} ) \ast x_{i}) \end{aligned}$

Setting these derivatives to zero yields a system of equations whose solution minimizes the $MSE$:

$\left\{ \begin{aligned} 0 &= -\frac{2}{n} \ast \sum\limits_{i=0}^{n}(y_{i} - w_{0} - w_{1} \ast x_{i}) \\ 0 &= -\frac{2}{n} \ast \sum\limits_{i=0}^{n}((y_{i} - w_{0} - w_{1} \ast x_{i}) \ast x_{i}) \end{aligned} \right.$

Expanding the sums and taking into account that $-\frac{2}{n}$ cannot be equal to zero, we set the second factors to zero:

$\left\{ \begin{aligned} 0 &= -w_{0} \ast n + \sum\limits_{i=0}^{n}y_{i} - w_{1} \ast \sum\limits_{i=0}^{n}x_{i} \\ 0 &= \sum\limits_{i=0}^{n}(y_{i} \ast x_{i}) + w_{0} \ast \sum\limits_{i=0}^{n}x_{i} - w_{1} \ast \sum\limits_{i=0}^{n}x_{i}^{2} \end{aligned} \right.$

We express $w_{0}$ from the first equation:

$\begin{aligned} w_{0} = \frac{\sum\limits_{i=0}^{n}y_{i}}{n} - w_{1}\frac{\sum\limits_{i=0}^{n}x_{i}}{n} \end{aligned}$

Substituting this into the second equation, we solve for $w_{1}$:

$\begin{aligned} 0 = \sum\limits_{i=0}^{n}(y_{i} \ast x_{i}) + (\frac{\sum\limits_{i=0}^{n}y_{i}}{n} - w_{1}\frac{\sum\limits_{i=0}^{n}x_{i}}{n}) \ast \sum_{i=0}^{n}x_{i} - w_{1} \ast \sum\limits_{i=0}^{n}x_{i}^{2} \end{aligned}$

$\begin{aligned} 0 = \sum\limits_{i=0}^{n}(y_{i} \ast x_{i}) + \frac{\sum\limits_{i=0}^{n}(y_{i}\sum\limits_{i=0}^{n}x_{i})}{n} - w_{1}\frac{\sum\limits_{i=0}^{n}(x_{i}\sum\limits_{i=0}^{n}x_{i})}{n} - w_{1} \ast \sum\limits_{i=0}^{n}x_{i}^{2} \end{aligned}$

By solving for $w_{1}$ from this equation, we obtain the analytical solution.

$\begin{aligned} w_{1} = \frac{\frac{\sum\limits_{i=0}^{n}(x_{i}\sum\limits_{i=0}^{n}y_{i})}{n} - \sum\limits_{i=0}^{n}(y_{i} \ast x_{i})}{\frac{\sum\limits_{i=0}^{n}(x_{i}\sum\limits_{i=0}^{n}x_{i})}{n} - \sum\limits_{i=0}^{n}x_{i}^{2}} \end{aligned}$

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License

MIT