# My numerical test of the Absolute Rates hypothesis

- Tutorial

Hello, Habr!

This publication seemed interesting to me: We get absolute exchange rates from paired cross-currency exchange rates and I wanted to test the ability to find this aaaabsolute exchange rate through numerical modeling, generally abandoning linear algebra.

The results were interesting.

The experiment will be small: 4 currencies, 6 currency pairs. For each pair, one course measurement.

The hypothesis is that the value of any currency can be expressed as a value that will take into account the value of other currencies in which it is quoted, while other currencies themselves will be expressed in the value of all other currencies. This is an interesting recursive task.

There are 4 currencies:

For them, the currency pairs were dialed:

Please note that if the number of currencies is n = 4, then the number of pairs is k = (n ^ 2 - n) / 2 = 6. It makes no sense to look for usdeur if eurusd is quoted ...

At time t, one of the providers measured the exchange rate of currency pairs :

Calculations will be performed for these values.

I solve the problem by analytically taking the gradient of the loss function, which is essentially a system of equations.

The experiment code will be in R:

R allows using stats :: D to take a derivative of a function. For example, if we want to differentiate by the USD currency, we get the expression:

The gradient descent step will be controlled by the parameter lr and all this is taken with a negative sign.

That is, in human words, we select the rates of 4 currencies so that all currency pairs in the experiment receive values equal to the initial values of these pairs. Mmm, let's solve the puzzle - in the forehead!

In order not to stretch, I’ll immediately inform you the following: the experiment as a whole was successful, the code worked, the error went close, close to zero. But then I noticed that the results are always different.

To verify the (un) stability of the solution, I simulated 1000 times without fixing the PRNG seed for the starting values of the currency values.

And here comes the picture from the kata: error reaches 0.00001 and less (the optimization is set this way) always, while the values of currencies float the hell out of where. It turns out that there is always a different decision, gentlemen!

Once again, this picture, y-axis in the original units (not log.):

So that you can repeat this, below I am attaching the full code.

Here is what remains unclear to me:

The whole idea seems very vague in the absence of any intelligible prerequisites and limitations. But it was interesting!

Well, I also wanted to say that you can do without OLS when the data is tricky, the matrices are singular, well, or when the theory is poorly known (ehh ...).

Thanks eavprog for the initial message.

Till!

This publication seemed interesting to me: We get absolute exchange rates from paired cross-currency exchange rates and I wanted to test the ability to find this aaaabsolute exchange rate through numerical modeling, generally abandoning linear algebra.

The results were interesting.

The experiment will be small: 4 currencies, 6 currency pairs. For each pair, one course measurement.

### So, let's begin

The hypothesis is that the value of any currency can be expressed as a value that will take into account the value of other currencies in which it is quoted, while other currencies themselves will be expressed in the value of all other currencies. This is an interesting recursive task.

There are 4 currencies:

- usd
- eur
- chf
- gbp

For them, the currency pairs were dialed:

- eurusd
- gbpusd
- eurchf
- eurgbp
- gbpchf
- usdchf

Please note that if the number of currencies is n = 4, then the number of pairs is k = (n ^ 2 - n) / 2 = 6. It makes no sense to look for usdeur if eurusd is quoted ...

At time t, one of the providers measured the exchange rate of currency pairs :

Calculations will be performed for these values.

### Maths

I solve the problem by analytically taking the gradient of the loss function, which is essentially a system of equations.

The experiment code will be in R:

```
#set.seed(111)
usd <- runif(1)
eur <- runif(1)
chf <- runif(1)
gbp <- runif(1)
# snapshot of values at time t
eurusd <- 1.12012
gbpusd <- 1.30890
eurchf <- 1.14135
eurgbp <- 0.85570
gbpchf <- 1.33373
usdchf <- 1.01896
## symbolic task ------------
express <- expression(
(eurusd - eur / usd) ^ 2 +
(gbpusd - gbp / usd) ^ 2 +
(eurchf - eur / chf) ^ 2 +
(eurgbp - eur / gbp) ^ 2 +
(gbpchf - gbp / chf) ^ 2 +
(usdchf - usd / chf) ^ 2
)
eval(express)
x = 'usd'
D(express, x)
eval(D(express, x))
```

R allows using stats :: D to take a derivative of a function. For example, if we want to differentiate by the USD currency, we get the expression:

2 * (eur / usd ^ 2 * (eurusd - eur / usd)) + 2 * (gbp / usd ^ 2 * (gbpusd -To reduce the value of the express function, we will perform gradient descent and it is immediately clear (we see square differences) that the minimum value will be zero, which is what we need.

gbp / usd)) - 2 * (1 / chf * (usdchf - usd / chf) )

```
-deriv_vals * lr
```

The gradient descent step will be controlled by the parameter lr and all this is taken with a negative sign.

That is, in human words, we select the rates of 4 currencies so that all currency pairs in the experiment receive values equal to the initial values of these pairs. Mmm, let's solve the puzzle - in the forehead!

### results

In order not to stretch, I’ll immediately inform you the following: the experiment as a whole was successful, the code worked, the error went close, close to zero. But then I noticed that the results are always different.

**A question for connoisseurs:**it seems that this task has an unlimited number of solutions, but in this I am a complete zero, I think they will tell me in the comments.To verify the (un) stability of the solution, I simulated 1000 times without fixing the PRNG seed for the starting values of the currency values.

And here comes the picture from the kata: error reaches 0.00001 and less (the optimization is set this way) always, while the values of currencies float the hell out of where. It turns out that there is always a different decision, gentlemen!

Once again, this picture, y-axis in the original units (not log.):

So that you can repeat this, below I am attaching the full code.

**The code**

```
# clear environment
rm(list = ls()); gc()
## load libs
library(data.table)
library(ggplot2)
library(magrittr)
## set WD --------------------------------
# your dir here ...
## set vars -------------
currs <- c(
'usd',
'eur',
'chf',
'gbp'
)
############
## RUN SIMULATION LOOP -------------------------------
simuls <- 1000L
simul_dt <- data.table()
for(
s in seq_len(simuls)
)
{
#set.seed(111)
usd <- runif(1)
eur <- runif(1)
chf <- runif(1)
gbp <- runif(1)
# snapshot of values at time t
eurusd <- 1.12012
gbpusd <- 1.30890
eurchf <- 1.14135
eurgbp <- 0.85570
gbpchf <- 1.33373
usdchf <- 1.01896
## symbolic task ------------
express <- expression(
(eurusd - eur / usd) ^ 2 +
(gbpusd - gbp / usd) ^ 2 +
(eurchf - eur / chf) ^ 2 +
(eurgbp - eur / gbp) ^ 2 +
(gbpchf - gbp / chf) ^ 2 +
(usdchf - usd / chf) ^ 2
)
## define gradient and iterate to make descent to zero --------------
iter_max <- 1e+3
lr <- 1e-3
min_tolerance <- 0.00001
rm(grad_desc_func)
grad_desc_func <- function(
lr,
curr_list
)
{
derivs <- character(length(curr_list))
deriv_vals <- numeric(length(curr_list))
grads <- numeric(length(curr_list))
# symbolic derivatives
derivs <- sapply(
curr_list,
function(x){
D(express, x)
}
)
# derivative values
deriv_vals <- sapply(
derivs,
function(x){
eval(x)
}
)
# gradient change values
-deriv_vals * lr
}
## get gradient values ----------
progress_list <- list()
for(
i in seq_len(iter_max)
)
{
grad_deltas <- grad_desc_func(lr, curr_list = currs)
currency_vals <- sapply(
currs
, function(x)
{
# update currency values
current_val <- get(x, envir = .GlobalEnv)
new_delta <- grad_deltas[x]
if(new_delta > -1 & new_delta < 1)
{
new_delta = new_delta
} else {
new_delta = sign(new_delta)
}
new_val <- current_val + new_delta
if(new_val > 0 & new_val < 2)
{
new_val = new_val
} else {
new_val = current_val
}
names(new_val) <- NULL
# change values of currencies by gradient descent step in global env
assign(x, new_val , envir = .GlobalEnv)
# save history of values for later plotting
new_val
}
)
progress_list[[i]] <- c(
currency_vals,
eval(express)
)
if(
eval(express) < min_tolerance
)
{
break('solution was found')
}
}
## check results ----------
# print(
# paste0(
# 'Final error: '
# , round(eval(express), 5)
# )
# )
#
# print(
# round(unlist(mget(currs)), 5)
# )
progress_dt <- rbindlist(
lapply(
progress_list
, function(x)
{
as.data.frame(t(x))
}
)
)
colnames(progress_dt)[length(colnames(progress_dt))] <- 'error'
progress_dt[, steps := 1:nrow(progress_dt)]
progress_dt_melt <-
melt(
progress_dt
, id.vars = 'steps'
, measure.vars = colnames(progress_dt)[colnames(progress_dt) != 'steps']
)
progress_dt_melt[, simul := s]
simul_dt <- rbind(
simul_dt
, progress_dt_melt
)
}
ggplot(data = simul_dt) +
facet_wrap(~ variable, scales = 'free') +
geom_line(
aes(
x = steps
, y = value
, group = simul
, color = simul
)
) +
scale_y_log10() +
theme_minimal()
```

*The code for 1000 simulations works for about a minute.*### Conclusion

Here is what remains unclear to me:

- Is it possible to stabilize the solution in a tricky mathematical way;
- whether there will be a convergence with more currencies and currency pairs;
- if there can be no stability, then for each new data snapshot, our currencies will walk as they like, if you do not fix the PRNG seed, and this is a failure.

The whole idea seems very vague in the absence of any intelligible prerequisites and limitations. But it was interesting!

Well, I also wanted to say that you can do without OLS when the data is tricky, the matrices are singular, well, or when the theory is poorly known (ehh ...).

Thanks eavprog for the initial message.

Till!