Zero sample correlation does not necessarily mean there is no linear relationship - page 56

New comment
 
Demi:
As a specialist in econometrics, I suggest you ignore all these frivolous and unprofessional remarks and get to the main thing - take the tools from MT4 and clearly demonstrate the power of econometrics on these rows by building your TS based on cointegration.

Is such a system possible? It needs the movement of the instrument to be divisible by a multiple of the spread. I thought that only trend trading is possible.
 
EconModel:
Is such a system possible? It needs the movement of an instrument to be a multiple of the spread. I thought only trend trading was possible.

This is what you call it.

You said that cointegration exists in the financial markets. Hence, arbitrage is possible - cointegrated instruments always "converge". What does "movement of an instrument is a multiple of the spread" have to do with it?

Or did I misunderstand?

 
Demi:

This is what you call it.

You said that cointegration exists in the financial markets. Hence, arbitrage is possible - cointegrated instruments always "converge". What does "movement of an instrument is a multiple of the spread" have to do with it?

Or did I misunderstand?

Unfortunately I can't find a residual chart for cointegrated series. As far as I remember, there is not more than 5 bars in the market. And often just one bar at all. When I saw that, I didn't continue. That's why I'm asking. I've been using EURUSD - GBPUSD hourly.
 
EconModel:
Unfortunately, I cannot find a residual chart for cointegrated series. As far as I remember, you can't get more than 5 bars in a market there. And often just one bar at all. When I saw that, I didn't continue. That's why I'm asking. I've been using EURUSD - GBPUSD hourly.

There are no cointegrated series in the market. This is why QC is used for price series
 
Demi:
there are no cointegrated rows on the market
I gave you a counter-example yesterday.
 
anonymous:
I gave you a counter-example yesterday.


There was no example. Someone wrote the name of two instruments - it was. There was no example.

An example of two instruments, with a cointegration test, describing the TS with the obligatory account of the spread

 
Demi:


There was no example. Someone wrote the name of two tools - there was. There was no example.

Example of two instruments, with a cointegration test, describing the TS with a mandatory spread

What do you mean by "no"?

We take the regression, look at the coefficients, if the probability is close to 0, then we check the residue for the unit root. That's it. Just threw the results away. I'm too lazy to do it all again. But in them (on my word) everything was fine. Co-integration was.

PS. Found a whole thread on cointegration. There are examples there, though on EViews (where did he get it?).

 
Demi:


There was no example. Someone wrote the name of two instruments - there was. There was no example.

Example of two instruments, with cointegration test

library(tseries)
library(zoo)

prices.brk.a <- get.hist.quote(
  instrument = 'BRK-A',
  start = '2010-01-01',
  end = '2013-04-15',
  provider = 'yahoo',
  quote = 'AdjClose'
  )
prices.brk.b <- get.hist.quote(
  instrument = 'BRK-B',
  start = '2010-01-01',
  end = '2013-04-15',
  provider = 'yahoo',
  quote = 'AdjClose'
  )

model <- lm(prices.brk.a ~ prices.brk.b)
spread <- residuals(model)
plot(spread)

summary(model)

adf.test(as.numeric(spread), alternative = 'stationary')



> summary(model)

Call:
lm(formula = prices.brk.a ~ prices.brk.b)

Residuals:
     Min       1 Q   Median       3 Q      Max 
-1405.40  -143.47   -44.62    83.85  1985.01 

Coefficients:
             Estimate Std. Error  t value Pr(>|t|)    
(Intercept)    92.353    110.787    0.834    0.405    
prices.brk.b 1499.520      1.353 1108.055   <2 e-16 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 

Residual standard error: 290.1 on 822 degrees of freedom
Multiple R-squared: 0.9993,     Adjusted R-squared: 0.9993 
F-statistic: 1.228 e+06 on 1 and 822 DF,  p-value: < 2.2 e-16 

> adf.test(as.numeric(spread), alternative = 'stationary')

        Augmented Dickey-Fuller Test

data:  as.numeric(spread) 
Dickey-Fuller = -6.1143, Lag order = 9, p-value = 0.01
alternative hypothesis: stationary 

Warning message:
In adf.test(as.numeric(spread), alternative = "stationary") :
  p-value smaller than printed p-value
with a description of the TS

Wouldn't it be too much for you?

with obligatory taking into account the spread.

You don't know how to trade on order-driver market without spread? Poor thing.

 
anonymous:

Won't you get fat?

Can't trade on the order-driver market without a spread? You poor thing.

Pardon me. Your probability = 0.405. The bias in the regression is unnecessary.
 
EconModel:
Pardon me. You have a probability = 0.405. The bias in the regression is unnecessary.

It gave an error in the hedge ratio of 0.07%: it should be 1500.6429 instead of 1499.520. How to go on living?! :(
1...495051525354555657585960
New comment