# Co-integration constraints of coint(X,Z) given coint(X,Y) and coint(Y,Z)?

The Augmented Dickey-Fuller Test can be used to measure how well ranked certain pairs are against others for co-integration.

So then say we have a known co-integration between X and Y and between Y and Z, is there a constraint to the range of co-integration between X and Z?

For example, if we know coint(X,Y) = -0.1 and coint(Y,Z) = -0.3 can we then use these in some formula which would then say with certainty that -x.x < coint(X,Z) < -x.x?

This is similar to the correlation constraint of the same scenario given by (Olkin, 1981).

EDIT - As Richard has pointed out, I may have misunderstood the ADF Test. I'll rephrase the question here:
If we know $X$ is cointegrated (via some test e.g. Engle Granger) with $Y$ and $Y$ with $Z$, is there some measure of how probable $X$ will be cointegrated with $Z$?

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With $coint(X,Y) = -0.1$ you mean the $\beta$ in the cointegration equation? Does this mean that $y_t +0.1 x_t$ is stationary? –  Richard Jan 21 at 8:42
Oops sorry, should have made it more clear. I am talking about the ADF Test Statistic. –  Ubobo Jan 21 at 9:02
So it is the $\gamma$ in your link, right? How does this relate to the basic idea of cointegration as stated e.g. here: en.wikipedia.org/wiki/Cointegration ? –  Richard Jan 21 at 9:38
As far as I understand this the test is just used to test whether the residuals are stationary. Do you think this helps? Wouldn't you rather want to know something about the regression coefficients? If you really speak about $\gamma$ then please clarify your question. –  Richard Jan 21 at 9:39
@Richard Re-reading the links now I've probably misunderstood. Originally thought that ADF gave a rank of how likely two variables were cointegrated! So I guess I'll rephrase my question: If we know X is cointegrated with Y and Y with Z, is there some measure of how probable X will be cointegrated with Z? Sorry for the confusion. –  Ubobo Jan 21 at 9:47

Regarding you comments, I'm adding an answer here because I will not have enough space to explain my point, so please forgive for this.

Lets start from the beginning, and assume :

(1) $X_t - \beta_1Y_t = \epsilon_t$ ($\epsilon_t$ is stationary)

(2) $Y_t - \beta_2Z_t = \eta_t$ ($\eta_t$ is stationnary)

then (1) + $\beta_1$(2) gives $X_t - \beta_1\beta_2Z_t = \epsilon_t+\beta_1\eta_t = \nu_t$

Even is $\epsilon_t$ and $\eta_t$ are stationary processes, the linear combination is stationary only if these processes are independent (which is a strong assumption), or in a weak assumption, they should be joint weak stationarity.

Indeed, to be stationary, $\nu_t$ should have an autocovariance which is independent of time $t$, while:

$Cov(\nu_{t+h}, \nu_t) = Cov(\epsilon_{t+h}+\beta_1\eta_{t+h}, \epsilon_t+\beta_1\eta_t) = [Cov(\epsilon_{t+h}, \epsilon_{t})+ \beta_1^2Cov(\eta_{t+h}, \eta_t)] + \beta_{1}[Cov(\eta_{t+h}, \epsilon_t)+Cov(\epsilon_{t+h}, \eta_t)]$

In general, $Cov(\eta_{t+h}, \epsilon_t)$ and $Cov(\epsilon_{t+h}, \eta_t)$ are not only functions of $h$ but also of $t$.

If $\epsilon_t$ and $\eta_t$ are independent, then these covariances are 0, the first ones are independent of $t$ which leads to the desired result. But, as I said, in reality you only need that $\epsilon_t$ and $\eta_t$ be Joint Weakly Stationary, which means that $\forall t_1, t_2~~ Cov(\epsilon_{t_1}, \eta_{t_2})=f(|t_2-t_1|)$ (or $h$ if $t_1=t+h$ and $t_2=t$)

http://math.stackexchange.com/questions/377333/sum-of-stationary-process

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Very good answer –  Richard Jan 22 at 14:56
@Richard: Thank you Richard! I hope this will help Ubobo. –  aajajim Jan 22 at 16:14
@aajajim Thanks, this really cleared things up! You too Richard. You both have been great. –  Ubobo Jan 23 at 0:39

I am trying to give a some comments (too much for a real comment) a start: $X$ cointegrated with $Y$ means there is a $\beta_1$ such that $$X_t - \beta_1 Y_t = u_t,$$ and $u_t$ is stationary. $\beta_1$ can be estimated by $$\beta_1 = \frac{Cov(X_t,Y_t)}{Var(Y_t)} = Cor(X_t,Y_t) \frac{\sqrt{Var(X_t)}}{\sqrt{Var(Y_t)}}.$$ For $Y$ and $Z$ we have $$Y_t - \beta_2 Z_t = v_t,$$ with $v_t$ stationary and $\beta_2 = Cor(Z_t,Y_t) \frac{\sqrt{Var(Y_t)}}{\sqrt{Var(Z_t)}}$.

Then the question is whether there is a $\beta_3$ such that $$X_t - \beta_3 Z_t = e_t$$ with $e_t$ stationary. We would estimate $$\beta_3 = Cor(X_t,Z_t) \frac{\sqrt{Var(X_t)}}{\sqrt{Var(Z_t)}}.$$ As linear combinations of stationary time series are stationary we get $$k_t = u_t - v_t = X_t - \beta_1 Y_t - (Y_t - \beta_2 Z_t) = X_t - (1+\beta_1) Y_t + \beta_2 Z_t$$ and we have trivially found a linear combination of the three that is stationary - this is multicointegration. At the moment I don't know whether this can be simplified - maybe someone else does.

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You prove there is multiple cointegration, but I'm not sure if that necessarily implies cointegration for the third pair. I looked for papers to check if 2 pair-wise cointegrations imply cointegration on the third pair, but didn't find anything. Typically Johansen's test is what you use to test for multiple cointegration. In practice, I don't really spend much time worrying about these issues as I can always just estimate whatever relationships I want. –  John Jan 21 at 15:29
@John I agree - I did not answer the question. I just thought writing things down helps. If a real answer comes in I could delete mine. –  Richard Jan 21 at 15:33
I think doing some kind of simulation might help clarify things, but I'm not sure how to get an analytical answer. –  John Jan 21 at 15:35
In the same manner as Richard did, I tried to write things down, and here is my findings: We assume: 1) $X_t - \beta_1 Y_t = \epsilon_t$ 2) $Y_t - \beta_2Z_t = \eta_t$ hence if you multiply equation (2) by $\beta_1$ and substract to equation (1) you end up with: $X_t + \beta_1\beta_2 Z_t = \epsilon_t - \beta_1 \eta_t = \mu_t$ If $\mu_t$ is stationary then i think that you have you co-intergration relation with exact value for co-integration paramater. $\mu_t$ would stationnary only if $\epsilon_t$ and $eta_t$ are independent, which isn't guaranted (problem of the dog and the drunk) –  aajajim Jan 21 at 18:29
That's where I got stuck on too. –  John Jan 21 at 19:37