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I am trying to prove that a long-short strategy invested according to the cointegrated relationship from Engle-Granger's. So essentially I'm trying to show that the return $r_{XY}$ of the portfolio (X long Y short) has zero $\beta$ (in other words it is market neutral in CAPM sense).

CAPM says that $r_Y = r_m\beta_Y + \alpha_Y$, $r_X = r_m\beta_X + \alpha_X$, hence $$r_{XY}=r_X - br_Y = \beta_Xr_m + \alpha_X - b\beta_Yr_m - b\alpha_Y = r_m(\beta_X-b\beta_Y) + \alpha_X - b\alpha_Y$$

In order to be market neutral we need that $\beta_X-b\beta_Y=0$ but I'm struggeling a bit to prove this.

Since $\beta$ is defined as $$\beta = \frac{Cov(r_p,r_m)}{Var(r_m)}$$ I firstly need to find $r_X$ and $r_Y$ to find the $\beta$'s.

This is were I get stuck. I have been trying to use that, assuming the stocks continues to be cointegrated, we need that $X_t(1+r_X) = bY_t(1+r_Y) + \mu + \epsilon_t$ $(\star)$. Then solve for $r_X$ or $r_Y$ and plug it into our equations, but it doesn't really work. For $\beta_X-b\beta_Y$ to be $0$ we need that $r_X = br_Y+constant$, but this doesn't work with $(\star)$$$$$Is it possible to do it like this, or is there some other way to prove that it is market neutral in CAPM sense? Or maybe it isn't market neutral according to CAPM?

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1 Answer 1

you get what should get. You can't prove that strategy long $X$ short $Y$ is market neutral: is strategy long EUR/USD short USD/CHF risk neutral? I wouldn't say that. It depends, on what? On relationship between these variables, so it is perfectly hedged only if dX=dY so your task is bad stated: it should be rather: what should be

$b$

to assure that

$r_{XY}=r_X - br_Y=0$

and answer is $b = \frac{Cov(r_X,r_Y)}{Var(r_X)}$ so regression coefficient, under assumption that errors are normally distributed, of course they might be not and then not linear combination is optimal but some else, nonlinear, it has just assure that if X moves dX then Y moves by dY so you just need to know what position to take. Since you want to determine constant $b$ size in asset then if they don't create linear combination you can't find linear market neutral combination and forget about CAPM and your $\beta$

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I am not sure if I completely follow you. Are you talking about 'in practice'? In one of your previous comments (another question): "great. "a" is not important, if you want to be "market neutral" you go 1 stock of X long/short and 2 stocks of Y in the opposite dir. you can choose any other size ofcourse assuring that relation X/Y will be 1/2 – cf16". I'm only concerned about the theory here. I think it seems kind of intuitive that it would be indeed market netural if you invest according to this relationship. However, I wanted to see if this is consistent with CAPM. –  Good Guy Mike Apr 17 '13 at 12:48
    
Of course this does not hold in practice since these are just models. Also, I find that from $\beta_X - b\beta_Y = 0$ you get that $b = \frac{\beta_X}{\beta_Y} = \frac{Cov(r_X,r_m)}{Cov(r_Y,r_m)} \neq$ what you said? Although if this is indeed true this means that the market neutrality from the cointegration is consistent with CAPM? If CAPM says that b needs to be what you said, and that is indeed the the regression coefficient you get from the cointegration. But of course, these are just models and in practice this falls apart for several reasons. –  Good Guy Mike Apr 17 '13 at 12:56
    
@GoodGuyMike no market neutrality in cointegration by definition, cointegration in definition has nothing related to "Market Neutrality", all cointegration tells us is that errors will be nice, which is also what we are interested in when searching for neutrality –  0d0a Apr 17 '13 at 13:30
    
Hmm, don't understand this last comment. What do you mean by "all cointegration tells us is that errors will be nice, which is also what we are interested in when searching for neutrality"? If cointegration doesn't imply market neutrality (by definition, as you say). Isn't a long/short strategy that uses the concept of cointegration, such as pairs trading, always market neutral? –  Good Guy Mike Apr 17 '13 at 18:16
    
yes it is: because errors e=y-bx are stationary (at least in a weak sense) –  0d0a Apr 17 '13 at 18:44

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