Is the average of betas from Y ~ X and X ~ Y valid?

Question

I am interested in the relationship between two time series variables: $Y$ and $X$. The two variables are related to each other, and it's not clear from theory which one causes the other.

Given this, I have no good reason to prefer the linear regression $ Y = \alpha + \beta X$ over $ X = \kappa + \gamma Y $.

Clearly there is some relationship between $\beta$ and $\gamma$, though I recall enough statistics to understand that $\beta = 1/ \gamma$ is not true. Or perhaps it's not even close? I'm a bit hazy.

The problem is to decide how much of $X$ one ought to hold against $Y$.

I'm considering taking the average of $\beta$ and $1/ \gamma$ and using that as the hedge ratio.

Is the average of $\beta$ and $1/ \gamma$ a meaningful concept?

And as a secondary question (perhaps this should be another post), what is the appropriate way to deal with the fact that the two variables are related to each other -- meaning that there really isn't an independent and dependent variable?

Answer 1

To see the connection between both representations, take a bivariate Normal vector: $$ \begin{pmatrix} X_1 \\ X_2 \end{pmatrix} \sim \mathcal{N} \left( \begin{pmatrix} \mu_1 \\ \mu_2 \end{pmatrix} , \begin{pmatrix} \sigma^2_1 & \rho \sigma_1 \sigma_2 \\ \rho \sigma_1 \sigma_2 & \sigma^2_2 \end{pmatrix} \right) $$ with conditionals $$X_1 \mid X_2=x_2 \sim \mathcal{N} \left( \mu_1 + \rho \frac{\sigma_1}{\sigma_2}(x_2 - \mu_2),(1-\rho^2)\sigma^2_1 \right)$$ and $$X_2 \mid X_1=x_1 \sim \mathcal{N} \left( \mu_2 + \rho \frac{\sigma_2}{\sigma_1}(x_1 - \mu_1),(1-\rho^2)\sigma^2_2 \right)$$ This means that $$X_1=\underbrace{\left(\mu_1-\rho \frac{\sigma_1}{\sigma_2}\mu_2\right)}_\alpha+\underbrace{\rho \frac{\sigma_1}{\sigma_2}}_\beta X_2+\sqrt{1-\rho^2}\sigma_1\epsilon_1$$ and $$X_2=\underbrace{\left(\mu_2-\rho \frac{\sigma_2}{\sigma_1}\mu_1\right)}_\kappa+\underbrace{\rho \frac{\sigma_2}{\sigma_1}}_\gamma X_1+\sqrt{1-\rho^2}\sigma_2\epsilon_2$$ which means (a) $\gamma$ is not $1/\beta$ and (b) the connection between the two regressions depends on the joint distribution of $(X_1,X_2)$.

Answer 2

Converted from a comment.....

The exact values of $\beta$ and $\gamma$ can be found in this answer of mine to Effect of switching responses and explanatory variables in simple linear regression, and, as you suspect, $\beta$ is not the reciprocal of $\gamma$, and averaging $\beta$ and $\gamma$ (or averaging $\beta$ and $1/\gamma$) is not the right way to go. A pictorial view of what $\beta$ and $\gamma$ are minimizing is given in Elvis's answer to the same question, and in the answer, he introduces a "least rectangles" regression that might be what you are looking for. The comments following Elvis's answer should not be neglected; they relate this "least rectangles" regression to other, previously studied, techniques. In particular, note that Moderator chl points out that this method is of interest when it is not clear which is the predictor variable and which the response variable.

Answer 3

$\beta$ and $\gamma$

As Xi'an noted in his answer the $\beta$ and $\gamma$ are related to each other by relating to the conditional means $X|Y$ and $Y|X$ (which in their turn relate to a single joint distribution) these are not symmetric in the sense that $\beta \neq 1/\gamma$. This is neither the case if you would 'know' the true $\sigma$ and $\rho$ instead of using estimates. You have $$\beta = \rho_{XY} \frac{\sigma_Y}{\sigma_X}$$ and $$\gamma = \rho_{XY} \frac{\sigma_X}{\sigma_Y}$$

or you could say

$$\beta \gamma = \rho_{XY}^2 \leq 1$$

See also simple linear regression on wikipedia for computation of the $\beta$ and $\gamma$.

It is this correlation term which sort of disturbs the symmetry. When the $\beta$ and $\gamma$ would be simply the ratio of the standard deviation $\sigma_Y/\sigma_X$ and $\sigma_X/\sigma_Y$ then they would indeed be each others inverse. The $\rho_{XY}$ term can be seen as modifying this as a sort of regression to the mean.

• With perfect correlation $\rho_{XY} = 1$ then you can fully predict $X$ based on $Y$ or vice versa. The slopes will be equal $$\beta \gamma = 1$$
• But with less than perfect correlation, $\rho_{XY} < 1$, you can not make those perfect predictions and the conditional mean will be somewhat closer to the unconditional mean, in comparison to a simple scaling by $\sigma_Y/\sigma_X$ or $\sigma_X/\sigma_Y$. The slopes of the regression lines will be less steep. The slopes will be not related as each others reciprocal and their product will be smaller than one $$\beta \gamma < 1$$

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Is a regression line the right method?

You may wonder whether these conditional probabilities and regression lines is what you need to determine your ratios of $X$ and $Y$. It is unclear to me how you would wish to use a regression line in the computation of an optimal ratio.

Below is an alternative way to compute the ratio. This method does have symmetry (ie if you switch X and Y then you will get the same ratio).

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Alternative

Say, the yields of bonds $X$ and $Y$ are distributed according to a multivariate normal distribution$^\dagger$ with correlation $\rho_{XY}$ and standard deviations $\sigma_X$ and $\sigma_Y$ then the yield of a hedge that is sum of $X$ and $Y$ will be normal distributed:

$$H = \alpha X + (1-\alpha) Y \sim N(\mu_H,\sigma_H^2)$$

were $0 \leq \alpha \leq 1$ and with

$$\begin{array}{rcl} \mu_H &=& \alpha \mu_X+(1-\alpha) \mu_Y \\ \sigma_H^2 &=& \alpha^2 \sigma_X^2 + (1-\alpha)^2 \sigma_Y^2 + 2 \alpha (1-\alpha) \rho_{XY} \sigma_X \sigma_Y \\ & =& \alpha^2(\sigma_X^2+\sigma_Y^2 -2 \rho_{XY} \sigma_X\sigma_Y) + \alpha (-2 \sigma_Y^2+2\rho_{XY}\sigma_X\sigma_Y) +\sigma_Y^2 \end{array} $$

The maximum of the mean $\mu_H$ will be at $$\alpha = 0 \text{ or } \alpha=1$$ or not existing when $\mu_X=\mu_Y$.

The minimum of the variance $\sigma_H^2$ will be at $$\alpha = 1 - \frac{\sigma_X^2 -\rho_{XY}\sigma_X\sigma_Y}{\sigma_X^2 +\sigma_Y^2 -2 \rho_{XY} \sigma_X\sigma_Y} = \frac{\sigma_Y^2-\rho_{XY}\sigma_X\sigma_Y}{\sigma_X^2+\sigma_Y^2 -2 \rho_{XY} \sigma_X\sigma_Y} $$

The optimum will be somewhere in between those two extremes and depends on how you wish to compare losses and gains

Note that now there is a symmetry between $\alpha$ and $1-\alpha$. It does not matter whether you use the hedge $H=\alpha_1 X+(1-\alpha_1)Y$ or the hedge $H=\alpha_2 Y + (1-\alpha_2) X$. You will get the same ratios in terms of $\alpha_1 = 1-\alpha_2$.

Minimal variance case and relation with principle components

In the minimal variance case (here you actually do not need to assume a multivariate Normal distribution) you get the following hedge ratio as optimum $$\frac{\alpha}{1-\alpha} = \frac{var(Y) - cov(X,Y)}{var(X)-cov(X,Y)}$$ which can be expressed in terms of the regression coefficients $\beta = cov(X,Y)/var(X)$ and $\gamma = cov(X,Y)/var(Y)$ and is as following $$\frac{\alpha}{1-\alpha} = \frac{1-\beta}{1-\gamma}$$

In a situation with more than two variables/stocks/bonds you might generalize this to the last (smallest eigenvalue) principle component.

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Variants

Improvements of the model can be made by using different distributions than multivariate normal. Also you could incorporate the time in a more sophisticated model to make better predictions of future values/distributions for the pair $X,Y$.

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^{$\dagger$ This is a simplification but it suits the purpose of explaining how one can, and should, perform the analysis to find an optimal ratio without a regression line.}

Answer 4

Perhaps the approach of "Granger causality" might help. This would help you to assess whether X is a good predictor of Y or whether X is a better of Y. In other words, it tells you whether beta or gamma is the thing to take more seriously. Also, considering that you are dealing with time series data, it tells you how much of the history of X counts towards the prediction of Y (or vice versa).

Wikipedia gives a simple explanation: A time series X is said to Granger-cause Y if it can be shown, usually through a series of t-tests and F-tests on lagged values of X (and with lagged values of Y also included), that those X values provide statistically significant information about future values of Y.

What you do is the following:

• regress X(t-1) and Y(t-1) on Y(t)
• regress X(t-1), X(t-2), Y(t-1), Y(t-2) on Y(t)
• regress X(t-1), X(t-2), X(t-3), Y(t-1), Y(t-2), Y(t-3) on Y(t)

Continue for whatever history length might be reasonable. Check the significance of the F-statistics for each regression. Then do the same the reverse (so, now regress the past values of X and Y on X(t)) and see which regressions have significant F-values.

A very straightforward example, with R code, is found here. Granger causality has been critiqued for not actually establishing causality (in some cases). But it seems that you application is really about "predictive causality," which is exactly what the Granger causality approach is meant for.

The point is that the approach will tell you whether X predicts Y or whether Y predicts X (so you no longer would be tempted to artificially--and incorrectly--compound the two regression coefficients) and it gives you a better prediction (as you will know how much history of X and Y you need to know to predict Y), which is useful for hedging purposes, right?