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$\beta$ 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 $$\beta \gamma < 1$$

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).

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.

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 for the pair $X,Y$.

^{$\dagger$ 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.}

$\beta$ 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 $$\beta \gamma < 1$$

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).

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.

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 $X,Y$.

^{$\dagger$ 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.}

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 $\beta = 1/\gamma$. 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 $$\gamma = \rho_{XY} \frac{\sigma_X}{\sigma_Y}$$

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. But with $\rho_{XY} < 1$ 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 $\sigma_X/\sigma_Y$.

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 $$\beta \gamma < 1$$

$\beta$ 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 $\beta = 1/\gamma$. 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 $$\gamma = \rho_{XY} \frac{\sigma_X}{\sigma_Y}$$

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. But with $\rho_{XY} < 1$ 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 $\sigma_X/\sigma_Y$.

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).

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

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 is in terms of the regression coefficients $\beta = cov(X,Y)/var(X)$ and $\gamma = cov(X,Y)/var(Y)$ can also be expressed 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.

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 for the pair $X,Y$.

^{$\dagger$ 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.}

$\beta$ 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 $\beta = 1/\gamma$. 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 $$\gamma = \rho_{XY} \frac{\sigma_X}{\sigma_Y}$$

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. But with $\rho_{XY} < 1$ 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 $\sigma_X/\sigma_Y$.

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).

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.

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 for the pair $X,Y$.

^{$\dagger$ 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.}