Approach: Write $\text{Var}(\text{PnL})$ as a quadratic in $h$, take the derivative, set to zero.

Formal Solution:

Your P\&L is $\text{PnL} = \mu - X + hY$. The mean is $\mu$ (since $E[X] = 0$ in a short-horizon returns model and $E[Y] = 0$). The variance is:

$\text{Var}(\text{PnL}) = \text{Var}(-X + hY) = \sigma_X^2 - 2h\rho\sigma_X\sigma_Y + h^2\sigma_Y^2$

This is a convex quadratic in $h$. Take the derivative and set to zero:

$\frac{d}{dh}\text{Var}(\text{PnL}) = -2\rho\sigma_X\sigma_Y + 2h\sigma_Y^2 = 0$

Solving:

$h^{*} = \frac{\rho\sigma_X\sigma_Y}{\sigma_Y^2} = \rho\frac{\sigma_X}{\sigma_Y}$

This is the classic minimum-variance hedge ratio -- exactly the slope coefficient from regressing $X$ on $Y$.

Substitute $h^{*}$ back into the variance:

$\text{Var}^{*}(\text{PnL}) = \sigma_X^2 - 2\rho\frac{\sigma_X}{\sigma_Y}\cdot\rho\sigma_X\sigma_Y + \rho^2\frac{\sigma_X^2}{\sigma_Y^2}\cdot\sigma_Y^2$

$= \sigma_X^2 - 2\rho^2\sigma_X^2 + \rho^2\sigma_X^2 = \sigma_X^2(1 - \rho^2)$

The factor $(1 - \rho^2)$ is precisely the fraction of unexplained variance -- what regression cannot remove.

For part 3, we need:

$\frac{\mu}{\text{Var}^{*}(\text{PnL})} > S_0 \implies \frac{\mu}{\sigma_X^2(1 - \rho^2)} > S_0$

Rearranging:

$1 - \rho^2 < \frac{\mu}{S_0\sigma_X^2} \implies \rho^2 > 1 - \frac{\mu}{S_0\sigma_X^2}$

Define $R = \mu / (S_0\sigma_X^2)$. If $R \geq 1$, the target is met for all $\rho$ (even unhedged). If $R < 1$, you need:

$|\rho| > \sqrt{1 - R} = \sqrt{1 - \frac{\mu}{S_0\sigma_X^2}}$

Answer:

1. $h^{*} = \rho\,\sigma_X / \sigma_Y$ (the regression beta of $X$ on $Y$).

1. $\text{Var}^{*}(\text{PnL}) = \sigma_X^2(1 - \rho^2)$.

1. The risk-adjusted ratio exceeds $S_0$ when $|\rho| > \sqrt{1 - \mu/(S_0\sigma_X^2)}$, provided $\mu < S_0\sigma_X^2$; otherwise, even the unhedged position meets the target.

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