- #1
Siron
- 148
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Hello!
In the topic I'm working on, I derived that the parameter $\sigma>0$ satisfies
$$\sigma = \sqrt{\frac{\nu}{T} - \lambda(\mu^2+\delta^2)},$$
where $\nu>0$ denotes the variance and $T>0$ denotes the time. The parameters $\mu \in \mathbb{R}, \delta>0$ and $\lambda>0$ are parameters that I have already computed before $\sigma$ (so I already know their values for different $T$). The problem is: how to guarantee that $\sigma>0$ is real? This can be stated equivalently as:
$$\mu^2+\delta^2 < \frac{\nu}{\lambda T} (*)$$
which means that the couples $(\mu,\delta)$ have to lie within a half circle with radius $\sqrt{\nu}/(\sqrt{\lambda T})$. Now, in most of my computations the derived couples $(\mu,\delta)$ do not satisfy $(*)$. What are some proper mathematical ways to make adjustments such that $(*)$ is satisfied? Obviously I don't want to adjust too many parameters.
Approach 1
My first approach was to find a couple $(\mu^{*},\delta^{*})$ within the half circle such that the distance between the derived couple $(\mu,\delta)$ and $(\mu^{*},\delta^{*})$ is minimal. To that end, I first computed the couple $(\mu',\delta')$ on the half circle, that is, $(\mu')^2+(\delta')^2 = \nu/(\lambda T)$, such that the distance between $(\mu,\delta)$ and $(\mu',\delta')$ is minimal whereafter I adjusted $\delta'$ by $\delta'-\epsilon$ for an arbitrary small $\epsilon>0$ to guarantee that the adjusted couple lies indeed within the half circle. Thus $\mu^{*} = \mu'$ and $\delta^{*} = \delta'-\epsilon$. Disadvantage is that I have to adjust two parameters.
Approach 2
My second thought was to increase the radius of the circle such that the derived couple $(\mu,\delta)$ lies within the half circle. To that end, I will adjust $\lambda>0$ such that $(*)$ is satisfied. Suppose that $(\mu,\delta)$ does not lie within the half circle, that is,
$$\mu^2+\delta^2 > \frac{\nu}{\lambda T},$$
which implies that
$$\lambda > \frac{\nu}{T(\mu^2+\delta^2)}.$$
Hence, replacing $\lambda$ by $\lambda = \nu/(T(\mu^2+\delta^2))-\epsilon$ guarantees that $\sigma$ is real.
NOTE: In the context of my work it's better to not adjust $\nu$ or $T$.
What are some other common approaches?
Many thanks!
In the topic I'm working on, I derived that the parameter $\sigma>0$ satisfies
$$\sigma = \sqrt{\frac{\nu}{T} - \lambda(\mu^2+\delta^2)},$$
where $\nu>0$ denotes the variance and $T>0$ denotes the time. The parameters $\mu \in \mathbb{R}, \delta>0$ and $\lambda>0$ are parameters that I have already computed before $\sigma$ (so I already know their values for different $T$). The problem is: how to guarantee that $\sigma>0$ is real? This can be stated equivalently as:
$$\mu^2+\delta^2 < \frac{\nu}{\lambda T} (*)$$
which means that the couples $(\mu,\delta)$ have to lie within a half circle with radius $\sqrt{\nu}/(\sqrt{\lambda T})$. Now, in most of my computations the derived couples $(\mu,\delta)$ do not satisfy $(*)$. What are some proper mathematical ways to make adjustments such that $(*)$ is satisfied? Obviously I don't want to adjust too many parameters.
Approach 1
My first approach was to find a couple $(\mu^{*},\delta^{*})$ within the half circle such that the distance between the derived couple $(\mu,\delta)$ and $(\mu^{*},\delta^{*})$ is minimal. To that end, I first computed the couple $(\mu',\delta')$ on the half circle, that is, $(\mu')^2+(\delta')^2 = \nu/(\lambda T)$, such that the distance between $(\mu,\delta)$ and $(\mu',\delta')$ is minimal whereafter I adjusted $\delta'$ by $\delta'-\epsilon$ for an arbitrary small $\epsilon>0$ to guarantee that the adjusted couple lies indeed within the half circle. Thus $\mu^{*} = \mu'$ and $\delta^{*} = \delta'-\epsilon$. Disadvantage is that I have to adjust two parameters.
Approach 2
My second thought was to increase the radius of the circle such that the derived couple $(\mu,\delta)$ lies within the half circle. To that end, I will adjust $\lambda>0$ such that $(*)$ is satisfied. Suppose that $(\mu,\delta)$ does not lie within the half circle, that is,
$$\mu^2+\delta^2 > \frac{\nu}{\lambda T},$$
which implies that
$$\lambda > \frac{\nu}{T(\mu^2+\delta^2)}.$$
Hence, replacing $\lambda$ by $\lambda = \nu/(T(\mu^2+\delta^2))-\epsilon$ guarantees that $\sigma$ is real.
NOTE: In the context of my work it's better to not adjust $\nu$ or $T$.
What are some other common approaches?
Many thanks!