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jak
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When we integrate a function, we must make some choice about how we approximate it before we take the limit.

enter image description here

In principle, we can choose $\tau_i$ to be any value between $t_{i-1}$ and $t_i$. But for an ordinary Riemann integral our choice doesn't matter since for any value of the intermediate point $\tau \equiv \frac{\tau_i}{t_i-t_{i-1}}$, we find the same value in the limit of vanishing box sizes.

enter image description here

For stochastic integrals, however, this is no longer the case. For example, for the Itô integral, we choose $\tau =0$, while for the Stratonovich integral we choose $\tau = 0.5$.

I'm wondering what feature of stochastic integrals leads to their dependence on the choice of $\tau$? (Since I'm a physicist by trade, a somewhat intutive argument would be great.)I'm wondering what feature of stochastic integrals leads to their dependence on the choice of $\tau$? (Since I'm a physicist by trade, a somewhat intutive argument would be great.)

When we integrate a function, we must make some choice about how we approximate it before we take the limit.

enter image description here

In principle, we can choose $\tau_i$ to be any value between $t_{i-1}$ and $t_i$. But for an ordinary Riemann integral our choice doesn't matter since for any value of the intermediate point $\tau \equiv \frac{\tau_i}{t_i-t_{i-1}}$, we find the same value in the limit of vanishing box sizes.

enter image description here

For stochastic integrals, however, this is no longer the case. For example, for the Itô integral, we choose $\tau =0$, while for the Stratonovich integral we choose $\tau = 0.5$.

I'm wondering what feature of stochastic integrals leads to their dependence on the choice of $\tau$? (Since I'm a physicist by trade, a somewhat intutive argument would be great.)

When we integrate a function, we must make some choice about how we approximate it before we take the limit.

enter image description here

In principle, we can choose $\tau_i$ to be any value between $t_{i-1}$ and $t_i$. But for an ordinary Riemann integral our choice doesn't matter since for any value of the intermediate point $\tau \equiv \frac{\tau_i}{t_i-t_{i-1}}$, we find the same value in the limit of vanishing box sizes.

enter image description here

For stochastic integrals, however, this is no longer the case. For example, for the Itô integral, we choose $\tau =0$, while for the Stratonovich integral we choose $\tau = 0.5$.

I'm wondering what feature of stochastic integrals leads to their dependence on the choice of $\tau$? (Since I'm a physicist by trade, a somewhat intutive argument would be great.)

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edit approved Jan 4, 2020 at 17:08
Oli
edit approved Jan 4, 2020 at 17:08
Oli
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Why do stochastic intergalsintegrals depend on the choice of partitioning points?

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jak
jak
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Why do stochastic intergals depend on the choice of partitioning points?

When we integrate a function, we must make some choice about how we approximate it before we take the limit.

enter image description here

In principle, we can choose $\tau_i$ to be any value between $t_{i-1}$ and $t_i$. But for an ordinary Riemann integral our choice doesn't matter since for any value of the intermediate point $\tau \equiv \frac{\tau_i}{t_i-t_{i-1}}$, we find the same value in the limit of vanishing box sizes.

enter image description here

For stochastic integrals, however, this is no longer the case. For example, for the Itô integral, we choose $\tau =0$, while for the Stratonovich integral we choose $\tau = 0.5$.

I'm wondering what feature of stochastic integrals leads to their dependence on the choice of $\tau$? (Since I'm a physicist by trade, a somewhat intutive argument would be great.)