Return to Answer

As mentioned in an answer here and in the blog here,

A sufficient condition for the integral $\int_0^t f(\omega, s)\, dB_s$ to be a martingale on $[0,T]$ is that

1. $f(\omega,s)$ is adapted, measurable in s, and
2. $\mathbb{E}\left(\int_0^T f^2(\omega,s)\,ds\right) < \infty$. In this case, indeed, $\mathsf{E} \left(\int_0^T f(\omega,s)\, dB_s\right)=0$.

So if those conditions are satisfied for the given $f,g$ (eg. the integrability condition), then yes.

If we don't have the integrability, that integral will not even be well-defined/finite and so we cannot compute its expectation. Also, by taking $f=x^{p}$ for some $0<p<1/3$, we then deal with rough-integrals.

As mentioned in an answer here and in the blog here,

A sufficient condition for the integral $\int_0^t f(\omega, s)\, dB_s$ to be a martingale on $[0,T]$ is that

1. $f(\omega,s)$ is adapted, measurable in s, and
2. $\mathbb{E}\left(\int_0^T f^2(\omega,s)\,ds\right) < \infty$. In this case, indeed, $\mathsf{E} \left(\int_0^T f(\omega,s)\, dB_s\right)=0$.

So if those conditions are satisfied for the given $f,g$ (eg. the integrability condition), then yes.

If we don't have the integrability, that integral will not even be well-defined/finite and so we cannot compute its expectation. Also, by taking $f=x^{p}$ for some $0<p<1/2$, we then deal with rough-integrals.

As mentioned in an answer here and in the blog here,

A sufficient condition for the integral $\int_0^t f(\omega, s)\, dB_s$ to be a martingale on $[0,T]$ is that

1. $f(\omega,s)$ is adapted, measurable in s, and
2. $\mathbb{E}\left(\int_0^T f^2(\omega,s)\,ds\right) < \infty$. In this case, indeed, $\mathsf{E} \left(\int_0^T f(\omega,s)\, dB_s\right)=0$.

So if those conditions are satisfied for the given $f,g$ (eg. the integrability condition), then yes.

If we don't have the integrability, that integral will not even be well-defined/finite and so we cannot compute its expectation. For example, taking $g=x^{p}$ for some $p<1/3$, we then deal with rough-integrals.

As mentioned in an answer here and in the blog here,

A sufficient condition for the integral $\int_0^t f(\omega, s)\, dB_s$ to be a martingale on $[0,T]$ is that

1. $f(\omega,s)$ is adapted, measurable in s, and
2. $\mathbb{E}\left(\int_0^T f^2(\omega,s)\,ds\right) < \infty$. In this case, indeed, $\mathsf{E} \left(\int_0^T f(\omega,s)\, dB_s\right)=0$.

So if those conditions are satisfied for the given $f,g$ (eg. the integrability condition), then yes.

If we don't have the integrability, that integral will not even be well-defined/finite and so we cannot compute its expectation. Also, by taking $f=x^{p}$ for some $0<p<1/3$, we then deal with rough-integrals.

As mentioned here,

A sufficient condition for the integral $\int_0^t f(\omega, s)\, dB_s$ to be a martingale on $[0,T]$ is that

1. $f(\omega,s)$ is adapted, measurable in s, and
2. $\mathbb{E}\left(\int_0^T f^2(\omega,s)\,ds\right) < \infty$. In this case, indeed, $\mathsf{E} \left(\int_0^T f(\omega,s)\, dB_s\right)=0$.

So if those conditions are satisfied for the given $f,g$ (eg. the integrability condition), then yes.

If we don't have the integrability, that integral will not even be well-defined/finite and so we cannot compute its expectation. For example, taking $g=x^{p}$ for some $p<1/3$, we then deal with rough-integrals.

As mentioned in an answer here and in the blog here,

A sufficient condition for the integral $\int_0^t f(\omega, s)\, dB_s$ to be a martingale on $[0,T]$ is that

1. $f(\omega,s)$ is adapted, measurable in s, and
2. $\mathbb{E}\left(\int_0^T f^2(\omega,s)\,ds\right) < \infty$. In this case, indeed, $\mathsf{E} \left(\int_0^T f(\omega,s)\, dB_s\right)=0$.

So if those conditions are satisfied for the given $f,g$ (eg. the integrability condition), then yes.

If we don't have the integrability, that integral will not even be well-defined/finite and so we cannot compute its expectation. For example, taking $g=x^{p}$ for some $p<1/3$, we then deal with rough-integrals.