I have seen a few textbooks that say $ \int_{0}^{t} W_{s}ds$ is NOT a martingale. I believe they are correct, but I'm confused what is wrong with the logic below that seems to show it is a martingale. Could anyone tell me where I went wrong?
By Ito's Lemma:
$$
\int_0^t W_s ds = tW_t – \int_0^t sdW_s
$$
Simplifying gives:
$$
\int_0^t W_s ds
= \int_0^t t dW_s – \int_0^t s dW_s
= \int_0^t (t-s) dW_s
$$
And $\int_{0}^{t} (t-s)dW_{s} $ is an Ito Integral which is a martingale, so $ \int_{0}^{t} W_{s}ds$ must be a martingale.
Best Answer
Take $s< t$ and write $$ \int_0^t W_r\,dr=\int_0^sW_r\,dr+\int_s^t W_r\,dr. $$ Then $$ \mathsf{E}\left[\int_0^t W_r\,dr\mid \mathcal{F}_s\right]=\int_0^sW_r\,dr+\mathsf{E}\left[\int_s^t W_r\,dr\mid \mathcal{F}_s\right], $$ where $\{\mathcal{F}_t\}$ is the underlying filtration, and $$ \mathsf{E}\left[\int_s^t (W_r-W_s+W_s)\,dr\mid \mathcal{F}_s\right]=(t-s)W_s\ne 0. $$