Integrability of linear SDE Consider a linear SDE of the form
$$dX_t = X_t(\alpha_tdt + \beta_t dW_t), \ X_0=x, $$
where $\alpha_t$, $\beta_t$ are $L^p$ integrable stochastic process:
$$\mathbb{E}\int_0^t |\alpha_s|^p ds , \mathbb{E}\int_0^t |\beta_s|^p ds  < \infty . $$
Can we say something about the existence/uniqueness of solution to $X$, and does $X$ inherit some integrability from the processes $\alpha$ and $\beta$?
 A: We can solve the SDE in a similar way to a Geometric Brownian Motion, see Wikipedia for example, up to time $\theta := \inf\left\{t : X_t = 0\right\}$. Next, if for $T > 0$ $$\mathbb{E}\left[\int_0^T\left|\alpha_s\right| ds + \int_0^T \left|\beta_s\right|^2 \right] < +\infty,$$ we have almost surely that
$$\left|\int_0^t \alpha_s ds \right| + \int_0^t \beta_s^2 ds + \left|\int_0^t \beta_s dW_s\right| < \infty$$
for all times $0 \leq t \leq T$ and that $ \theta > T$. As result we have unique explicit solution to SDE given by
$$ X_t = x \exp\left(\int_0^t \alpha_s ds + \int_0^t \beta_s dW_s - \frac{1}{2}\int_0^t \beta_s^2 ds\right), \quad t \in \left[0, T\right].$$
For the integrability of $X$, we need that $\int_0^T \left|a_s\right| ds$ and $\int_0^T \beta_s^s ds$ admit exponential moments; specifically we need there to exist $\lambda > 0$ such that
$$\mathbb{E}\left[\exp\left(\lambda \int_0^T \left|\alpha_s\right|ds + \lambda^2 \int_0^T \beta_s^2ds\right)\right] < \infty.$$
To simplify things, we suppose that $x = 1$. Let $k \geq 1$, using the the explicit formula for $X$ we get that
\begin{align}
\mathbb{E}\left[X_t^k\right] &= \mathbb{E}\left[\exp\left(k\int_0^t \alpha_s ds + k\int_0^t \beta_s dW_s - \frac{k}{2}\int_0^t \beta_s^2 ds\right)\right]\\
&=\mathbb{E}\left[\exp\left(k\int_0^t \alpha_s ds + \left(k^2 - \frac{k}{2}\right)\int_0^t \beta_s^2 ds + k\int_0^t \beta_s dW_s - k^2\int_0^t\beta_s^2 ds\right)\right]\\
&\leq \mathbb{E}\left[\exp\left(2k\int_0^t \alpha_s ds + 2\left(k^2 - \frac{k}{2}\right)\int_0^t \beta_s^2 ds\right)\right]^{1/2}\mathbb{E}\left[\exp\left(2k\int_0^t \beta_s dW_s - 2k^2\int_0^t\beta_s^2 ds\right)\right]^{1/2}
\end{align}
The last inequality is due to Cauchy-Schwarz. If $2k \leq \lambda$, then what is inside the second expectation satisfies Novikov's condition and thus is a martingale and the expectation is equal to 1. The first expectation is bounded by $\mathbb{E}\left[\exp\left(\lambda \int_0^T \left|\alpha_s\right|ds + \lambda^2 \int_0^T \beta_s^2ds\right)\right]^{1/2}$ which is finite by assumption and therefore $\mathbb{E}\left[X_t^k\right]$ is also finite.
We could probably refine the above argument to get weaker hypotheses on the coefficients $\alpha$ and $\beta$, but we need them to satisfy some kind of exponential moment condition. To see this, take $\alpha_t = Y $ and $\beta \equiv 0$, with $Y$ a log-normal random variable. For any $p \geq 1$,
$$\mathbb{E}\left[\int_0^T \left|\alpha_s\right|^p\right] = T^p \mathbb{E}\left[Y^p\right] < \infty,$$
since log-normal random variables admit moments of all orders. However for an $k > 0$,
$$\mathbb{E}\left[X_t^k\right] = \mathbb{E}\left[\exp\left(kt Y\right)\right] = \infty, $$
since the Laplace transform, i.e. moment generating function, of the log-normal distribution is infinite for positive arguments.
