# How to prove that a probability density has second moment?

The setting of my question is the following. We have a diffusion process

$$dS(t) = \mu S(t) \; dt + v(t,S(t)) \; dW(t)$$

where $W$ is a standard Brownian motion under an equivalent martingale measure $Q$ and $v$ satisfies all necessary regularity conditions for there to exist a unique solution (i.e. linear growth, local Lipschitz etc.). The Linear growth condition is the one important for my question:

$$v^2(t,y) \leq L(1+y^2) \text{ for every }t \in [0,T]$$

I'm reading an article where it's claimed that due to the linear growth assumption the marginal density function $p_t(\cdot) = \frac{d}{dy}Q[S(t) \leq y]$ must have second moment i.e. $E[S^2(t)]<\infty$.

It's just mentioned in a footnote so I get the feeling that it's a well known result, but I cannot prove it! Any help is appreciated!

$v(t,S(t))$ needs to have at most linear growth. For instance $v(t,S(t)) = \left( 1+ S(t)^2 \right)^{1/4}$ is regular, but has sub-linear growth. –  Sasha Feb 21 '12 at 15:35
Since $\mathrm d\langle S\rangle(t)=v^2(t,S(t))\mathrm dt$, Itô's formula for $X(t)=S(t)^2$ reads $$\mathrm dX(t)=2\mu X(t)\mathrm dt+2v(t,S(t))S(t)\mathrm dW(t)+v^2(t,S(t))\mathrm dt.$$ Hence $x(t)=\mathrm E(X(t))=\mathrm E(S(t)^2)$ is such that $$x'(t)=2\mu x(t)+\mathrm E(v^2(t,S(t))),$$ as long as $x(t)$ stays finite. Now, $v^2(t,s)\leqslant L\cdot(1+s^2)$ for some finite $L$, hence $$x'(t)\leqslant2\mu x(t)+L\cdot(1+x(t)).$$ This implies that $y(t)=\mathrm e^{-(2\mu+L)t}x(t)$ is such that $y'(t)\leqslant L\mathrm e^{-(2\mu+L)t}$ for every $t\geqslant0$, from which the finiteness of $y(t)$, hence of $x(t)$, is direct. To be more specific, $$x(t)\leqslant \mathrm e^{(2\mu+L)t}\left(x(0)+\frac{L}{2\mu+L}\right)-\frac{L}{2\mu+L}.$$
When going through the details I'm not sure that we know that the expectation of the Ito integral is zeros. So this approach basically boils down into showing that $E[\int_0^t S^2(u)v^2(u,S(u)) dt] < \infty$ –  DoubleTrouble Feb 21 '12 at 19:39