# Two identities in probability

I ma reading the book An introduction to stochastic processes in physics by Don Stephen Lemons. I have a question on two identities.

One identity is the identity (B.3) in Appendix b page 102. How to show that $$\int\int f(x,v)\frac{\partial p}{\partial t} dx\;dv =\left\langle \frac{df(X,V)}{dt} \right\rangle.$$ Here $X$ is a random variable (position) and $V$ (velocity) is a function of $X$.

The second identity is the first identity of (B.4) which is $$df = \frac{\partial f}{\partial X} dX + \frac{\partial f}{\partial V}dV + \frac{\partial^2 f}{\partial V^2}\frac{(dV)^2}{2}.$$ Why do we have the term $\frac{\partial^2 f}{\partial V^2}\frac{(dV)^2}{2}$? Thank you very much.

Edit: $X(t)$ is a random variable and $p(x,t)$ is its probability density: the probability of $X(t)$ takes value in $[x, x+dx]$ is $p(x,t)dx$. $dX=\sqrt{\delta^2dt}N_t(0,1)$. $dV=a(X,V)dt + \sqrt{b^2(X,V)dt}N_t(0,1)$. Here $a,b$ are two functions. $N_t(0,1)$ is the random variable with normal distribution. $\langle X \rangle$ is the expectation of X. $f(X,V)$ is a smooth function.

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Since it's physics, would I be right in guessing that $\langle Y\rangle$ means the expected value of $Y$? – Michael Hardy Sep 29 '11 at 21:36
....and that maybe $p$ means momentum? – Michael Hardy Sep 29 '11 at 21:36
The Google Books page you link to doesn't allow me to view this part of the book. Independent of that, questions should ideally be self-contained. I don't think there's enough context to answer your questions. – joriki Sep 29 '11 at 21:40
Cliche to say, but second identity reminds Ito lemma. Do you know what is it? – Ilya Sep 29 '11 at 22:00
@user9791 : the first identity it looks like Integration by Parts to me. – TheBridge Sep 30 '11 at 6:52