Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Here is my story, I have the following function :
$$ g(x)=(1+x)\cdot\exp\left(-\frac{(\log(x+a)+c)^2}{2\sigma^2}\right)1[x\ge y]=f(x)\cdot1[x\ge y] $$
with $a,c,\sigma$ being "good" reals so that $g$ stays well defined and let $X_t$ be a geometric Brownian motion.

Here function $g$ is neither a convex function nor a difference of two convex functions because of the indicator function at $x=y$ where continuity is badly broken.

An "illegal" move then is to apply blindly Itô-Tanaka formula to $g(X_T)$ and get : $$ g(X_T)=g(X_0) +\int_0^T D^-g(X_t)dX_t+ \int_{\mathbb{R}}\Lambda_T(a)\mu(da) $$

Where $D^-$ is the left derivtive operator and $\mu$ is the "second derivative measure" (see for example theorem 7.1 page 218 in Karatzas and Shreve's book "Brownian Motion and Stochastic calculus").

Following this formula blindly I would get (weakly) :
$$ D^-g(x)=f'(x)\cdot1[x>y]+f(y)\cdot\delta_y(x) $$

Now getting $\mu$ seems to formally go like:
$$ \mu(dx)=f''(x)\cdot1[x\ge y]dx+f'(y)\cdot\delta_y(dx) +f(y)(\delta_y)'(dx) $$

So we get at the (probably wrong but appealing) formula : $$ \begin{align} g(X_T)&=g(X_0)+\int_0^T \left(f'(X_t)\cdot 1[X_t\ge y]+f(y)\delta_y(X_t)\right) dX_t+\int_y^{\infty}\Lambda_T(x)f''(x)dx \\ &+f'(y)\Lambda_T(y)-f(y)\partial_y\Lambda_T(y) \end{align} $$

Here many terms seem to be not well defined so I was deriving a heuristic (and terribly bad) calculation only to see where it was leading to. Anyway I am now wondering what is the correct result in this case.

When I say correct I mean that would make explicit the compensator of the $g(X_t)$ process using local time, because in the end I would like to take the expectation of $g(X_t)$, get rid of the local martingale parts and get the expectation of $g(X_t)$ in the form of the expectation of the compensator expressed in local time of $X_t$ + $g(X_0)$.

Best regards

PS: here the function $f$ was chosen as it seemed simple enough to be Itô differentiable but with reasonable properties so that expectation might exists.

share|cite|improve this question
I fixed formulas - hope that you're ok about it. – Ilya Mar 12 '12 at 16:58
@ Ilya : No probs it seems fine to me, thank's for the fixes. Best regards – TheBridge Mar 12 '12 at 17:02
Maybe you could define $h(x)=g(x)+f(y)1_{\{x<y\}}$. Then $h$ should satisfy the conditions of Problem 6.24 in Karatzas and Shreve, and so you could do what you had in mind for $h$. Then $E[g(X_T)]=E[h(X_T)]-f(y)P(X_T < y)$. – Jason Swanson Mar 12 '12 at 19:56
@TheBridge: correct me if I am wrong, from the OP I understood that you are interested in $\mathsf E g(X_t)$. Since $X_t$ is a Markov process which semigroup I think is known, why don't you use this semigroup to find the expectation a-la Black-Scholes formula? – Ilya Mar 12 '12 at 21:35
@TheBridge: I see - the different use of adjectives leads to a nice duality :) – Ilya Mar 20 '12 at 17:07
up vote 1 down vote accepted

Let $h=g + f(y)1_{(-\infty,y)}$. Then $h$ satisfies the conditions of Problem 6.24 in Karatzas and Shreve, and so can be written as a difference of two convex functions. We therefore have \[ h(X_T) = h(X_0) + \int_0^T D^-h(X_t)\,dX_t + \int_{\mathbb{R}} \Lambda_T(x)\mu(dx). \] In this case, $D^-h = 1_{(y,\infty)}f'$ and \[ \mu(dx) = (1_{(y,\infty)}f'')(x)\,dx + f'(y)\delta_y(dx). \] Hence, \[ h(X_T) = h(X_0) + \int_0^T f'(X_t)1_{\{X_t>y\}}\,dX_t + \int_y^\infty \Lambda_T(x)f''(x)\,dx + f'(y)\Lambda_T(y). \] Since $h=g + f(y)1_{(-\infty,y)}$, this gives \begin{multline*} g(X_T) = g(X_0) + f(y) 1_{\{X_0 < y\}} - f(y)1_{\{X_T < y\}}\\ + \int_0^T f'(X_t)1_{\{X_t>y\}}\,dX_t + \int_y^\infty \Lambda_T(x)f''(x)\,dx + f'(y)\Lambda_T(y). \end{multline*}

share|cite|improve this answer
@ user11867 : Very nice idea thanks. – TheBridge Mar 12 '12 at 21:38

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.