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Consider a martingale $M_s$ and a function $f(M_s) = \frac{1-e^{-kM_s}}{M_s}$. If $M_s$ is an Ito process, it is obvious from Ito's Lemma that $f(M_s)$ is not a Martingale. Does the restriction hold more generally (for non-Ito processes), or can $f(M_s)$ be a martingale under some conditions?

More generally, I want to find $E_t[f(M_s)]$ and was wondering whether $E_t[f(M_s)] = f(M_t)$ under some assumptions on $M_s$ that could be reasonable in my application.

Giorgi Chavchanidze
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  • To get started assume $f$ is a little more: $C^2$ and apply Ito's formula. Then you see what $f$ must be so that only a $dW$-term (martingale) remains. – Kurt G. Nov 06 '23 at 18:42
  • What have you tried? – Jose Avilez Nov 06 '23 at 19:12
  • What is the context in which you find yourself asking this question? Right now, it seems both far too general to get any sense of concrete answers and $f$ seems too specialized for most common applications (i.e. in the context of Wiener processes, or more generally Ito processes, we'd often want to consider $f(s,M_s)$) As long as $f$ is allowed to depend on $M$ and $M$ can be as crazy as a general martingale with time domain $[0,N]$ can be, then this problem seems entirely intractable. – Brian Moehring Nov 06 '23 at 19:33
  • Thank you for your answers, I have clarified the question. – Giorgi Chavchanidze Nov 06 '23 at 20:42
  • Is $M$ a martingale or more generally an Ito process $X=M+A:$? In the latter case it is by far not obvious that $f(X_s) = \frac{1-e^{-kX_s}}{X_s}$ is not a martingale. By Ito's lemma, $$ df(X_s)=f'(X_s),dM_s+f'(X_s),dA_s+\frac12f''(X_s),d\langle M\rangle_s,. $$ What is obvious is that $f(X_s)$ is a local martingale when the sum of the last two terms vanishes. – Kurt G. Nov 07 '23 at 05:21
  • Thanks for the comment. Since I defined $M_s$ as a martingale from the beginning, what I meant by "if $M_s$ is an Ito process" is "if $M_s$ is a zero-drift Ito process". In that case, $f(M_s)$ is obviously not a martingale by your argument. – Giorgi Chavchanidze Nov 07 '23 at 20:52

1 Answers1

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Suppose $M_1 = x_1$. Since $X = M_2 - M_1$ can be an arbitrary centred random variable, and the identity must hold for this, we can see that for every centred (and integrable etc.) random variable $X$, and real $x_1$, $$\mathbb{E}[f(X + x_1)] = f(x_1).$$

This clearly holds if $f$ is an affine function. We shall argue that it can hold only if $f$ is affine.

Working first in $\mathbb{R},$ suppose for the sake of contradiction that $f$ is not affine. Then there exist some $a < b < c$ such that $$ f(b)(c-a) \neq f(a)(c-b) + f(c)(b-a).$$ Take $x_1 = b$, and let $X = c-b $ with probability $(b-a)/(c-a)$ and $a-b$ with probability $(c-b)/(c-a)$. By direct computation, $$ \mathbb{E}[f(X+x_1)] = \frac{(c-b) f(a) + (b-a) f(c)}{c-a},$$ but by assumption this is not equal to $f(b) = f(x_1)$. It follows that $f$ must be affine everywhere.

The same argument works in $\mathbb{R}^d$ as well: $f$ is not affine iff there exist some $d$ linearly independent points $a_1,\cdots a_d,$ and a point $b = \sum \lambda_i a_i$ for $\lambda_i \ge 0, \sum \lambda_i = 1$, such that $f(b) \neq \sum \lambda_i f(a_i).$ Then again taking $x_1 = b,$ and $X = a_i - b$ with probability $\lambda_i$ serves as a witness violating the equality for nonaffine functions.

stochasticboy321
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  • I doubt that the same argument works for a function $f:\mathbb R^d\to\mathbb R,,d>1,.$ The function $f(x,y)=x^2-y^2$ is harmonic in $\mathbb R^2$ and by Ito's formula $f(W_t)$ is a martingale when $W_t$ is a two-dimensional Brownian motion. – Kurt G. Nov 06 '23 at 20:32
  • @KurtG. On its face, the answer given here implies there is some 2d martingale $M$ such that $f(M_t)$ isn't a martingale, not that $f(W_t)$ isn't a martingale. – Brian Moehring Nov 06 '23 at 21:16
  • @KurtG. I guess I interpreted the quesiton differently from what was intended. In the original writing, there was no information about $M$, and as I understood it, the question was what kinds of functions admit that (effectively) $E[f(X)] = f(E[X[)$ for all random variables $X$. This is somewhat obviously only affine maps, but I hadn't explicitly worked out a proof before, and found it neat enough to write down :) – stochasticboy321 Nov 06 '23 at 23:55