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Published online by Cambridge University Press: 14 July 2016
The purpose of this note is to point out some difficulties which arise in connection with the stochastic calculus used in the previous paper and to suggest some ways in which these difficulties may be overcome. It turns out that if one wishes to restrict the application of the stochastic calculus to stochastic processes whose sample functions are step-functions with probability one, then difficulties of the following sort arise. Suppose X(ω, t) is a stochastic process whose sample functions are, with probability one, right-continuous step-functions in t on the interval [a, b). Then for every ω outside a null set N and every t ɛ [a, b), there is a δ(ω, t) such that < 5(w, t) implies X(ω, t + h) = X(ω, t) which in turn implies (X(ω, t + h) — X(ω, t))/h → 0 with probability one as Since convergence with probability one implies convergence in probability, it follows that the right-hand derivative of X(ω, t) in probability is zero at every t ɛ [a, b). Consequently, if the left-hand t ɛ [a, b), then it is also zero, but in general left-hand derivatives in probability on [a, b) may not even exist as finite-valued random functions. It is also worthwhile to observe that if a process is not differentiable in probability, then it is not differentiable in quadratic mean. From these observations it follows that the claims in Section 6 of the previous paper regarding the existence of non-zero derivatives in quadratic mean are in error, for if a derivative in quadratic mean exists, then it is a derivative in probability and is therefore zero. Incidentally, it turns out that the Poisson process has a zero derivative in probability but is not differentiable in quadratic mean. The fact that the Poisson process has a derivative in probability seems to depend rather crucially on the fact that it is also a process with independent increments.