In Salih Neftci's book on stochastic calculus, on pg 130/131 he considers example 1, a generalized Brownian motion process X_t with small (independent) incremental changes distributed as:He demonstrates that X_t is not a martingale, but on pg 132 he transforms it into a martingale, by subtracting off the deterministic drift:and Z_t is a martingale. I think I understand this. What I don't understand is on pg 133 he writes:QuoteThe 3rd example is more complicated and will only be partially dealt with here. Again assume that X_t is defined [as above] and consider the transformation:where \alpha is any real number. Suppose the mean of X_t is zero. Does this transformation result in a martingale?The answer is yes. We shall prove it in later chapters. However, notice something odd. The X_t is itself a martingale. Why is it that one still has to subtract the function of time g(t),I have two questions about this passage.1. How can increments of X_t be normally distributed as \Delta X_t \sim N(\mu\Delta t, \sigma^2\Delta t) and still have mean zero? Assuming \mu > 0, the expectation of X_t should become arbitrarily large the larger t becomes. How can we suppose the mean of X_t is zero?2. In example 1, Neftci explicitly writes that X_t is *not* a martingale (before we transform it). Yet in the 3rd example, he explicitly calls X_t a martingale. Is this a typo?

He says "suppose the mean of X_t is zero", so do that.A more pedantic presentation would have been:Define the "class of processes" which have mean mu. Then, in the 3rd example, he could have said: now we are talking about a specific process from that class, namely . Sometimes probabilists do that -- they put the parameters as subscripts or superscripts on the X. As you can see, while it is more explicit, it can typeset awkwardly -- so Neftci just said it in words.

Ahhh. OK, that makes *perfect* sense. Thanks for the explanation!