Hi, within Tuckman's lognormal model, dr = a r dr + (vol) r dw. To clarify, is a equivalent to lambda (drift) and r equivalent to r(0), that is, the initial rate?
I am struggling a bit between the initial formula for the lognormal model (Tuckman 10.5) and the formula which we'd likely use on the exam, which seems to be Tuckman 10.12.
Is the connection here that if: dr = a r dr + (vol) r dw, and d(ln[r]) = a(t) dt + (vol) dw, then to get the rate in any period of the binomial tree, r(i) = r(0) * e^(d(ln[r]), which would be equivalent to r(i) = r(0) * dr = r(0) * (a dr + (vol) dw) = a r dr + (vol) r dw?
In other words, a r dr + (vol) r dw = r(0) * e^(d(ln[r])?
I am struggling a bit between the initial formula for the lognormal model (Tuckman 10.5) and the formula which we'd likely use on the exam, which seems to be Tuckman 10.12.
Is the connection here that if: dr = a r dr + (vol) r dw, and d(ln[r]) = a(t) dt + (vol) dw, then to get the rate in any period of the binomial tree, r(i) = r(0) * e^(d(ln[r]), which would be equivalent to r(i) = r(0) * dr = r(0) * (a dr + (vol) dw) = a r dr + (vol) r dw?
In other words, a r dr + (vol) r dw = r(0) * e^(d(ln[r])?
