For the Dothan model和問[B(t)]=∞EQ[B(t)]=∞E^Q[B(t)]=infty?
How can I show that for the Dothan short rate model We have $ E^Q[B(t)]=\infty $ ?
Where Dothan short rate model is " $ dr_t=ar_tdt+\sigma r_tdW_t $ “.
I appreciate any help.
Thanks.
I have to correct myself: Looking at the integral it is clear that $ E[\exp(\exp(Y))] $ is infinite for Gaussian (and most other) $ Y $ . The approximative argument can be found here: $ E[\exp(\int_0^{dt} r_u du)] \approx E[(r_0 + r_{dt})/2 dt] $ thus it is the expectation of the exponential of a log-normal (= exponential of a normal)..
First I must appreciate the @Richard’s help that cause to solved this question.
The Dothan model with this dynamic " $ dr_t=ar_tdt+\sigma r_tdW_t $ " is easily integrated
$ r(t)=r(s)exp ( \mu (t-s)+\sigma (W_t-W_s)) $
Where $ \mu=a-\frac{\sigma^2}{2} $
so We have
$ E^Q[B_t]=E^Q[exp(\int_0^t r(u)du)]\approx E^Q[e^{e^y}] $
Where $ y $ is Gaussian distributed so the expectation equals to infinite.
Please excuse my brevity.