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.
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Sign up to join this communityHow 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.