# For the Dothan model $E^Q[B(t)]=\infty$?

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.

• What is $B_t$? the bank account $\exp(\int_0^t r_s ds)$ ? Please add moredetais. Where do you have the claim from? – Richard Dec 30 '14 at 9:11
• Hi @Richard , yes $B_t$ is the bank account. Sorry for my defect. – Roozbe Dec 30 '14 at 9:18
• The short rate in this model is geometric Brownian motion, thus the bank account is the exponential of the integral of GBM. We have to check why this is/could be of infinite expectation ... – Richard Dec 30 '14 at 9:25
• Yes @Richard , I think that I solved it .Thanks friend. – Roozbe Dec 30 '14 at 9:28
• @Roozbe Could you post your solution here for sake of completeness? – vanguard2k Dec 30 '14 at 9:39

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.