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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1$\begingroup$ What is $B_t$? the bank account $\exp(\int_0^t r_s ds)$ ? Please add moredetais. Where do you have the claim from? $\endgroup$ – Ric Dec 30 '14 at 9:11
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1$\begingroup$ Hi @Richard , yes $B_t$ is the bank account. Sorry for my defect. $\endgroup$ – Roozbe Dec 30 '14 at 9:18
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1$\begingroup$ 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 ... $\endgroup$ – Ric Dec 30 '14 at 9:25
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$\begingroup$ Yes @Richard , I think that I solved it .Thanks friend. $\endgroup$ – Roozbe Dec 30 '14 at 9:28
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$\begingroup$ @Roozbe Could you post your solution here for sake of completeness? $\endgroup$ – vanguard2k Dec 30 '14 at 9:39
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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)..
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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.
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2$\begingroup$ You can find the same argument in Brigo/Mercurio - Interest Rate Models - Chapter 3.2.2. $\endgroup$ – vanguard2k Dec 30 '14 at 14:40
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$\begingroup$ Hi @vanguard2k , Thank you so much for your help, I'll check it. $\endgroup$ – Roozbe Dec 30 '14 at 14:53
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1$\begingroup$ The last formula is not trivial. You need the expectaion of the exponential of the integral of geometric Brownian motion. Is the integral of GBM log-normally distributed? The product yes, the integral/sum? It is not clear. $\endgroup$ – Ric Jan 2 '15 at 11:37
