Why Expected Accumulated Dothan Short Rates Can Be Infinite

Summary

The document asks why the expected value of an accumulated quantity based on the Dothan short-rate model may be infinite. It gives the model’s geometric Brownian motion solution: the short rate is lognormally distributed at each time, with drift adjusted by half the volatility squared. The answers then connect the accumulated rate to an exponential of an integral and use an approximation involving the exponential of a Gaussian variable to argue that the expectation diverges.

The key mathematical point is that exponentiating a lognormal variable can produce an infinite expectation, even though the underlying variable has finite moments. However, the document’s approximation is not a complete proof for the time-integrated rate: it does not establish that the integral has the same expectation behavior as the proposed approximation. The cited links contain no additional proof in the text provided, so the argument should be treated as intuition rather than a rigorous derivation.

Key ideas

• The Dothan model makes the short rate a geometric Brownian motion.
• The solution has a lognormal distribution at each fixed time.
• Exponentials of lognormal variables can have infinite expectations.
• The document’s approximation does not by itself prove the claim for the full time integral.

Tags

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


# 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.

## Answer by Richi Wa (score 1, accepted)

https://quant.stackexchange.com/a/16054

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)..

## Answer by Roozbe (score 2)

https://quant.stackexchange.com/a/16018

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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