Dothan's model - expected value of $B_t$

Question

Let's consider Dothan's model of the short-time interest rate:

$$d r_t = \mu r_t dt + \sigma r_t dW_t,$$ where

$r_0 = r$, $\sigma>0$, $\mu \in \mathbb{R}$.

Prove that: $\mathbb{E}(B_t)= \infty$, where $B_t$ is the banking account process.

My first idea was to find the solution of SDE $d r_t = \mu r_t dt + \sigma r_t dW_t$, which is: $r_t = r_0 \exp((\mu-\frac{\sigma^2}{2})t + \sigma W_t)$. Knowing that $W_t \sim N(0,t)$ and $B_t = \exp(\int\limits_0^t r_u du$), I tried to calculate $E(B_t)$, but I failed.

Please, give a reference for "Dothan's model" (name unknown to most users). — Jean Marie, Dec 21 '16 at 12:46
Of course, for example: https://web.mst.edu/~bohner/fim-10/fim-chap4.pdf section 4.3 — Elizabeth_Banks, Dec 21 '16 at 12:56
$B_t \ne \exp\left(-\int\limits_0^t r_u du\right)$ Indeed $B_t = \exp\left(\int\limits_0^t r_u du\right)$ — Behrouz Maleki, Dec 21 '16 at 13:05

Behrouz Maleki · Accepted Answer · 2016-12-21T13:25:12.943

One of the big drawback of lognormal models is the explosion of the bank account. Let $\Delta t$ be small, then $$\mathbb{E}[B_{\Delta t}]=\mathbb{E}\left[\exp\left(\int_0^{\Delta t}r_u du\right)\right]\approx\mathbb{E}\left[\exp\left(\frac{r_0+r_{\Delta t}}{2}\Delta t\right)\right]$$ We have $$\mathbb{E}[B_{\Delta t}]\approx\mathbb{E}[\exp(\exp(Y))]$$ where Y is Gaussian distributed. But such an expectation is infinite. This means that in arbitrarily small time the bank account growths to infinity in average.

Dothan's model - expected value of $B_t$

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