I wonder would anybody tell me how to calculate $\mathbb{E}(B_3)$ Assuming that $\int_0^{t}r_s\,ds\sim N(0.03t,0.25t)$, then is


I have similar problem solved:

Assuming that $\int_0^t r_s ds \sim N(0.01t, 0.2t)$, then compute the value of $\mathbb{E}(D( 0,1))$.

here is my solution:

$D(t,T):=\frac{B_t}{B_T}=e^{\int_t^T r_s ds}$.

And, as the exponent is a normal distribution, we can apply the following,

$\mathbb{E} (D(0,1))= e^{μ + \sigma^2 / 2} = e^{-0.01+1(0.2) /2} = e^{0.09}$.

but what about $\mathbb{E}(B_3)$ from previous problem


1 Answer 1


This is rather similar to the solution you mentioned in your question :)

Let $(r_t)$ be the short rate with $\int_0^{t}r_s\mathrm{d}s\sim N(0.03t,0.25t)$ and $B_t$ the value of the bank account. Recall that by definition $\mathrm{d}B_t=r_tB_t\mathrm{d}t$ and thus $B_t=B_0\exp\left(\int_0^t r_s\mathrm{d}s\right)$. Thus, $(B_t)$ is for every time point $t$ log-normally distributed with \begin{align*} \mathbb{E}[B_3] &= B_0\mathbb{E}\left[\exp\left(\int_0^t r_s\mathrm{d}s\right)\right] \\ &= B_0 \mathbb{E}\left[e^{0.03t+\sqrt{0.25t}Z}\right] \\ &= B_0 e^{0.155t}, \end{align*} where $Z\sim N(0,1)$. I used that $\mathbb{E}\left[e^{\mu+\sigma Z}\right]=e^{\mu+\frac{1}{2}\sigma^2}$.

  • $\begingroup$ Thank you very much $\endgroup$ Nov 27, 2019 at 21:19

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