# Compute value of $\mathbb{E}(B_3)$

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

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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

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 $$\mathbb{E}\left[e^{\mu+\sigma Z}\right]=e^{\mu+\frac{1}{2}\sigma^2}$$.