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