Calculating futures price

Consider a world as follows:

$$\frac{dB}{B} = r_tdt$$ $$\frac{dS}{S} = r_tdt - 0.05dW_1 + 0.5dW_2$$ $$dr_t = 0.2 dW_1$$

where $$r_0=0$$. The Wiener processes $$W_1$$ and $$W_2$$ are independent. The price of any asset in this world is $$P_0 = E_0\left[\exp\left(-\int_0^T r_t dt\right)P_T\right ]$$

Calculate the futures price of a two-year futures contract on $$S$$.

My questions:

The futures price is just given by: $$E_0\left[S_T\right ]$$

But I am having trouble computing the above expression for the futures price.

• I'm sorry, maybe I'm missing something but how did you get the last term with lognormal $\sigma^2 = (0.15^2+0.5^2)T$? We have $\int_0^T r_tdt \sim N(0, 0.2^2T^3/6)$ and $-0.05W_{1,T} + 0.5W_{2,T} \sim N(0,(0.05^2+0.5^2)T)$. But the sum of these two normally distributed random variables should have a covariance? May 7, 2021 at 9:08
• Ah OK sorry, I read $r_t = 0.2 W_1(t)$ in your question. I'll rephrase my answer. May 7, 2021 at 10:36

So the first thing is to note that using Fubini (see here) $$\int_0^T r(t) dt = \int_0^T \int_0^t dr(u) dt = \int_0^T \int_u^T dt dr(u) = 0.2 \int_0^T (T-u) dW_1(u)$$ such that $$\int_0^T r(t) dt \sim \mathcal{N}\left( 0, 0.2^2 \, \int_0^T (T-u)^2 du = 0.2^2 \frac{T^3}{3} \right)$$ From that observation, in the expression $$S_T = S_0\exp\left(- (0.05^2+0.5^2)\frac{T}{2}\right) \exp\left( \int_0^T r_t dt - 0.05W_{1}(T) + 0.5W_{2}(T) \right)$$ The last term on the RHS is a lognormal with mean $$\mu = E_0\left[ \int_0^T r_t dt - 0.05W_1(T) + 0.5W_2(T) \right] = 0$$ and variance (Itô isommetry + independence of $$W_1$$ and $$W_2$$) \begin{align} \sigma^2 &= \Bbb{V}_0 \left[ \int_0^T (0.2(T-u)-0.05) dW_1(u) + 0.5W_2(T) \right] \\ &= \int_0^T (0.2(T-u)-0.05)^2 du + 0.5^2 T \\ &= 0.2^2 \frac{T^3}{3} + 0.01 T + 0.05^2 T + 0.5^2 T \end{align} Now using the fact that the expectation of a lognormal with parameters $$(\mu,\sigma^2)$$ is $$\exp(\mu+\sigma^2/2)$$ you get \begin{align} F(0,T) &= \Bbb{E}_0[S_T] \\ &= S_0\exp\left(- (0.05^2+0.5^2)\frac{T}{2}\right) \exp\left( 0.2^2 \frac{T^3}{6} + (0.01 + 0.05^2 + 0.5^2)\frac{T}{2} \right) \\ &= S_0\exp\left(0.005 T + 0.04 \frac{T^3}{6}\right) \end{align}