Suppose the diffusion process for the short rate $r_t$ under the risk-neutral measure $Q$ is given by: $$ dr_t = \theta(t)dt+\sigma dZ_t $$
where $Z_t$ is a Brownian motion.
I am trying to show that the parameter $\theta(t)$ is related to the slope of forward rate curve: $$ \theta(t) = \left. \frac{\partial F}{\partial T}(0,T) \right|_{T=t} + \sigma^2 t $$ where: $$ \partial F(0,T) = -\frac{\partial \log P(0, T)}{T} $$
I am lost as to where to start. Any help would be appreciated
You can apply this approach for your model (it's the Ho-Lee model) but also to other short rate models such as Hull-White model.
(1) First, integrate twice the short rate SDE to get $\int_0^t r_u du$, you will find out that it's gaussian with this distribution: $$ \begin{aligned} -\int_0^t r_udu &\sim \mathcal{N}(m_t = -r(0)t - \int_0^t \int_0^u \theta(v)dv, v_t = \frac{\sigma^2t^3}{6}) \end{aligned} $$ This might help you get there: Integral of Brownian motion w.r.t. time
(2) compute the zero-coupon bond price given by the model: $$ P(0,t) = \mathbb{E} \left[e^{-\int_0^t r_udu}\right]=e^{m_t + \frac{v_t}{2}} $$
(2) Then, take the logarithm and differentiate twice to get the desired expression: $$ \theta(t) = \left. \frac{\partial F(0,T) }{\partial T} \right|_{T=t} + \sigma^2 t $$
