# Derivation of continuous Euler-Maruyama approximation

I am currently going through the Euler-Maruyama approximation and have question regarding the linear interpolation of it.

For the stochastic differential equation $$dX_t=a(t,X_t)dt+\sigma(t,X_t)dW_t \quad \ X_0=x_0 \quad \ t \in [0,T],$$ using discretization points $0=t_0\le \dots \le t_k \le \dots t_n=T, \quad k=0,\dots,n$ and timestep $\Delta=T/n$ the Euler-Maruyama scheme is given by

$$x_{k+1}=x_k+a(x_k)\Delta+\sigma(x_k)\Delta_kW \ \ \text{with} \ \ \Delta_kW=W_{t_{k+1}}-W_{t_k}.$$

Using piecewise linear interpolation we obtain the time continuous Euler-Maruyama scheme $$x^{c}_t=x_k+a(x_k)(t-t_k)+\sigma(x_k)(W_t-W_{t_k}) \ \ \text{for} \ \ t \in[t_k,t_{k+1}).$$

Up to this point everything is clear and I can verify that.

But now the biggest discretization point smaller or equal to $t$ denoted by $\lfloor t\rfloor=\{t_k:t_k\le t\}$, is used to define the scheme on $[0,T]$ and noticing $x^{c}_{t_k}=x_k$ we get the representation $$x^{c}_t=x_0 +\int_{0}^{t}a(x^{c}_{\lfloor u \rfloor})\ du+\int_{0}^{t}\sigma(x^{c}_{\lfloor u \rfloor})\ dW_u.$$ I don't understand how to get this representation.

I tried for $t\in[0,T]$ to interpolate between $\lfloor t \rfloor$ and $(\lfloor t\rfloor +\Delta)$ yielding: $x^{c}_t=x_k+ \frac{t-\lfloor t \rfloor }{\lfloor t \rfloor +\Delta-\lfloor t \rfloor}(x_{k+1}-x_k)$ and plugging in the above gives $x^{c}_t=x_k+a(x^{c}_{\lfloor t \rfloor})(t- \lfloor t \rfloor)+ \sigma(x^{c}_{\lfloor t \rfloor})(W_t-W_{\lfloor t \rfloor})$.

But how do I get to the integral representation from here ? I would appreciate any help or hints on how to get there. Many thanks in advance.