I have a discrete stochastic process that can be described with the following equation
$$y_{(i+1)\tau}=k y_{i\tau}+w\xi_{i+1}$$
where $i\in\Bbb{N}$, $(\xi_i)_i$ is an IID sequence with $\xi_i \sim N(0,1)$, $0\ll\tau\in\Bbb{R}$, $0<k<1$ and $0<w\in\Bbb{R}$
And I want to approximate it with a Ornstein-Uhlenbeck Process. My final derivation is
$$dy=\frac{\ln{(k)}}{\tau}ydt+\frac{w}{\sqrt{\tau}}dW_t$$
where $W_t$ is the usual Wiener Process.
Is this correct?
---EDIT---
Following Calculon suggestions I came up with the following derivation.
Starting with the SDE. $$dy_t = ay_tdt + bdW_t$$
The solution is $$y_t = y_0e^{at} + be^{at}\int_0^te^{-as}dW_s$$
So we can see that $$y_{t + \Delta t} = y_te^{a\Delta t} + be^{a\Delta t}\int_{0}^{\Delta t}e^{-as}dW_{t+s}$$ Thus it follows that $$y_{(i+1)\tau} = y_{i\tau}e^{a\tau} + be^{a\tau}\int_{0}^{\tau}e^{-as}dW_{t+s}$$
So we note that $$be^{a\tau}\int_{0}^{\tau}e^{-as}dW_{t+s}\sim N\left(0,b^2e^{2a\tau}\int_{0}^{\tau}e^{-2as}\,ds\right)=N\left(0,b^2\frac{e^{2a\tau}-1}{2a}\right)$$
So we rewrite $$y_{(i+1)\tau} = e^{a\tau}y_{i\tau} + p\xi_{i+1}$$ where $(\xi_i)_i$ is an IID sequence of standard normal random variables and $p = \sqrt{\frac{b^2}{2a}(e^{2a\tau}-1)}$
So we match the coefficients $a$ and $b$ to $k$ and $w$ in my discrete-time model: $e^{a\tau} = k$ and $p = w$. Hence, $$a = \frac{\log{k}}{\tau}$$ and $$b = \frac{w}{\sqrt{\tau}}\sqrt{\frac{2\ln{k}}{k^2-1}}$$
First of all, start with this SDE. $$dy_t = ay_tdt + bdW_t \qquad \triangledown$$
The solution to this SDE is given by $$y_t = y_0e^{at} + be^{at}\int_0^te^{-as}dW_s$$
We can write $$y_{(i+1)\tau} = y_0e^{a(i+1)\tau} + be^{a(i+1)\tau}\int_{0}^{(i+1)\tau}e^{-as}dW_s \qquad \square$$ $$y_{i\tau} = y_0e^{ai\tau} + be^{ai\tau}\int_{0}^{i\tau}e^{-as}dW_s \qquad \triangle$$ Multiplying ($\triangle$) by $e^{a\tau}$ we get $$e^{a\tau}y_{i\tau} = y_0e^{a(i+1)\tau} + be^{a(i+1)\tau}\int_{0}^{i\tau}e^{-as}dW_s \qquad \blacktriangle$$ Decomposing the stochastic integral in ($\square$) we get $$y_{(i+1)\tau} = y_0e^{a(i+1)\tau} + be^{a(i+1)\tau}\int_{0}^{i\tau}e^{-as}dW_s+be^{a(i+1)\tau}\int_{i\tau}^{(i+1)\tau}e^{-as}dW_s \qquad \blacksquare$$
Now by $(\blacktriangle)$ and ($\blacksquare$) it follows that $$y_{(i+1)\tau} = e^{a\tau}y_{i\tau}+be^{a(i+1)\tau}\int_{i\tau}^{(i+1)\tau}e^{-as}dW_s \qquad \star$$ Note that $$be^{a(i+1)\tau}\int_{i\tau}^{(i+1)\tau}e^{-as}dW_s\sim N(0,b^2e^{2a(i+1)\tau}\int_{i\tau}^{(i+1)\tau}e^{-2as}\,ds)$$ You can check yourself that $$b^2e^{2a(i+1)\tau}\int_{i\tau}^{(i+1)\tau}e^{-2as}\,ds = \frac{b^2}{2a}(e^{2a\tau-1)}$$ Now rewrite ($\star$) as $$y_{(i+1)\tau} = e^{a\tau}y_{i\tau} + p\xi_{i+1} \qquad \bigstar$$ where $(\xi_i)_i$ is an IID sequence of standard normal random variables (why independent?) and $p = \sqrt{\frac{b^2}{2a}(e^{2a\tau}-1)}$
It is now possible to match the coefficients $a$ and $b$ to $k$ and $w$ in your discrete-time model. Clearly, $e^{a\tau} = k$ and $p = w$. Hence, $$a = \frac{\log{k}}{\tau}$$ and $$\lvert b\rvert = w\sqrt{\frac{2a}{e^{2a\tau}-1}}$$ Without loss of generality you can pick $b > 0$ (why?). Hence,
$$b = w\sqrt{\frac{2\log{k}}{\tau(k^2-1)}}$$
Note that if you start with the O-U process in ($\triangledown$) then the corresponding AR(1) process with the parameters prescribed above will be an exact discretization of the O-U process for any $\tau$.