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To simulate an Ito diffusion, $$dX_t = f(t,X_t)dt + g(t,X_t)dW_t,$$ you can discretize this in time by using an equidistant partition $(t_0^N,t_1^N,\dots,t_N^N)$ of the time interval $[0,T]$. The Euler-Maruyama method then uses $$X_{i+1}^{(N)} = X_{i}^{(N)} + f(t_i^N,X_i^{(N)}) \Delta t+ g(t_i^N,X_{i}^{(N)}) \Delta W_i,$$ for $i=0,1,\dots,N-1$. The value of $\Delta W_i$ can be sampled from a normal distribution $\mathcal N(0,\Delta t)$. Under regularity conditions on $f$ and $g$, the process $X^{(N)}$ will converge to $X$ for $N\rightarrow\infty$ (see Kloeden and Platen, Theorem 9.6.2, for a proof of the time-homogeneous case).

For efficiency, it is tempting to replace the value $\Delta W_i=W_{t_{i+1}}-W_{t_i}$ by a discrete random variable that equals $$\pm\sqrt{\Delta t}$$ with probabilities 0.5. This random variable has the same mean and variance as $W_{t_{i+1}}-W_{t_i}$. Call the resulting distribution $\tilde{X}^{(N)}$. Do the processes $\tilde{X}^{(N)}$ also converge to $X$? If so, what can be said about the rate of convergence?

Notes:

  1. On the Wikipedia page of the Milstein method there is a brief discussion of the different rates of convergence.
  2. I have simulated a call option price using this method and it gave the correct result.
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  • $\begingroup$ It's very easy to make a normal from a standard random variable: if $U,V$ are Uniform[0,1]., then $\sqrt(-2\log U)\sin(2\pi V)$ is standard normal. See the Box-Muller transformation. $\endgroup$ Commented Apr 16 at 14:58
  • $\begingroup$ @AnthonyQuas the Box-Muller transformation creates 2 normally distributed values from 2 uniform values in [0,1]. But from 1 uniform value in [0,1], you can extract at least 32 random bits using its binary expansion, making it 32 times faster. $\endgroup$ Commented Apr 16 at 16:49
  • $\begingroup$ However, I did my simulation in Python, and Python is very slow at putting random bits in an array. Therefore it was not faster than the usual method. Even in an optimized environment, my idea will only be significantly faster if extracting the random bits is the bottleneck of the proces. $\endgroup$ Commented Apr 16 at 17:05
  • $\begingroup$ only the first two moments of the noise are relevant for Brownian motion, provided that all moments are finite, so you can be confident on the convergence to the same process; I don't know about the rate of convergence, I would not expect it to depend sensitively on how you model the noise; we typically take Gaussian for convenience. $\endgroup$ Commented Apr 19 at 15:28
  • $\begingroup$ @CarloBeenakker do you have a source/proof for this claim? $\endgroup$ Commented Apr 19 at 15:39

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