# Tag Info

## Hot answers tagged euler

7

the LIBOR market model the Heston model -- Euler and Milstein are actually bad for this and much more sophisticated methods are necessary local volatility models

2

I would proceed as follows: \begin{align*} x_t &= x_{t-\delta t} + \alpha (\beta - x_{t-\delta t}) \delta t + \sigma x_{t-\delta t}^\gamma (w_t - w_{t-\delta t}),\\ x_t &= \max (x_t, \ 0), \end{align*} where $w_t - w_{t-\delta t}$ is a normal random variable with mean $0$ and variance $\delta t$, which can be obtained by an independent draw for each ...

2

For GBM you can show it by theory (see Kloeden) or show it empirically as follows. You can observe that EM has strong order of convergence equal to 0.5, whereas Milstein's is 1.0. Not always the latter is superior. E.g. simple EM beats simple Mistein for Heston. In this case is good to use adapted schemes (Giles) for instance, especially dealing with jumps. ...

1

In arithmetic brownian, drift does not depend on the previous price, so it is simply $\mu \Delta t$ as you have done. It depends on the previous price in geometric brownian though. Let’s recall the GBM equation: $dS_t=\mu S_t dt +\sigma S_t dB_t$ Discretising: $\Delta S_t=\mu S_t \Delta t + \sigma S_t \sqrt{\Delta t} N[0,1]$ $S_{t+1}-S_t=\mu S_t \Delta t ... 1 I figured it out. The problem is just, that I am not taking the same driving brownian motion. That would mean, if I am calculating$E[\bar X^{\delta}]$and$E[\bar X^{1/2\delta}]$. The sample of path is complete different and thus not comparable. For weak convergence, one need to be sure, that the sample path are almost surely the same. As a result, the ... 1 first, there is a formula for the continuously monitored case. second, if you use log coordinates the Euler discretization is exact so this should be done. third, the convergence for discretely monitored to continuously is actually very slow so you will need a lot of steps. fourth, it's actually better to draw the hitting time to the barrier rather than ... 1$r-\frac{\sigma^2}{2}\$ for the drift only applies to the log-returns. The Euler discretisation simply discretises the SDE directly. You'd use the risk-free rate for you drift under the risk-neutral measure for your question. For your reference: Please read the wikipedia for more details.

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