New answers tagged brownian-motion
1
No need to use ito's lemma. You can simulate your process directly from the equation:
$$dS_t = \mu S_t dt + \sigma S_t^{\beta/2} dW_t$$
which means that:
$$S_{t_{i+1}}=S_{t_i}+\mu S_{t_i}(t_{i+1}-t_i)+\sigma S_{t_i}^{\beta/2}\sqrt{t_{i+1}-t_i}Z_{i}$$
where $Z_i$ is a realization of normal distribution with mean 0 and variance equal to 1.
1
Your equation looks ok. If interest rates are deterministic then forwards (being the same as futures) are driftless under the risk neutral measure. Otherwise, Forwards are driftless (i.e. martingales) under the corresponding forward measure while futures are martingales under the risk neutral measure.
2
Risk-neutral pricing is to help with relative value type questions: If I know the value of this what should the value of that be if it depends in some way on this. It doesn't help with absolute value type questions: Should I buy this or that, is the implied volatility too low or high etc. Those are generally "real world measure" type questions.
1
Let $$Z_t=Y_t\int_0^t\frac{a}{Y_s}ds$$
Then $Z_0=0$.
We differentiate $Z_t$ and obtain
$$dZ_t=\int_0^t\frac{a}{Y_s}dsdY_t+Y_t\frac{a}{Y_t}dt=\int_0^t\frac{a}{Y_s}ds(rY_tdt+\sigma Y_td\tilde{W_t})+adt$$
$$=rY_t\int_0^t\frac{a}{Y_s}dsdt+\sigma Y_t\int_0^t\frac{a}{Y_s}dsd\tilde{W_t}+adt$$
Then
$$dZ_t=rZ_tdt+\sigma Z_td\tilde{W_t}+adt$$
We have
$$d(e^{-rt}Z_t)=...
2
Let's take a standard Brownian motion $(B_t)$ and let's try to compute $\int_0^t B_s\mathrm{d}B_s$ in the Riemann-Stieltjes sense.
Let $0=t_0<t_1<...<t_n=t$ be a partition and let $y_i=t_{i-1}$ or $y=t_i$ for $i=1,...,n$ be two intermediate partitions. Thus,
\begin{align*}
S^1_n(t) &= \sum_{i=1}^n B_{t_{i-1}}(B_{t_i}-B_{t_{i-1}}), \\
S^2_n(t) &...
2
When you simulate a sample path of a standard Brownian motion, you are generating a sequence $(B_t)_{t \in \mathbb{\Pi}}$ where $\mathbb{\Pi} := \{t_0, ..., t_n\}$ is your time partition. You can view that sequence as $n$ draws of the same random variable, although no one could say that this isn't also 1 draw each of $n$ independent normal random variables.
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