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## New answers tagged stochastic-processes

0

By using the fact that the brownian integral has expected value $0$, we find \begin{eqnarray*} & & \left(1-e^{-\rho dt}\right)v(X_{t}) \\ &=& E_{t}\left\{ \int_{t}^{t+dt}e^{-\rho(s-t)}u(X_{s})ds+e^{-\rho dt}\left[\int_{t}^{t+dt}v'(X_{s})dX_{s}+\frac{1}{2}\int_{t}^{t+dt}\sigma(X_{s},s)^2 v''(X_{s})ds\right]\right\} \\ &= & E_{t}\left\{ ...

1

This is a special case of the question of why $$\int_0^T f(t) dW_t$$ is normally distributed for a continuous function $f(t).$ This Ito integral can be approximated by a sum $$\sum_{i=0}^{N-1} f(i T/N) (W_{(i+1)T/N} - W_{i T/N}) .$$ The Brownian increments $(W_{(i+1)T/N} - W_{i T/N})$ are independent normally distributed random variables. The key point ...

3

For a Brownian motion, if you wait $dt$, the variance will grow linearly with (proportionally to) $dt$. For a fractional Brownian motion, it will grow with a power law of $dt$, in fact in $dt^{H}$, where $H$ is the Hurst exponent. See wikipedia for more details. It means the fBM will somehow keep memory of the past. When $H$ is lower than 1/2, it will mean ...

3

The more phenomenological definitions in his books are probably more helpful. Whether one uses the fractal dimension, Hurst coefficient, or exponential coefficient alpha, there is a value that corresponds to pure Brownian motion, a regime relative to this value that corresponds to persistence of motion, and the opposite regime that corresponds to ...

1

For simplicity, We assume that $\alpha$ is a positive constant. You need to show that, for any $t>0$, \begin{align*} \int_0^t e^{\alpha u} dW_u \end{align*} is normally distributed. Consider the process $\{X_t, t \geq 0\}$, where \begin{align*} X_t = \frac{1}{\sqrt{\frac{1}{t}\int_0^t e^{2\alpha u} du}}\int_0^t e^{\alpha u} dW_u, \end{align*} for ...

0

I think that you are a bit confused: the support of the Black-Scholes model is $(0,+\infty)$, that is to say the underlying asset price is non-negative, like a stock. The Vasicek model has an OU process whose support is $(-\infty,+\infty)$, that is to say the underlying can be negative. Therefore all equivalent measures (of which the martingale is one) ...

0

It might be easier to go the other way: start with $$d\mathcal{E}_t = \mathcal{E}_t dX_t$$ apply Ito to the $\log$ function d\log(\mathcal{E})_t = \frac{1}{\mathcal{E}_t}d\mathcal{E}_t - \frac{1}{2} \frac{1}{\mathcal{E}_t^2}d\langle\mathcal{E},\mathcal{E}\rangle_t = \frac{1}{\mathcal{E}_t}\mathcal{E}_tdX_t - \frac{1}{2} ... 0 For question I, the identity \begin{align*} \rho_t = \exp\big(-\lambda_t W_t - \frac{1}{2} \lambda_t^2t\big) \end{align*} does not appear correct, unless \lambda_t is a constant. For question II, yes. If X_t = -\int_0^t \lambda_s dW_s, then \langle X \rangle_t = \int_0^t \lambda_s^2 ds. For question III, you need to note that \begin{align*} \langle X ... 3 There is a shortcut around the Forward Equation when you are looking for the stationary distribution. Let me write dX = \mu(X)dt +\sigma(X)dW $$for$$ \mu(x)=b(1-x)-ax\ \text{ and }\ \sigma^2(x)=x(1-x) $$The Forward Equation indeed states that the stationary distribution p(x) will be satisfied for \partial p/\partial t = 0, therefore$$ ...

0

So I have a "+" sign for the second term (not negative) dr =r[(θ(t)+ d(lnσ)/dt * lnr + 1/2*σ^2)dt + σdW] I left out the subscript t's..... You can let V = log r then Apply Ito and solve for A and B... where B = r*σ

0

You shall distinguish between the random variables and their distributions. That is, when we talk about Brownian motion, we often define it in terms of finite-dimensional distributions (that is, conditional distribution is normal with the current value as the mean and the variance of $1$). By Kolmogorov's Extension Theorem (KET), there exists a unique ...

3

Let's consider the following example: the process is initialized randomly with $\pm1$ and then stays there forever. Seems stationary to me, but it would never cross its mean.

2

The portfolio is self-financing. You simply forgot a term in $b$ and a $-t$ term in $V$: \begin{eqnarray} V_t &=& a_t S_t + b_t \beta_t = (2B_t ) (10+ B_t) + (- t - B_t^2 - 20B_t)1 \\ &=& 20B_t + 2B_t^2 - t - B_t^2 - 20B_t \\ &=& B_t^2 - t \end{eqnarray} Applying Ito's lemma \begin{eqnarray} dV_t &=& (2B_t dB_t + ...

1

As per your comments, this is the Kunita Watanabe decomposition. See the post at http://math.stackexchange.com/questions/413103/kunita-watanabe-decomposition and the presentation http://www.eurandom.nl/events/workshops/2011/ISI_MRM/Presentation/Vanmaele.pdf

0

Your choice of $a_t$ and $b_t$ is feasible. For a self-financing portfolio, the units invested should be static within an infinitesimal time interval, that is, no extra investing or withdrawing during this period. In other words, the portfolio value changes only through its underlyings. For a further discussion, See the article ...

0

For simplicity, we assume the necessary positivity, and then we can ignore the absolute signs. Note that \begin{align*} \big(C_1 - a e^{-bt} \big) d\omega = \big(C_1 - a e^{-bt} \big) f(t) dt + cbe^{-bt} \omega dt. \end{align*} That is, \begin{align*} \big(C_1 - a e^{-bt} \big)\,d\omega - cbe^{-bt} \omega \, dt= \big(C_1 - a e^{-bt} \big) f(t)dt. ...

0

User9403 nails it! Intuitively higher expected rate of return => higher stock price => higher (call) option price!

-2

"Why the expected return rate of a stock has nothing to do with its option price?" It has everything to do with the option price! The option price is a function of the stock price. If the expected rate of return on the stock price declines, the stock price will decline as will the option price. "Suppose I have two stocks A and B, the price is the same ...

2

I think to gain intution you have to understand that the same agents that value the stocks will value the options. And agents compensate for volatility by demanding higher expected returns. Therefore you should ask: Why are stocks priced as they are in the first place? In your example, the stock with higher volatility has much lower expected return. This ...

2

Yes and No. In the absence of arbitragers, the price of the option will be different for each speculator based on their drift expectations (and each speculator has a risk in his position and will limit his ability to trade large sizes to avoid bankruptcy) and the option price will converge to priced off a supply-and-demand driven drift expectation. ...

-1

The answer is no, because although a mean-reverting process has necessarily to be stationary, it is not true the opposite, that is a stationary process has to be mean-reverting, as you stated in the question. Look at this article for a formal definition of a mean-reverting process. Now, think about one of the most famous mean-reverting process: the ...

2

Because you can hedge. Once you have delta hedged, the pay-off is symmetric about up and down moves so drift doesn't matter. Also the delta-hedged call and the delta hedged put have to have the same value since they have the same pay-off. (Put-call parity) Yet any argument that the call should be worth more because of drift says that the put should be ...

1

Practically, it is very difficult to get a measurement of a stock's true drift while there are very well-documented processes to estimate volatility. It is therefore very convenient mathematically to select the risk neutral pricing measure that eliminates idiosyncratic drift. At its heart, Black Scholes constructs a dynamic, replicating portfolio for an ...

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