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let $Y_t=\ln X_t$ by application of Ito lemma we have \begin{align} dY_t=\frac{1}{X_t}dX_t-\frac{1}{2X_t^2}d[X,X](t)=(v-\frac{1}{2}\sigma^2)dt+\sigma\,dW_t \end{align} by integration on $[t_1,t_2]$, we have \begin{align} Y_{t_2}=Y_{t_1}+(v-\frac{1}{2}\sigma^2)(t_2-t_1)+\sigma\,(W_{t_2}-W_{t_1}) \end{align} then \begin{align} ...


1

There are three main issues. As per my comment, one is the lack of specification for the distribution of the jumps (I'll assume that there is a $J_0 = 0$ at time 0 (otherwise, the process doesn't account for no jumps). Unless $P (J \leq -1) = 0$, your price process is problematic, and the Girsanov theorem is not applicable. To see why: $S_t = S_0 e^{\sigma ...


0

You have to make further assumptions on the distribution of $J_i$s. For example, if $J_i$s are iid normal, your option pricing problem becomes that of Merton (1976) and the solution to it is an infinite sum. If $J_i$s are assumed to be double exponential, you end up with Kou (2004) model and it has an analytical solution. Furthermore, there are three ...


4

You know that Brownian motion {W(t)} is a stochastic process with the following properties: (Independence of increments) W(t) − W(s) , for t > s , is independent of the past, that is, of W(u) , 0 ≤ u ≤ s, or of $F_s$ , the σ-field generated by W(u), u ≤ s. (Normal increments) W(t) − W(s) has Normal distribution with mean 0 and variance t − s. This implies ...


4

Let M denote the number of underlying traded assets in the model excluding the risk free asset, and let R denote the number of random sources. Generically we then have the following relations: 1. The model is arbitrage free if and only if M ≤ R. 2. The model is complete if and only if M ≥ R. 3. The model is complete and arbitrage free if and only if M = R. ...


4

The uniqueness of the risk-neutral measure comes from the abundance of tradable assets. Let $B_t$ be the money-market account at time $t$. Let $Q_1$ and $Q_2$ be two risk-neutral measures. Then, for any tradable asset $X$ with maturity $T$, \begin{align*} E^{Q_1}\left(\frac{X_T}{B_T}\right) &= E^{Q_2}\left(\frac{X_T}{B_T}\right)\\ &=\frac{X_0}{B_0}. ...


2

essentially it comes down the fact that the dyadic quadratic variation of $W_t$ is $t$ with probability 1 and any measure change has to preserve this fact. Changing volatility would violate this invariance.


5

You ask 2 questions and I try to answer: 1) Why do we use geometric Brownian motion ($\ln S_t-\ln S_0$ is normally distributed)? In this case you have $$ S_t = S_0 \exp( (\mu-\sigma^2/2) t + \sigma B_t), $$ which means that you model positive prices. Furthermore the log-return $$ \ln(S_t/S_0) = (\mu-\sigma^2/2) t + \sigma B_t, $$ is normally distributed. ...


1

Given its price today, the stock price at time T is lognormally distributed, whereas $lnS_T$ is normally distributed, that is $lnS_T$ ~ $N \Bigr(lnS_0 + (\mu- \frac{\sigma^2}{2}T),\sigma^2T \Bigl)$ see for example Hull - Options, Futures, and other Derivatives. Plugging in the numbers you get $lnS_T$ ~ $N(3.981291519,0.16875)$ Then the probability you ...


2

As the stock price process $S$ follows a geometric Brownian motion, we have that \begin{align*} S_T &= S_0 e^{(\mu-\frac{1}{2}\sigma^2)\, T + \sigma\, W_T}\\ &= S_0 e^{(\mu-\frac{1}{2}\sigma^2)\, T + \sigma\, \sqrt{T}\, \xi}, \end{align*} where $\xi$ is a standard normal random variable. Then, we have the probability \begin{align*} P(S_T > 95) ...


1

This is the Black Scholes Call Price: \begin{align} C(S, t) &= N(d_1)S - N(d_2) Ke^{-r(T - t)} \\ d_1 &= \frac{1}{\sigma\sqrt{T - t}}\left[\ln\left(\frac{S}{K}\right) + \left(r + \frac{\sigma^2}{2}\right)(T - t)\right] \\ d_2 &= \frac{1}{\sigma\sqrt{T - t}}\left[\ln\left(\frac{S}{K}\right) + \left(r - \frac{\sigma^2}{2}\right)(T - ...


0

Maybe this is rather a comment then an answer but three points: GBM is a stochastic model for stock prices. It is used to price derivatives in an arbitrage free setting. In this case you look at a process whose expected return is just the risk free rate (due to no-arbitrage). Forecast this price is trivial. It is debatable how forecastable stock prices ...


1

There is probably nothing wrong with your code although I did not check it in Mathematica. Normally, Geometric Brownian motion is just a model. Here, you simulate lots of paths and then average over it. The first plot gives something like $$ E(S_t) = S_0*\exp(\mu t) $$ with $S_0$ the initial stock price. However, because of the simulation, you do not get ...


1

In the Black-Scholes framework, we assume the log returns are normally distributed. This is equal to saying the underlying is log-normally distributed. If you look at Geometric Brownian Motion on wikipedia, you'll see this: The above solution S_t (for any value of t) is a **log-normally distributed** random variable The wikipedia is correct.



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