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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 ...


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) ...


In this case it is just the notion that your payoff function should not explode at some point - made mathematically rigorous. Have a look at the following picture from wikipedia: Intuitively the Lipschitz condition (or Lipschitz continuity) ensures that your payoff function always remains entirely outside the white cone, so it cannot e.g. become ...


I think that this code solves your problems. In your case h0 is zero while lag can be set equal to 6 (or 5) function y=NWtest(ret,lag,h0) T=size(ret,1); vv=var(ret); for l=1:1:lag cc=cov(ret(1:end-l),ret(l+1:end)); vv=vv+2*(1-l/lag)*cc(1,2); end y=(mean(ret)-h0)/sqrt(vv)*sqrt(T); end


Write out the simple equations $$\begin{align} Y_j &= a_0 Z_j + a_1 Z_{j-1} + a_2 Z_{j-2}\\ Y_{j-1} &= a_0 Z_{j-1} + a_1 Z_{j-2} + a_2 Z_{j-3} \end{align}$$ There are some very simple cases that make $Y_j \perp Y_{j-1}$ due to the independence assumption of the random variables $\{Z_i\}_{i\in\mathbb{Z}}$. An example is $a_0 \in \mathbb{R}\setminus ...


My first idea would be to try writing up the characteristic functions and use the theorem stated in the bottom answer about independence here: http://math.stackexchange.com/questions/376511/a-criterion-for-independence-based-on-characteristic-function

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