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4

I'm not sure I understand, but if you want to compute the variance of $exp(X)$, where $X$ is normally distributed with mean $\mu$ and variance $\sigma^2$, that variance is (from Wikipedia): $$\left(\exp{(\sigma^2)} - 1\right) \exp{(2\mu + \sigma^2)}$$


3

Here couple pointers to push you back on the right path (so I hope): Start with the payoff function and hence $S(T)$, which consists of $(W(T)-W(t))$ , $W$ being a Brownian Motion under the risk neutral measure) you can greatly simplify by working with a standard normal random variable: $$Y = \frac{-(W(T)-W(t))}{\sqrt{T-t}}$$, which helps to get rid of ...


3

The distribution of the log of a stock price in n days is a normal distribution with mean of $\log(current_price)$ and standard deviation of $volatility*\sqrt(n/365.2425)$ if you're using calendar days, and assuming no dividends and 0% risk-free interest rate. Note that the standard deviation is independent of the current_price: if ...


3

As @Rustam notes, "correlation" of deterministic functions in the sense you describe is a special case of allowing $\mu$ and $\sigma$ to have a term structure of arbitrary shape. Since the latter is easy to treat, no one bothers with restricted forms of it. Now, there quite a few people who deal with models that let $\sigma$ change with $S$. I am thinking ...


2

What you have to start with is: $$dS_t=\mu S_t dt + \sigma S_t dW_t$$ where $W_t$ is a standard brownian motion (SBM). You want to solve for $S_t$, so how would you proceed? If you integrate both sides of the equation between 0 and $T$, you get: $$S_T - S_0= \mu \int_0^T S_t dt + \sigma \int_0^T S_t dW_t$$ Okay and then what? The fact that you have ...


1

Your formula, as it stands, is incorrect, at least is if $E$ means the "expected value under real-world probabilities". I wrote a blog post explaining the basic rationale behind risk-neutral pricing where you will see that if the Fundamental Theorem of Asset Pricing theorem holds, you can write: Let $X_t=S_{1,t}-S_{2,t}$ $$e^{-rt} X_t = ...



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