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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): $$\left(\exp{(\sigma^2)} - 1\right) \exp{(2\mu + \sigma^2)}$$


4

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 ...


4

You're forgetting that -2.52 is still in natural logarithm terms. So the correct answer is 2.71828183 raised to the -2.52 power which equals 0.08. Your ending portfolio value is 8% of what it was a year ago.


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

There are many ways answering this, here is one: We assume the asset price at $t=T$, $S_T = S_{T-1} \times (S_T / S_{T-1})$. Assuming continuous compounding, we can write, $S_T = S_{T-1} \times \exp(R_{T-1})$. Working the same way for the previous period, we get $S_{T} = S_{T-2} \times \exp(R_{T-1}+R_T)$. Working all the way back to the initial value of ...


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 = ...


1

One way to start thinking about this is to work out a couple of Discrete versions of Ito's lemma ├śksendal (6th edition) Example 3.1.9: almost surely, $$ B_t^2 - t = \int_0^t 2B_s dB_s $$ This has a discrete version which holds everywhere: let $X_n=\pm 1$ and $S_n=\sum_{i=1}^n X_i$, then $$ S^2_n-n = 2\sum_{i=0}^{n-1} S_i X_{i+1} $$ To verify ...


1

I doubt you can do this. Correction term appears in Ito because Brownian motion has infinite variation (non zero quadratic variation). In discrete and therefore finite models you cannot observe this phenomenon.



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