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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. ...

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

if they are stocks, this problem is called pricing a Margrabe option and it is generally solved by change of numeraire. Take $S_2$ to be the numeraire. Then the value of the option is $$S_2(0) \mathbb{E}_{S_2}( (S_1(T)/S_2(T)-1)_+)$$ where the expectation is taken in the measure that has $S_1/S_2$ as a martingale. Since it's a martingale and log-normal at ...

1

Well, log-normality does not allow you to have negative earnings and companies do have negative earnings. I suggest you to download the earnings data and perform a Jarque-Bera test for normality.

1

One way to start thinking about this is to work out a couple of Discrete versions of Ito's lemma Øksendal (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 ...

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