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I'll outline how you can estimate the (implied) real-world density function from (observed) option prices. Having found this real-world density, you can then compute all sorts of probabilities and quantify the market's expectation of future prices. Recall firstly that (European-style) options are priced as risk-neutral expectation of the discounted payoff. ...


1

This is a case of bivariate normal as there are two normal variables (as opposed to two variables driven by the same common random factor). The answer to your question is the conditional distribution of one of the variables given the other variable: $Y|X \sim N\left(\mu_y+\rho \sigma_y \frac{x-\mu_x}{\sigma_x}, \sigma_y^2 \left(1-\rho^2\right) \right)$ For ...


1

I think in your book they prove that $\mathbb{P}(S_n \leq a)=\frac{a^n}{n!}$ with $0 \leq a \leq 1$, and $a=1$ is the particular case. $n=0$ is trivial. By induction, we assume that $\mathbb{P}(S_n \leq y)=\frac{y^n}{n!}$ $ \forall y \in [0,1]$ Let $a \in [0,1]$, we calculate $\mathbb{P}(S_{n+1} \leq a)$. We use the independence between $S_n$ and $X_{n+1}$...


3

On average half the time you will win 0.5 times your current bankroll and half the time you will lose 0.5 times your current bankroll. Over N plays your expected growth will be (0.5)^(N/2)(1.5)^(N/2) and you will tend to lose money over time and in the limit since 0.5*1.5 = 0.75 < 1. This happens because gaining and losing 50% are not equivalent. Think ...


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