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2

Generally speaking, in the Black-Scholes model, the stock price process $\{S(t), t \geq 0\}$ is modelled as $$S(t) = S(0) e^{\displaystyle \left(r-\frac{1}{2}\sigma^2\right)t + \sigma W(t)}.$$ Now, comparing the above with $S(t) = 6e^{-2t+2W(t)}$ (note that I wrote $-2t$ instead of $2t$, otherwise it does not hold, so I assume it is a typo) and given that ...


2

Let $N$ take values in the natural numbers and $X_i$ be iid as the other variable; you want the variance of $Y=\sum_{i=1}^N X_i$. The conditional variance formula gives $$Var(Y)=Var(E(Y|N))+E(Var(Y|N))\\ =Var(NE(X_1))+E(NVar(X_1))\\ =(EX_1)^2Var(N)+Var(X_1)EN.$$ In case $N, X_i$ are taken from dice rolls, I get $EX_1=EN=\sum_{i=1}^6i/6=7/2,$ ...


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As commented on the question, the assumption has to be made that Kyle is the same number of standard deviation units from the mean. Therefore my calculation is correct. If that assumption cannot be made then the probably is not possible to solve with the given information.


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I think using the central limit theorem is a good approach, but you can also use Hoeffding's inequalities to get a bound that's easier to compute. As mentioned in the other answer, it's wise to approximate with a Binomial distribution since $2^{105}$ is much larger than $2^{95}$. I'm going to start out without this assumption though to illustrate how good ...


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Since $2^{95}$ is small compared to $2^{105}$, we can make the approximation that we are sampling with replacement: let $$X_1,...,X_n \overset{iid}{\sim}\text{Bernoulli}(p)$$ where $n=2^{95}$, $p=\frac{1}{20}$. Then their sample mean $\bar{X}:=\frac{1}{n} \sum_{i=1}^n X_i$ has mean and variance: $$E(\bar{X})=p, \sigma^2 := Var(\bar{X})=\frac{p(1-p)}{n}$$ ...


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My answer based on the second version of the original post: According to the central limit theorem, the standard deviation of the sample mean of $n$ data from a population is $\sigma_{\overline{X}}=\sigma_X/\sqrt{n}$, where $\sigma_X$ is the population standard deviation. In your case, $\sigma_{\overline{X}}=40/\sqrt{100}=4$. My answer based on the first ...



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