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Under GBM $$\frac {dS_t}{S_t} = \mu dt + \sigma dW_t$$ we get $$S_T = S_0 e^{(\mu - \frac{1}{2}\sigma^2)T + \sigma W_T}$$ suggesting that $$S_T \sim \text{ln}\mathcal {N} ( \tilde {\mu}, \tilde {\sigma})$$ where \begin{align} \tilde {\mu} &= \ln S_0 + (\mu - \frac{1}{2}\sigma^2)T \\ \tilde {\sigma} &= \sigma \sqrt {T} \end{align} Now if X \... 3 Note that \begin{align*} \frac{S_T-S_t}{S_t} &= \frac{S_T-K +K-S_t}{S_t}\\ &=\frac{(S_T-K)^+-(K-S_T)^+ +K-S_t}{S_t}. \end{align*} Then, \begin{align*} E\left(\frac{S_T-S_t}{S_t} \mid \mathcal{F}_t \right) &= \frac{e^{rT}}{S_t}(C_t-P_t)+ \frac{K-S_t}{S_t}. \end{align*} where \begin{align*} C_t &= e^{-rT} E\left((S_T-K)^+ \mid \mathcal{F}_t \... 2 Each of these can be used, but each has serious drawbacks. No. 1 is inaccurate unless you useN>>10$years of data. But decades of data may not be available or may no longer be relevant to today's economy. No. 2 is good except that the CAPM has been rejected by empirical tests. More advanced models from Asset Pricing Theory may be helpful (FF3, FF5, ... 2 Yes, for a short time horizon like 1 - 10 days, assuming$\mu = 0$is fine. As you'd correctly pointed out, for 1 - 10 days (and referring to the link you'd referenced to), it scales linearly by$T$(recall that$T$is an annual number, so convert to a % number in reference to days), but volatility scales by$\sqrt{T}$and so it is much larger than$T$for ... 1 Here's a try/start: Let$A,B$, and$C$be three possible events, and let$U(event)$be the utility derived from each event. For example, if event$A$corresponds to the event of winning the lottery, then$U(A)$will presumably be a very large value. By contrast, if event$C$corresponds to the event of falling off a ladder and breaking an arm,$U(C)$will ... 1 Yes leverage amplifies the exposure of equity to systematic risks. Just consider the standard textbook formula (Modigliani-Miller):$\beta_e = \beta_a \times (1+\frac{D(1-\tau)}{V})$where$\beta_e$is the sensitivity of the stock to systematic risk,$\tau$is the tax-rate and$D/V$is the leverage ratio. So beta (i.e. the exposure to systematic risk) ... 1 I can't comment yet on the topic due to my reputation level (so I will throw an answer up) but having just done my MFE capstone research on EVT implementation for VaR. According to my advisor who was a director of a quant research group at Citi before returning to academia, not many people are doing this. My research was to start collecting data comparing ... 1 Financial markets & Corporate Strategy - Grinblatt & Titman The book is very intuitive, but as a consequence less comprehensive than ex. Options, Futures, and other Derivatives by Hull (which is seen as the basic foundation of everything quant in some parts of the industry.) A great entry level book to finance, and is publically avaliable here: ... 1 If I had to give only one title this would be it: FT Guide to Understanding Finance by J. Estrada (Second Edition published 2011) It explains all of the above concepts (and more) in a very accessible, yet mathematically correct manner. A sample can be found: Here The only thing is that it is not really short (the first part, i.e. up to p. 150, is ... 1 In case of 2 securities, each and every combination of portfolio lies on efficient frontier. In your question, you have given to achieve expected return of exactly 10%. So, we have $$E(R_p)=w_1E(K_1) + w_2E(K_2)=0.10 \tag{1}$$ subject to: $$w_1 + w_2=1 \tag{2}$$ Solve your equation 1 and 2 to get$w_1$and$w_2$. Resulting weights would lead to minimum ... 1 It is really simple and probably not a question for this forum. You just need:$\alpha * .08 + (1-\alpha) * .02 = .12$. Solve for alpha and then check the standard deviation that should be .25. 1 This will depend on the definition of "return on the long run". If we define the annualized return on the long run by$\frac{1}{T}\ln \frac{S_T}{S_0}$for a certain time$Tin the future, then \begin{align*} E\left( \frac{1}{T}\ln \frac{S_T}{S_0} \right) = \mu-\frac{1}{2}\sigma^2, \end{align*} as claimed. Note that\mu$is the instant, or instantaneous, ... 1 Trying to shed some light here: What we also see using this here, is that if returns are log-normally distributed, ie. $$1 + r = \exp(\mu + \sigma Z),$$ with$Z$standard-normal, then $$E[1+r] = \exp(\mu + \frac 12 \sigma^2)$$ holds. But the geometric mean$GM$is given by$\exp(\mu)\$ and we have $$\log(GM) = \mu = \log(E[1+r]) - \sigma^2 /2$$ and ...