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The value in domestic currency of your (unhedged) investment is the product of the S&P500 and the USD exchange rate. But on any day the logarithmic return of your investment is the sum of the logarithmic return on the S&P500 index and the logarithmic return on the USD. If you want to use arithmetic (rather than logarithmic) returns then it is a ...


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Another take on the question which uses stochastic calculus [Digression] Assume deterministic and constant rates without loss of generality. Also assume the absence of arbitrage opportunities and market completeness Let $B_t$ denote the time-$t$ value of a risk-free money market account in which 1 unit of currency $C$ has been invested at $t=0$: ...


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Let $\tau = T-t$. Then \begin{align*} S_T = S_t e^{(\mu - \frac{1}{2}\sigma^2) \tau + \sigma \sqrt{\tau}\, Z}, \end{align*} where $Z$ is a standard normal random variable, independent of $\mathcal{F}_t$. Moreover, \begin{align*} E\left(S_T 1_{\{S_T >K\}}\mid \mathcal{F}_t \right) &= E\left(S_t e^{(\mu - \frac{1}{2}\sigma^2) \tau + \sigma \sqrt{\tau}\, ...


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This question is extremely interesting and not that straightforward. See answer here. From a financial perspective this is very much like pricing an American call (stopping rule = intrinsic value from exercice (i.e. current cash earned) > continuation value (i.e. what you can expect to gain). Note that you can never win more than 13 nor lose (at worst you ...


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Any position that is long the market. Eg long stocks, short puts on stocks etc, is being compensated for taking risk. Any position that is bearish eg short the market, or short calls on the market, is being penalized for taking the risk. There's no contradiction. Investors overall are long stocks, and they need to get paid to take the risk. That's what ...


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A Brownian bridge can be built simply from a forced process. For example, if we define the process $Z$ by $$ Z_{t}=\left(\dfrac{T-t}{T-t_0}\right)\left(a-W_{t_0}\right)+\left(\dfrac{t-t_0}{T-t_0}\right)\left(b-W_{T}\right)+W_{t}\;\;\;; t\in[t_0,T] $$ Where $W$ is a standard Brownian motion. The process $Z$ is a Brownian bridge satisfying the following ...


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I edit this answer to give more details. The process for $r$ above is the geometric Brownian motion (GBM) used to model stock prices in the Black-Scholes framework. Thus the question is about (the expectation of the) exponetial of the integral of GBM. The intergral of GBM is closely connected to Asian options. Thus one can study the literature about this ...


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I don't think you can have an explicit form. Let $Y_t= e^{at}X_t$ then : $$ Y_t -Y_0 =\sum_{i=1}^{N_t}e^{aT_i} $$ where $(T_i)_{i=1...N_t}$ are the jump times of your poisson process. then $$P(Y_t\leq x)=\sum_{n\geq 0}\frac{(mt)^n}{n!}e^{-mt}P(\sum_{i=1}^{N_t}e^{aT_i}\leq x|N_t=n)$$ $$P(\sum_{i=1}^{N_t}e^{aT_i}\leq x|N_t=n) ...



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