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3
votes
Let $$X_t = \int_0^t \sigma(s) dW_s$$ denote a stochastic integral in the Itô sense. In that case one can write $r_t = f(t,X_t)$ where $$f:(t,x) \to c + \int_0^t \sigma^2(s)(t-s) ds + x \tag{1}$$ and …
answered Oct 20 '16 by Quantuple
5
votes
Answer Assuming the Poisson process $N_t$ is independent from the Brownian motions $(W_{1,t},W_{2,t})$, you'll have \begin{align} df(X_{1,t},X_{2,t}) &= \frac{\partial f}{\partial X_{1,t}} dX_{1,t}^ …
answered Feb 15 '18 by Quantuple
4
votes
Your logic is fine $$ X_t \sim \mathcal {N}(X_0+\mu t, \sigma^2 t) $$ Thus, $\left (\frac {X_t}{\sigma\sqrt {t}}\right)^2 $ indeed exhibits a non central chi-squared distribution $$ \left (\frac {X_ …
answered Mar 29 '16 by Quantuple
4
votes
This is merely a mathematical trick. You cannot easily integrate $dS_t = S_t(\mu dt + \sigma dW_t)$ over time because the RHS depends on $S_t$. Using Ito's lemma on the log price gets you: $d\ln(S_ …
answered Aug 24 '17 by Quantuple
7
votes
Let $$Z_t = \exp(-X_t)$$ with $$X_t = \sigma(T-t)W_t+\sigma\int_0^tW_sds+\int_0^Tf(0,u)du+\int_0^t\int_s^T\alpha(s,u)du ds $$ and $W_t$ a standard Brownian motion, along with the usual assumptions. W …
answered Oct 19 '16 by Quantuple
3
votes
It is a complete solution. Bearing in mind the SDE verified by $(X_t)_{t \geq 0}$, applying Itô's lemma to compute the (stochastic) differential of $f(X_t)$ yields \begin{align} df(X_t) &= \underbrac …
answered Jan 5 '18 by Quantuple
5
votes
You seem to use the term "volatility" to describe two very different quantities: (1) the diffusion coefficient of your SDE and (2) the standard deviation of the log-returns under your modelling assump …
answered Jul 18 '16 by Quantuple
2
votes
Assume that under the real world measure $$ dS_t/S_t = (\alpha-\delta) dt + \sigma dZ_t^\Bbb{P} \tag{1} $$ Under the EMM $\Bbb{Q}$ one then needs to have (fundamental theorem of asset pricing: in the …
answered Mar 7 '17 by Quantuple
8
votes
$X_t$ being a stochastic process, one cannot use ordinary calculus to express the differential of a (sufficiently well-behaved) function $f$ of $t$ and $X_t$. Instead one should turn to Itô's lemma, …
answered Apr 26 '16 by Quantuple
12
votes
Let $$ dS_t = \mu S_t dt + \sigma S_t dW_t + S_{t^-} dJ_t $$ where $$ J_t = \sum_{j=1}^{N_t} (V_j - 1) $$ is a compound Poisson process, with $V_j$ i.i.d. jump sizes (positive random variables) whose …
answered Mar 24 '16 by Quantuple
1
vote
Note that because the integrand is deterministic this Itô integral is normally distributed with parameters (cf. Itô isometry) $$ I_t := \int_0^t \sqrt{s} dW_s \sim N(0, t^2/2) $$ Now you can just use …
answered Oct 9 '18 by Quantuple
6
votes
Quantiles are preserved under monotonic transformations, hence the quantile for $Y$ is simply the exponential of the quantile of $X$, no need for corrections whatsoever (see here for instance). Put …
answered Jul 27 '17 by Quantuple