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7

It's a lemma! Ito's Lemma gives the change of coordinates rule for stochastic calculus. The multiplication rule is a shorthand way of expressing it.


7

Stochastics are usually applied in the field of derivatives pricing. In this setting the task is to price a derivative such that it fits into the landscape of tradable instruments (no-arbitrage). We work using the risk-neutral measure - usually denoted by $Q$. The measure is derived from other traded instruments. In risk analysis (e.g. calculate the VaR, ES ...


7

A stochastic differential equation is nothing more than a short-hand notation for a corresponding integral equation. So the initial SDE you provided actually means $$ \int_0^t d S_u = \int_0^t \mu(S_u, u) du + \int_0^t\sigma(S_u, u) dW_u$$ This is how the SDE is defined (see e.g. here). The reason is that you cannot differentiate a Brownian motion. It does ...


6

As @SRKX suggested, full answers are provided below. (a). Using Ito's lemma \begin{align*} d\left(W_t^3\right) &= 3W_t^2 dW_t +3W_t dt. \end{align*} Integrating on both sides, we obtain that \begin{align*} W_t^3 = 3\int_0^t W_s^2 dW_s +3\int_0^t W_s ds. \end{align*} Consequently, \begin{align*} \int_0^t W_s^2 dW_s = \frac{1}{3}W_t^3 -\int_0^t W_s ...


5

The problem is equivalent to given to 2 independent standard normals $W$ and $Z$ the probability of $$ W > 0, \text{ and } W+Z<0. $$ or $$ W > 0, \text{ and } Z<-W. $$ Plotting this set we see it is the bottom half of the lower right quadrant. The probability of being in the lower right quadrant is clearly $0.25$ by symmetry. The probability ...


5

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 $$ \left (\frac {X_t}{\sigma\sqrt {t}}\right)^2 \sim \chi^2\left(k=1,\lambda=\left (\frac {X_0+\mu t}{\sigma\sqrt {t}}\right)^2\right) $$ whence the law of $S_t := X_t^2$. As ...


5

What can be shown is that the above expressions are equal in probability. First check the distribution. As any linear combination of a Gaussian is Gaussian the right hand side is Gaussian - the left hand side too. Then we need the 2 moments: The expected values - it is zero ... easy to see. Next what you did not specify is that the correlation between ...


4

You know that $E\left[\int_{0}^{s}W_udu\right]=E\left[\int_{0}^{t}W_vdv\right]=0$. By definition \begin{align} & Cov\left(\int_{0}^{s}W_u\,du\,\,,\,\int_{0}^{t}W_v\,dv\right)=E\left[\int_{0}^{s}W_u\,du\int_{0}^{t}W_v\,dv\right]-0 \end{align} then \begin{align} & ...


4

In this case it is just the notion that your payoff function should not explode at some point - made mathematically rigorous. Have a look at the following picture from wikipedia: Intuitively the Lipschitz condition (or Lipschitz continuity) ensures that your payoff function always remains entirely outside the white cone, so it cannot e.g. become ...


4

Formally, this is a shorthand for the quadratic variation. For a more rudimentary definition, $\langle W, W\rangle$ is a process such that $W^2-\langle W, W\rangle$ is a martingale. Moreover, $\langle W, W\rangle_t$ is a limit, in probability, of the variation \begin{align*} \sum_{i=1}^n|W_{t_{i}}-W_{t_{i-1}}|^2, \end{align*} over the partition ...


4

Maybe I'm missing something? Given $f:\mathbb{R}^n \rightarrow \mathbb{R}^m$, you can write $f = (f_1,\ldots,f_m)$, where each $f_i:\mathbb{R}^n \rightarrow \mathbb{R}$. Apply Ito to each $f_i$ separately.


4

B1~N(0,1) and B2=B1+Z, for Z~N(0,1). From that E(B1*B1)=E(B1*B2)=1, E(B2*B2)=2. Therefore they are bivariate Gaussian with covariance matrix (1,1;1,2) therefore probability is around 12%, which is the volume over the bottom-right quadrant.


4

Let $\{P_t \mid t \geq 0\}$ be a compound Poisson process, where \begin{align*} P_t = \sum_{i=1}^{N_t} (V_i -1), \end{align*} and $N_t$ is a Poisson process with intensity $\lambda$ and jump times $\tau_i$, $i = 1, \ldots, \infty$. Let $Y_i=\ln V_i$ and $f(x)$ be the density function. Then \begin{align*} P_t - \lambda t E(V_1) &= P_t - \lambda t ...


4

Feynman–Kac Theorem: Assume that $F$ is a solution to the boundary value problem \begin{align} &F_t+\mu(t,x)F_x+\frac{1}{2}\sigma^2(t,x)F_{xx}-rF=0\\ &F(T,x)=\Phi(x), \end{align} Assume furthermore that the process $e^{-r_s}\sigma(s,X_s)F_s$ is in $\mathcal L^2$ where \begin{align} dX_s=\mu(s,x)ds+\sigma(s,x)dW_s, \end{align} then $F$ has the ...


4

I think all they are doing is integrating and estimating $$P(|W_t| \leq 2) = \int_{-2}^{2} \frac{d}{dr} P(W_t \leq r) dr $$ so $$ P(|W_t| \leq 2) \leq 4 \sup \limits_{r \in [-2,2]} \frac{d}{dr}P(W_t \leq r) $$ The normal density is maximal at zero and we are done.


4

We consider the case where the Novikov condition is satisfied, that is, \begin{align*} E\left[\exp\left(\frac{1}{2}\int_0^T \theta^2_s ds \right)\right] < \infty. \end{align*} Then $\{L_t \mid t \ge 0\}$ is a $(\mathscr{F}_t, \mathbb{P})$-martingale. On $\mathscr{F}_T$, we define the probability measure $Q$ by \begin{align*} ...


3

Note that, for $0 \leq s < t$, \begin{align*} W_t^3 &= (W_t-W_s+W_s)^3\\ &= (W_t-W_s)^3 + 3(W_t-W_s)^2 W_s + 3 (W_t-W_s) W_s^2 + W_s^3. \end{align*} Moreover, \begin{align*} E\big( (W_t-W_s)^3 \mid \mathcal{F}_s\big) &= E\big( (W_t-W_s)^3\big)\\ &= 0,\\ E\big((W_t-W_s)^2 W_s \mid \mathcal{F}_s\big) &= W_s E\big( (W_t-W_s)^2\big)\\ ...


3

It is nearly a Bronwian motion. Just the variance is not correct: The question is more tricky than it seems. A Brownian motion has the distribution properties stated below, so does a linear combination of BMs. But after all it is a martingale in a certain filtration (set of information) which has to be defined. $B_t$ is a BM in its own filtration, so is ...


3

Your notations are really hard to follow as you define $\mathbb{P}$ twice at the beginning. The notation $\mathbb{P} = \mathbb{\hat{P}}$ and $\mathbb{P} =\mathbb{\tilde{P}}$ is not meaningful as the probability measure $\mathbb{P}$ is already fixed and used for the real world probability measure. I think that this is the reason why you are getting confused. ...


3

There is somewhere summary of methods that can be used to estimate parameters of SDE? If you'd like a brief survey, consider the following packages as well as the accompanying papers (note: you may want to follow the citations listed therein): https://cran.r-project.org/web/packages/Sim.DiffProc/ "Estimation of Stochastic Differential Equations with ...


3

Given efficient markets, asset prices should be unpredictable in the sense that any upcoming returns are uncorrelated with current or past returns. Hence for traded assets the price should follow something more similar to a GBM than an O-U process. However, many financial metrics are not prices; for example interest rates or volatility. O-U processes may ...


3

Let \begin{align*} L(t; T, T + \Delta) = \frac{1}{\Delta} \left[ \frac{P(t,T)}{P(t, T+\Delta)} - 1 \right] \end{align*} be the forward Libor rate at time $t$ for the period $[T, T+\Delta]$. Consider a caplet with payoff at $T+\Delta$ of the form \begin{align} \Delta\max\big(L(T; T, T + \Delta) -K, \, 0 \big) &= \Delta\max\big((L(T; T, T + ...


3

Below I assume that you meant: $\psi (T) = \max (S_t - S_T, 0) $ which constitutes the payout of a forward start rather than a lookback option. If not please clarify your question... If you are looking for the option price $V_0$, assuming a Black-Scholes diffusion (GBM + constant interest rates), you have \begin{align*} V_0 &= P(0,T) E[ \psi (T) \vert ...


3

Apply Ito's lemma to $f(W_t) = W_t^2$ then $$ f(W_T) = f(W_0) + \int_0^T f'(W_t) dW_t + \frac{1}{2} \int_0^T f''(W_t) dt. $$ Thus $$ W_T^2 = 2 \int_0^T W_tdW_t + \frac12 2 T = 2 \int_0^T W_tdW_t + T. $$ If we rearrange terms then we get $$ \int_0^T W_tdW_t = (W_T^2-T)/2. $$


3

While Richard's answer is technically correct, just saying the result can be obtained using Ito's formula doesn't make the issue much clearer. So let me go into the microscopics of the issue. The Ito integral is defined in the following way. Suppose we divide the time interval $[0,t]$ into $n$ pieces with $t_i = i~dt$ where $dt=\frac{t}{n}$ then we define ...


3

I would not say there is no link to what you say but here would be my view. Intuitive explanation If you wait for a delay $h$ before exercising, you lose your exercise right between $t$ and $t+h$, this leads to a loss in value. Supermartingale property proof (to apply it in your case : $\phi_t=e^{-rt}(L-S_t)^+$) If we denote $\phi$ the obstacle, and ...


2

Note that $X$ is a continuous martingale. Moreover, the quadratic variation is given by \begin{align*} \langle X_t, \, X_t\rangle = \int_0^t |\sigma_u|^2 du = c^2 t. \end{align*} That is, \begin{align*} \langle X_t/c, \, X_t/c\rangle = t. \end{align*} From Levy's characterization, $X/c$ is by law a Brownian motion, which we denote by $\beta$. Then, by law, ...


2

It appears that we need only to observe the following: \begin{align*} \lim_{\lambda\rightarrow 0}\frac{1}{\lambda}\int_0^{\lambda t}\sigma^2_u du &= \lim_{\lambda\rightarrow 0}\int_0^{ t}\sigma^2_{\lambda u} du\\ &= \int_0^{ t}\sigma^2_{0} du \\ &=\sigma^2_{0} t. \end{align*}


2

This is a special case of the question of why $$ \int_0^T f(t) dW_t $$ is normally distributed for a continuous function $f(t).$ This Ito integral can be approximated by a sum $$ \sum_{i=0}^{N-1} f(i T/N) (W_{(i+1)T/N} - W_{i T/N}) .$$ The Brownian increments $(W_{(i+1)T/N} - W_{i T/N})$ are independent normally distributed random variables. The key point ...



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