# Tag Info

8

From $(2)$, \begin{align*} \ln S_t &=\ln F_{t, t} \\ &= \ln F_{0, t}-\frac{1}{2}\int_0^t\sigma^2 e^{-2\lambda (t-s)}ds+\int_0^t \sigma e^{-\lambda(t-s)} dB_s\\ &=\ln F_{0, t}-\frac{\sigma^2}{4\lambda} \left(1-e^{-2\lambda t}\right)+e^{-\lambda t}\int_0^t \sigma e^{\lambda s} dB_s. \end{align*} Then, \begin{align*} \lambda e^{-\lambda t}\int_0^t \...

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 ds. \...

6

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} & Cov\left(\int_{0}^{s}W_u\,du\,\,,\,\int_{0}^{t}W_v\,dv\right)=\int_{0}^{s}\int_{0}^{t}E\,...

6

The dynamics \begin{align*} \frac{dS_t}{S_t} =\mu dt + \sigma dW_t. \end{align*} is under the real-world measure $\mathbb{P}$. Then, \begin{align*} d\ln S_t =\Big(\mu-\frac{1}{2}\sigma^2 \Big) dt + \sigma dW_t. \end{align*} Therefore, \begin{align*} \ln S_T = \ln S_t + \Big(\mu-\frac{1}{2}\sigma^2 \Big)(T-t) + \sigma \big(W_T-W_t\big).\tag{1} \end{align*} ...

5

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 ...

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 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 Let$\{P_t \mid t \geq 0\}be a compound Poisson process, where \begin{align*} P_t = \sum_{i=1}^{N_t} (V_i -1), \end{align*} andN_t$is a Poisson process with intensity$\lambda$and jump times$\tau_i$,$i = 1, \ldots, \infty$. Let$Y_i=\ln V_i$and$f(x)be the density function. Then \begin{align*} P_t - \lambda t E(V_1) &= P_t - \lambda t \int_{\... 4 We consider the case where the Novikov condition is satisfied, that is, \begin{align*} E\left[\exp\left(\frac{1}{2}\int_0^T \theta^2_s ds \right)\right] < \infty. \end{align*} Then\{L_t \mid t \ge 0\}$is a$(\mathscr{F}_t, \mathbb{P})$-martingale. On$\mathscr{F}_T$, we define the probability measure$Qby \begin{align*} \frac{dQ}{dP}\big|_{\mathscr{F}... 4 In generaldX^2$is an ad-hoc or heuristic form of$d\langle X, X\rangle_t$, where$\langle X, X\rangle_tis the quadratic variation, which is defined by \begin{align*} \langle X, X\rangle_t = \lim_{\pi\rightarrow 0} \sum_{i=1}^n (X_{t_i}-X_{t_{i-1}})^2. \end{align*} Here,0=t_0 < \cdots < t_n = t$, and$\pi = \max\{ t_i-t_{i-1}, i=1,\ldots, n\}$. ... 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 Sim....

3

For a time interval $[0,T]$, Girsanov theorem states that given a process $\lambda$ such that process $U$, defined by $$dU_t = -\lambda_tU_tdW_t, \; U_0=1,$$ is a $P$-martingale, then one can define a new measure $Q$ equivalent to $P$ by $$\frac{dQ}{dP} = U_T,$$ and a standard Brownian motion under $Q$, $W^\star$, by $$dW^\star_t = dW_t + \lambda_tdt.$$ In ...

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 + \Delta)-\alpha)...

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 $\... 3 Assume deterministic and constant interest rates. For an investor in the foreign economy i.e. a market participant that can only trade assets delivering a payout in the foreign currency, let us define $$\tilde{X}_t = \tilde{X}_0 \exp \left(\left(r_f-r_d-\frac{\sigma_\tilde{X}^2}{2}\right)+\sigma_\tilde{X} W_t^{\tilde{X},\mathbb{Q}^f} \right)$$ $$Y_t =... 3 Let$$ f_{\lambda}(t,r)=E^{(t,r)}\left[e^{-\lambda r_{T}}\right] $$where E^{(t,r)} denotes the expectation conditional on r_{t}=r. We assume f is smooth for the remainder. Let \theta=T\wedge\inf\left\{ s>t\colon\left|r_{s}-r\right|>1\right\} . By the Markov property of \{r_{t}\},$$ f_{\lambda}(t,r)=E^{(t,r)}\left[f_{\lambda}(\left(t+h\right)... 3 Here's my 2 cents: a) Conditional expectations can always be seen as martingales (this is a direct consequence of the tower property). Thus, we here have that $$M_t := E^*[e^{-\lambda {r_{T}}}|r_t]$$ is a martingale. Applying Itô's lemma to$M_t = f_{\lambda}(t,r_t)$as you did is a good starting point. But doing this, leaves you with an SDE, not a PDE.... 3 Your problem probably comes from the notations used. Let the Moment Generating Function (MGF) of a random variable$X$be defined as $$M_X(u) := E[e^{uX}]$$ From this definition, it entails that $$E(X^n) = M_X^{(n)}(u=0) = \frac{d^{n} M_X}{ d u^{n}}(u=0)$$ Knowing this, the function $$f_{\lambda}(t,r)=E[e^{-\lambda {r_{T}}}|r_t=r]$$ can be ... 3 We assume that$\gamma(s, t)$is differentiable with respect to$t\$. Then, \begin{align*} dx_t = \left(\int_0^t \frac{\partial\gamma(s, t)}{\partial t} dW_s \right)dt + \gamma(t, t) dW_t. \end{align*}

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