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6

$$d Y \left(t\right) := d \left[\int_0^t{a \left(s\right)\mathrm{d}W_s}\right] = a \left(t\right) dW_t$$ Note that since $Y$ is a driftless process, it is a local martingale, and because $a$ is bounded, a true martingale. Its quadratic variation is given by $$\langle Y \left(\cdot\right)\rangle_t = \int_0^t{a^2 \left(s\right)\mathrm{d}s}$$ by definition ...

5

For part 1 of your question, the short answer is no, calculating conditional density is a looong way of doing it. Possible but not the easiest. Here is the sketch for a shorter version. We note that $(X_{T/2},X_{T})$ is a jointly Gaussian vector with mean $\mu = (X_0 + aT/2,X_0 + aT)$, and the variance-covariance matrix $$\begin{pmatrix} b^2 T/2 & b^2 T/... 5 An alternative way is using the Stratonovich integral. By definition, we have$$\int_0^t X_s \, \circ dW_s = \lim_{n\rightarrow \infty} \sum_{i=1}^n \frac{X_{t_i} +X_{t_{i-1}}}{2}\left( W_{t_i} -W_{t_{i-1}}\right) \; \; (1)$$One can then show that for a deterministic smooth functions f and g we have:$$ \int_0^t g'(W_s)\, \circ dW_s = g(W_t)- g(W_0)\; \;...

5

Integrating $W_t$ Consider the partition $t_i=it/n$ with $t_0=0$ and $t_n=t$. Then, by definition, \begin{align*} \int_0^t W_s\text{d}W_s &= \lim_{n\to\infty} \sum_{i=0}^{n-1} W_{t_i}\left(W_{t_{i+1}}-W_{t_i}\right). \end{align*} You can do the limit by using the identity $$W_{t_i}\left(W_{t_{i+1}}-W_{t_i}\right)=\frac{1}{2}\left(W_{t_{i+1}}^2-W_{t_i}^2-... 5 As shown in Credit Risk Modeling Notes (Bielecki, Jeanblanc, Rutkowski), Corollary 1.3.1, for t < s, we have:$$ P(\tau \leq s | {\cal F}_t) = N\left( -Y_t \sigma^{-1}(s-t)^{-1/2}- \nu(s-t)^{1/2}\right ) + {\rm e}^{-2\nu \sigma^{-2}Y_t} N\left( -Y_t \sigma^{-1}(s-t)^{-1/2}+ \nu(s-t)^{1/2}\right ),$$where$$ Y_t = y_0+ \nu t +\sigma W_t, \: \sigma >0,...

5

As stated here, for $f = f(t, x) ∈ C^{1,2}(\mathbb{R}^2)$ a deterministic function and Ito process $$X_t = W_t^2,$$ the stochastic process $$Y_t = f(t,X_t)$$ is an Ito process and we have $$df (t,X_t) = \partial_tf(t,X_t)\,dt + \partial_xf(t,X_t)\,dX_t + \frac{1}{2} \partial_{xx}^2f(t,X_t)(dX_t)^2.$$ Since $$dX_t = 2W_t dW_t + dt$$ and $$(dX_t)^2 = ... 5 By the stationarity of the increments of Brownian motion, we have the following equality in law and an almost sure equality coming from a simple change of variable$$\sigma \int_t^T e^{au} \left(W_u - W_t\right)du \overset{law}{=} \sigma \int_t^T e^{au}W_{u-t}dt = \sigma \int_0^{T-t} e^{a\left(u + t\right)}W_{u}du = \sigma e^{at}\int_0^{T-t}e^{au}W_udu$$To ... 5 Once we get to the point of$$\int_0^{T-t} e^{au}W_udu$$(thanks Shiva), we could conclude in the following way: it is Gaussian since limit of Gaussians (the step-wise constant approximating functions are clearly Gaussians); its mean is$$ \mathbb E \Big[\int_0^{T-t} e^{au}W_udu\Big] = \int_0^{T-t} e^{au}\mathbb E[W_u]du = 0;$$and its variance is$$\mathbb ...

4

With your SDE for $F$, I get: $$dXdF = -a^2XFdt$$ $$FdX = rFdt + aXFdW$$ $$XdF = a^2XF dt -aXF dW$$ So, adding up: $$d(XF) = rF dt,$$ giving $$X_t = F^{-1}_t X_0 + rF^{-1}_t \int_0^t F_u du$$

4

Alternatively, we can use Ito isometry ($X$'s integrability and adaptability are assured by $a$'s boundness and adaptability, respectively): $$E[X_t|{\cal F}_s] = E[X_s\big|{\cal F}_s] + E\left[\left(\int_s^t a_udW_u\right)^2 - \int_s^t a_u^2du \big|{\cal F}_s \right]$$ = X_s + E\left[\left(\int_s^t a_udW_u\right)^2\big|{\cal F}_s\right] - E \left[ \... 4 Since \text{Cov}(X, Y) = E(XY) - EX EY, we have \begin{align} \text{Cov}(tB_{3t} - B_{2t} + 5, B_s - 1) &= E[tB_{3t}B_s - tB_{3t} - B_{2t}B_s + B_{2t} + 5B_s - 5] - (5)(-1) \\ &= tE[B_{3t}B_s] - E[B_{2t}B_s] \\ &= ts - s \end{align} where the first equaltiy is just mutliplying out the product, the second equality comes from discarding zero ... 4 In all honesty, Quadratic Variation for Stochastic Processes is an advanced topic, and computing it rigorously from first principles is a graduate-level probability question. Part 1: Quadratic Variation: Informal "proof" First, how is Quadratic Variation Defined? For a stochastic process X_t, the quadratic variation, denoted <X_t>, is ... 4 Answering the title question: Let f(t,W_t)=W_t^2-t, then it is easier to derive the dynamics using the "general formula" for Itô's lemma (reference, see eq. 10):df(t,W_t)=\frac{\partial f}{\partial t} dt + \frac{\partial f}{\partial W_t} dW_t + \frac{1}{2}\frac{\partial^2f}{\partial W_t^2} dW_t^2$$where,$$\frac{\partial f}{\partial t} = -1, \...

4

I'd argue as follows. Let's simplify and assume $\mu_i=0,\sigma_i=1$ and let us set \begin{align} dW_t^{X}&=dW_t^{(1)}\\ dW_t^{Y}&=\rho dW_t^{(1)}+\sqrt{1-\rho^2}dW_t^{(2)} \end{align} Using Ito's lemma, $$dF(x,y)=F_xdx+F_ydy+\frac{1}{2}\left(F_{xx}dx^2+F_{yy}dy^2+2F_{xy}dxdy\right)$$ In our case: $$d\pi = \rho f_xg_y dt+\frac{1}{2}(gf_{xx}+... 3 Providing only a sketch here, using Ito integral definition (and commuting limit, summations and expectation), the result boils down to studying the expectation term:$$ E\left[ f_{t_{i-1}} (W_{t_i}-W_{t_{i-1}}) \cdot g_{t_{j-1}} (W_{t_j} - W_{t_{j-1}}) \right]. If the intervals don't overlap, i\not= j, and say t_i \leq t_{j-1}, then f_{t_{i-1}} g_{... 3 Since the process e^{-\mu t} is continuously differntiable, then it has finite variation. Thus, Ito's lemma essentially implies the 'normal' product rule: \begin{align} dU_t &= d(S_te^{-\mu t}) \\ &= e^{-\mu t}dS_t + S_td(e^{-\mu t}) \\ &= e^{-\mu t}\mu S_t dt + e^{-\mu t}\sigma S_t dw_t - \mu e^{-\mu t} S_t dt \\ &= e^{-\mu t}\sigma S_t ... 3 (As said in the comments, you need to put down some of your thoughts regarding the question too, like specifying the tools/theorems you would use or actual attempts to apply them, even if you can only cover early steps, not just the question itself.) Hints: We are givenX_t^\theta:=\exp \left(\theta W_t-\frac{1}{2} \theta^{2} t\right)=\sum_{n=0}^{\infty} \...

3

We need to show that $$\lim_{t\to1^-} (1-t)\int_0^t\frac1{1-s}dW_s \stackrel{\text{a.s.}}= 0.$$ $(M_t)_{t<1}=\Big(\int_0^t\frac1{1-s}dW_s\Big)_{t<1}$ is a martingale, so we can use Dambis-Dubins-Schwarz and say that $M_t = B_{\langle M\rangle_t}$ for a Brownian motion $B$ (with a different filtration of course). However, $\langle M\rangle_t = \int_0^t \... 3 Hints: Use Ito theorem for$\ln Z_t$to get: $$dZ_t = -\theta_t Z_t dW_t$$ Then use the product rule to compute$d(U_tZ_t)$. I introduced$U_t:=V_t B_t^{-1}$to keep things cleaner. $$dU_t = h'_t B_t^{-1}\sigma_t dW_t + h'_t B_t^{-1}\sigma_t \theta_t dt$$ $$dU_t\cdot dZ_t = -h'_t B_t^{-1}\sigma_t \theta_t Z_t dt$$ $$d(U_tZ_t) = U_tdZ_t + Z_tdU_t +dU_t\... 3$$ X_t = B_t 1_{t<0.5} + (x+ B_t) 1_{t\geq 0.5} = B_t + x1_{t\geq 0.5} [X, X]_t = [B, B]_t + x^2 1_{t\geq 0.5} = t+ x^2 1_{t\geq 0.5}$$(the author probably intended to use 0.5 as jump size too) 2 For a Normally distributed random variable, X, with mean \mu, and variance V, the following is true: $$\mathbb{E} \{\exp(\theta X)\} = \exp\left(\theta\mu+\frac{1}{2}\theta^2V\right)$$ In your example, conditional upon time 0, and assuming B(0)=0, B(t) is Normally distributed variable with zero mean and variance t. ... 2 To model long-term prices (20 years ahead), I would not utilize this approach. While it is resourceful to study the characteristics of the energy prices and see how well they can fit into a particular distribution/process, my experience is that it only takes you that far. I have come across plenty of energy price forecasting consultants (Baringa, Wattsight, ... 2 Hints: Show first that$$E[((W^1_t + W^2_t)-(W^1_s + W^2_s))^2] = (2+2\rho)(t-s) $$Then conclude that$$ [(2+2\rho)^{-1/2} (W^1 + W^2)]_t =t $$On the other hand, show (using bilinearity of quadratic covariation) that$$ [(2+2\rho)^{-1/2} (W^1 + W^2)]_t = [(2+2\rho)^{-1/2} (W^1 + W^2), (2+2\rho)^{-1/2} (W^1 + W^2) ]_t  = (1+\rho)^{-1} (t+ [W^1, W^2]_t)... 2 Note that SDE (4) does have a "closed-form" representation. Let$Xbe $$X := S^p,$$ so (4) is a geometric Brownian motion SDE $$dX = (p\alpha + 2^{-1}p(p-1) \sigma^2) X dt + p \sigma X dW,$$ which, again due to Ito Lemma, is equivalent to $$d \ln X = (p\alpha + 2^{-1}p(p-1) \sigma^2 - 2^{-1}p^2 \sigma^2) dt + p \sigma dW$$ or d \ln X = ... 2 Lets assume the price of the underlying equals the strike at some point prior to expiry. Then the probability of the price being still greater or equal the strike at expiry is 0.5. So the probability of the European option paying out is exactly half of the probability for the American option. 2 There are two questions that you are asking: How to prove \text{Var}[I] = \frac{1}{3}? and What is \text{Cov}[I, W_1]? For Question 1, you absolutely can use Ito isometry. First, note that we can use integration by parts to obtain the formula: \begin{align} \int_0^1 W_t dt &= W_1 - \int_0^1tdW_t \\ &= \int_0^1(1-t)dW_t \end{align} So we can ... 2 For the first question, equality\mathbb{E}\left[\int_{[0,1]\times[0,1]} W_sW_tdsdt\right] = \int_{[0,1]\times[0,1]}\mathbb{E}[W_sW_t]dsdt \left(= \int_{[0,1]\times[0,1]}\min(s,t)dsdt\right) $$is due to commuting expectation and integral (not to Ito isometry), which in turn is allowed by Fubini's theorem condition being met:$$ \left(\int_{[0,1]\times[0,1]... 2 I found this to be a very interesting question, and I took a different approach to your working. Here's my attempt: Instead of considering the integral\int_s^t W_u du \rvert W_s=x, W_t=y$, we can consider the integral$\int_s^tB_u du$where$B_u$is a Brownian bridge process with$B_s = x$,$B_t = y$. Furthermore, we can shift the limits of the integral ... 2 To answer your question, few steps are necessary: I assume your are working under the BS model (from your code). Therefore we have that $$dS_t = \mu S_t dt + \sigma S_t dW_t$$ where$W_t$is a SBM. Using Ito's lemma we can recover: $$\log(S_t) - \log(S_0) \sim N \left((\mu - \frac{1}{2}\sigma^2)t, \ \ \sigma^2t \right)$$ Now, observe how$\exp(\log(S_t) - \...

2

If $s>0$, and the integral runs from $u=s$, then the integral only makes sense if we condition on what we know as of time $s$: we can write $W(s)=k$, where $k$ is some constant known at time $s$, i.e. the value of the Brownian motion $W_u$ known at time $u=s$ (can be zero, but doesn't need to be). Then, we have: \mathbb{E}\left[\int_{u=s}^{u=t}W_udu|\...

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