# Tag Info

0

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 ...

0

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]... 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 ... 1 So the first thing is to note that using Fubini (see here) \int_0^T r(t) dt = \int_0^T \int_0^t dr(u) dt = \int_0^T \int_u^T dt dr(u) = 0.2 \int_0^T (T-u) dW_1(u) $$such that$$ \int_0^T r(t) dt \sim \mathcal{N}\left( 0, 0.2^2 \, \int_0^T (T-u)^2 du = 0.2^2 \frac{T^3}{3} \right) $$From that observation, in the expression$$ S_T = S_0\exp\left(- (0.05^2+...

1

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: $$\mathbb{E}\left[\int_{u=s}^{u=t}W_udu|\... 3 (As said in the comments, you need to put down some of your thoughts regarding the question too, like specifying the tools/theorems you would use or actual attempts to apply them, even if you can only cover early steps, not just the question itself.) Hints: We are given$$X_t^\theta:=\exp \left(\theta W_t-\frac{1}{2} \theta^{2} t\right)=\sum_{n=0}^{\infty} \...

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 $$[(2+2\rho)^{-1/2} (W^1 + W^2)]_t = [(2+2\rho)^{-1/2} (W^1 + W^2), (2+2\rho)^{-1/2} (W^1 + W^2) ]_t$$ $$= (1+\rho)^{-1} (t+ [W^1, W^2]_t)... 2 Note that SDE (4) does have a "closed-form" representation. Let X be$$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 = ...

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): $$E[X_t|{\cal F}_s] = E[X_s\big|{\cal F}_s] + E\left[\left(\int_s^t a_udW_u\right)^2 - \int_s^t a_u^2du \big|{\cal F}_s \right]$$ $$= X_s + E\left[\left(\int_s^t a_udW_u\right)^2\big|{\cal F}_s\right] - E \left[ \... 0 Are W1 and W2 independent? I would assume there is some correlation structure? Cholesky decomposition would help in generating the path. It's very similar to heston model where Vt is volatility. https://en.wikipedia.org/wiki/Heston_model 6$$ 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 ...

Top 50 recent answers are included