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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 ...


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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]...


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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 ...


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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 ...


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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+...


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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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(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} \...


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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)...


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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 = ...


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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 $$ X_t = F^{-1}_t X_0 + rF^{-1}_t \int_0^t F_u du $$


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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[ \...


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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


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$$ d Y \left(t\right) := d \left[\int_0^t{a \left(s\right)\mathrm{d}W_s}\right] = a \left(t\right) dW_t $$ 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 ...


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