New answers tagged brownian-motion
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 ...
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