New answers tagged brownian-motion
1
The proofs seem good to me. An alternative answer to the first question is, with $s<t$:
$$\begin{align}
E\left(B_tW_t|\mathscr{F}_s\right)&= E\left((B_t-B_s+B_s)(W_t-W_s+W_s)|\mathscr{F}_s\right)
\\
&= E\left((B_t-B_s)(W_t-W_s)+(B_t-B_s)W_s+B_s(W_t-W_s)+B_sW_s|\mathscr{F}_s\right)
\\
&= E((B_t-B_s)(W_t-W_s))+W_sE(B_t-B_s)+B_sE(W_t-W_s)+B_sW_s
...
3
To prove that $Y(T)=0$, you integrate the SDE :
$$\int_{0}^{T}dY(t) = \int_{0}^{T}\nu dW(t) - \frac{\nu}{T} \int_{0}^{T}W(T) dt$$
$$Y(T)-Y(0)=\nu\left( W(T)-W(0)\right)- \frac{\nu}{T}W(T)\left(T-0\right)$$
Therefore $$Y(T)=Y(0)$$
You forgot to write the initial conditions but I guess $Y(0)=0.$
EDIT :
The dependency in $\nu$ is irrelevant in pricing ...
2
Best visualised in 3D. 2D works for Riemann (when the integrator is x, as in dx) but for Riemann-Stieltjes (when the integrator is a function of x, e.g., $\int{f(x)dg(x)}$), visualisation in 3D is more revealing. You can also then interpret the 3D chart in terms of its projection in 2D.
When the integrator is Brownian, as @ilovevolatility pointed out it, ...
4
Is $f$ a deterministic and differentiable function function of time? If so, write
$$
fdX = d(fX) - X (df/dt) dt
$$
The integral of the first term on the right is just the terminal value, the second term looks like your graph but it will be jagged because $X$ has jagged paths.
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