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I'm trying to calculate the distribution of

$\int_{t_{n}}^{t_{n+1}}\sigma_{1}(\tau)\int_{t_{n}}^{\tau}\sigma_{2}(t)dW_{t}d\tau$

where $W_{t}$ is a brownian motion, that is

$W_{t}|W_{t_{n}}\sim\mathcal{N}(W_{t_{n}},t-t_{n})\quad\forall t \ge t_{n}$ and $W_{t}$ is continuous for each possible outcome.

Note that we know that $\int_{t_{n}}^{\tau}\sigma_{2}(t)dW_{t}\sim\mathcal{N}(0,\int_{t_{n}}^{\tau}\sigma_{t}(t)^{2}dt)$.

How can this be done? For example, we could say that we have $\sigma_{i}(t)=e^{\alpha_{i}(t-t_{n})}$ for $i=1,2$.

user87889
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1 Answers1

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Considering $$ \sigma_3(t)=\sigma_2(t)\int_t^{t_{n+1}}\sigma_1(\tau)\mathrm d\tau, $$ this is the stochastic integral $$\int_{t_n}^{t_{n+1}}\sigma_3(t)\mathrm dW_t, $$ which, thanks to Itô's isometry, is centered gaussian with variance $$ \int_{t_n}^{t_{n+1}}\sigma_3(t)^2\mathrm dt, $$ and the rest depends on the exact functions $\sigma_1$ and $\sigma_2$ but should not cause major conceptual headaches.

Did
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