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I have to prove that the following vector has normal distribution:

$$ \left( W_t, \int_0^t W_s \, ds\right) $$

where $W_t$ is a Brownian motion. I tried just by a forward computation on the characteristic function using the Ito derivative. In order to make an easier computation let $$ \Gamma_t=exp\left( i \xi W_t + i \eta \int_0^t W_s \, ds \right) $$ and $$ Y_t = \int_0^t W_s \, ds $$ We have: $$ d \Gamma_t = i\xi \Gamma_t dW_t+i \eta\Gamma_t dY_t + \frac{1}{2}\left( -|\xi|^2\Gamma_t \, dt\, - i\eta \Gamma_t \, d\langle Y_t\rangle -2\xi \eta \Gamma_t d \langle Y_t, W_t\rangle\right) $$

Now, since we have that $$ dY_t = W_t dt $$ it follows that $$ d \langle Y_t \rangle = 0 $$

and

(this is a standard formal computation) $$ d \langle Y_t ,W_t\rangle= dY_t dW_t = W_t dt dW_t = 0 $$

Passing to the average: \begin{align} \mathbb{E} \left[ \Gamma_t \right] &=1 +i \eta \int_0^t \mathbb{E} \left[\Gamma_s dY_s \right]-\int_0^t\frac{|\xi|^2}{2} \mathbb{E} \left[ \Gamma_s \right] \, ds \\ &= 1 + i \eta \int_0^t \mathbb{E} \left[ \Gamma_s W_s \right] ds -\int_0^t\frac{|\xi|^2}{2} \mathbb{E} \left[ \Gamma_s \right] \, ds \end{align}

At this point I'm trying to get an ODE for $\mathbb{E}[\Gamma_t]$ but I'm a little bit stuck

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  • $\begingroup$ $d \langle Y_t W_t\rangle = d \langle Y_t ,W_t\rangle$ ? $\endgroup$ – Behrouz Maleki Dec 14 '16 at 18:11
  • $\begingroup$ $\int_{0}^{t}W_sds\sim\mathcal{N}\left(0,\frac 13 t^3\right)$ $\endgroup$ – Behrouz Maleki Dec 14 '16 at 18:13
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    $\begingroup$ $\mathbf E[\Gamma_t dY_t]= E[\Gamma_t W_t]dt\neq 0$. But this is a nice try. $\endgroup$ – zhoraster Dec 14 '16 at 18:24
  • $\begingroup$ @zhoraster You are right, that is a huge mistake... $\endgroup$ – clarkmaio Dec 14 '16 at 18:34
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At the end I think I've found the solution. Of course the problem is the non linearity of the term: $$ \mathbb{E}[\Gamma_t W_t] $$ In this case one can use this trick: $$ \partial_\xi \Gamma_t = i \Gamma_t W_t. $$ This comes in particoular from the exponential structure of the charateristic function so that it seems like to be a quite usefull trick for the future.

However let me continue the computation: \begin{align} \mathbb{E}[\Gamma_t] & = 1+i \eta\int_0^t \frac{1}{i}\partial_\xi\mathbb{E}[\Gamma_s] \, ds -\frac{|\xi|^2}{2}\int_0^t\mathbb{E}[\Gamma_s] \, ds \\ & = 1+\int^t_0 \eta \partial_\xi\mathbb{E}[\Gamma_t]- \frac{|\xi|^2}{2}\mathbb{E}[\Gamma_t] \, ds \end{align}

This is a good formulation for us since it means that the left side of the equation is a solution for the following PDE: $$ \partial_t V(t, \xi , \eta)=\eta \partial_\xi V(t,\xi,\eta)-\frac{|\xi|^2}{2}V(t,\xi,\eta) $$ with initial condition $V(0,\xi,\eta)=1$.

Now we are at "home" since we can use all our PDE-weapon to win. For example by choosing the solution of the form: $$ V=\exp(P(t,\xi,\eta)). $$

where $$ P(t,\xi,\eta)=a t^A \xi^2+b t^B \eta^2+c t^C \eta \xi. $$ Now the idea is just make a computation and choose suitable parameters. By appling the differentials operators we get the relation: $$ a A t^{A -1} \xi^2 + b B t^{B-1} \eta^2 + c C t^{C-1} \eta \xi = 2 a A t^A \xi \eta + c t^C \eta^2-\frac{\xi^2}{2} $$

This equation, as equivalence between two polynomes in the $\xi$ and $\eta$ variable, leads us to the following system for the coefficients:

$$ \left\{ \begin{array}{rcl} A a t^{A-1} & = & -\frac{1}{2} \\ Bbt^{B-1} & = & c t^C \\ 2 a A t^A & = & cCt^{C-1} \end{array} \right. $$

Ok...now this gives us (finally)

$$ \left\{ \begin{array}{rcl} A & = & 1 \\ a & = & -\frac{1}{2} \\ B & = & 3 \\ b & = & -\frac{1}{6} \\ C & = & 2\\ c & = & -\frac{1}{2}. \end{array} \right. $$

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  • $\begingroup$ I would like to stress an huge problem I found while doing these computations: the homogeneous solution is not unique since there is also the variable $\eta$ and since we haven't initial condition on $u$. For example we could choose as solution a function $\psi(\eta)$ or $\exp(\xi / t \eta)$ $\endgroup$ – clarkmaio Dec 20 '16 at 23:31
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Recall the following statement:

Let $X: \Omega \to \mathbb{R}^n$ be a Gaussian random vector and $A \in \mathbb{R}^{k \times n}$ a (deterministic) matrix. Then $Y := A \cdot X$ is Gaussian.

Since $(W_t)_{t \geq 0}$ is a Brownian motion, we know that $X:=(W_{t_0},W_{t_1},\ldots,W_{t_n})$ is Gaussian for any $t_0,\ldots,t_n$. If we choose $t_j := \frac{j}{n} t$, $n \in \mathbb{N}$, and define

$$A := \begin{pmatrix} n^{-1} & \ldots & n^{-1} & 0 \\ 0 & \ldots & 0 & 1 \end{pmatrix} \in \mathbb{R}^{2 \times (n+1)}$$

then, by the above statement, the vector

$$A \cdot X = \left( \frac{1}{n} \sum_{j=0}^{n-1} W_{t_j}, W_{t_n} \right) = \left( \sum_{j=0}^{n-1} W_{t_j} (t_{j+1}-t_j), W_t \right)$$

is Gaussian. Consequently, we find that

$$\left( \int_0^t W_s \, ds, W_t \right) = \lim_{n \to \infty} \left( \sum_{j=0}^{n-1} W_{t_j} (t_{j+1}-t_j), W_t \right)$$

is Gaussian as a pointwise limit of Gaussian random variables (see e.g. this question).

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