Exponential Moments of Correlated Brownian Motions

# Expectation of exponential of 3 correlated Brownian Motion


# Expectation of exponential of 3 correlated Brownian Motion












Consider, are correlated Brownian motions with a given

I want to calculate the, ,

I can't think of a way to solve this although I have solved an expectation question with only a single exponential Brownian Motion

## Answer by Kermittfrog (score 12, accepted)

https://quant.stackexchange.com/a/59941

Besides @StackG's splendid answer, I would like to offer an answer that is based on the notion that the multivariate Brownian motion is of course multivariate normally distributed, and on its moment generating function.

We know that

$$ \mathbb{E}\left(W_{i,t}W_{j,t}\right)=\rho_{i,j}t $$

i.e. an $N$-dimensional vector $X$ of correlated Brownian motions has time $t$-distribution (assuming $t_0=0$:

$$ X_t\sim \mathbb{N}\left(\mathbf{\mu},\mathbf{\Sigma}\right)=\mathbb{N}\left( \begin{bmatrix}0\\ \ldots \\\ldots \\ 0\end{bmatrix}, t\times\begin{bmatrix}1 & \rho_{1,2} & \ldots & \rho_{1,N}\\ \rho_{1,2} & 1 & \ldots & \rho_{2,N}\\ \ldots & \ldots & \ldots & \ldots \\ \rho_{1,N}&\rho_{2,N}&\ldots & 1 \end{bmatrix}\right) $$

The MGF of the multivariate normal distribution is

$$ M_X(\mathbf{t})\equiv\mathbb{E}\left( e^{\mathbf{t}^T\mathbf{X}}\right)=e^{\mathbf{t}^T\mathbf{\mu}+\frac{1}{2}\mathbf{t}^T\mathbf{\Sigma}\mathbf{t}} $$

In your case, $\mathbf{\mu}=0$ and $\mathbf{t}^T=\begin{pmatrix}\sigma_1&\sigma_2&\sigma_3\end{pmatrix}$. Hence,

$$ \begin{align} M_X(\begin{pmatrix}\sigma_1&\sigma_2&\sigma_3\end{pmatrix})&=e^{\frac{1}{2}\begin{pmatrix}\sigma_1&\sigma_2&\sigma_3\end{pmatrix}\mathbf{\Sigma}\begin{pmatrix}\sigma_1 \\ \sigma_2 \\ \sigma_3\end{pmatrix}}\\ &=e^{\frac{1}{2}t\left(\sigma_1^2+\sigma_2^2+\sigma_3^2+2\sigma_1\sigma_2\rho_{1,2}+2\sigma_1\sigma_3\rho_{1,3}+2\sigma_2\sigma_3\rho_{2,3}\right)} \end{align} $$

## Answer by StackG (score 12)

https://quant.stackexchange.com/a/59936

You need to rotate them so we can find some orthogonal axes.

A simple way to think about this is by remembering that we can decompose the second of two brownian motions into a sum of the first brownian and an independent component, using the expression \begin{align} W_{t,2} = \rho_{12} W_{t,1} + \sqrt{1-\rho_{12}^2} \tilde{W}_{t,2} \end{align} where $\tilde{W}_{t,2}$ is now independent of $W_{t,1}$

If we apply this expression twice, we get \begin{align} W_{t,2} &= \rho_{12} W_{t,1} + \sqrt{1-\rho_{12}^2} \tilde{W}_{t,2} \\ W_{t,3} &= \rho_{13} W_{t,1} + \sqrt{1-\rho_{13}^2} \tilde{W}_{t,3} \end{align}

We still don't know the correlation of $\tilde{W}_{t,2}$ and $\tilde{W}_{t,3}$ but this is determined by the correlation $\rho_{23}$ by repeated application of the expression above, as follows \begin{align} \rho_{23} &= \rho_{12}\rho_{13} + \sqrt{(1-\rho_{12}^2)(1-\rho_{13}^2)} \rho(\tilde{W}_{t,2}, \tilde{W}_{t,3}) \\ \rho(\tilde{W}_{t,2}, \tilde{W}_{t,3}) &= {\frac {\rho_{23} - \rho_{12}\rho_{13}} {\sqrt{(1-\rho_{12}^2)(1-\rho_{13}^2)}}} = \tilde{\rho} \end{align} so we can re-express $\tilde{W}_{t,3}$ as \begin{align} \tilde{W}_{t,3} &= \tilde{\rho} \tilde{W}_{t,2} + \sqrt{1-\tilde{\rho}^2} \tilde{\tilde{W}}_{t,3} \end{align}

Now we can express your expectation as the sum of three independent terms, which you can calculate individually and take the product: \begin{align} & {\mathbb E}[e^{\sigma_1 W_{t,1} + \sigma_2 W_{t,2} + \sigma_3 W_{t,3}}] \\ &= {\mathbb E}[e^{(\sigma_1 + \sigma_2 \rho_{12} + \sigma_3 \rho_{13}) W_{t,1} + (\sqrt{1-\rho_{12}^2} + \tilde{\rho})\tilde{W}_{t,2} + \sqrt{1-\tilde{\rho}} \tilde{\tilde{W_{t,3}}}}] \\ &= {\mathbb E}[e^{(\sigma_1 + \sigma_2 \rho_{12} + \sigma_3 \rho_{13}) W_{t,1}}] {\mathbb E}[e^{(\sigma_2\sqrt{1-\rho_{12}^2} + \sigma_3\tilde{\rho})\tilde{W}_{t,2}}]{\mathbb E}[e^{\sigma_3\sqrt{1-\tilde{\rho}} \tilde{\tilde{W_{t,3}}}}] \end{align} So it's just the product of three of your single-Weiner process expectations with slightly funky multipliers

## Answer by p.vitzliputzli (score 2)

https://quant.stackexchange.com/a/59958

One can also apply Ito's lemma (for correlated Brownian motion) for the function $$ f(I_1, I_2, I_3) = e^{I_1+I_2+I_3}.$$ The resulting SDE for $f$ will be of the form (with explicit t as an argument now) $$f(t) = f(0) + \frac{1}{2}k\int_0^t f(s) ds + \int_0^t \ldots dW_1 + \ldots$$ in which $k = \sigma_1^2 + \sigma_2^2 +\sigma_3^2 + 2 \rho_{12}\sigma_1\sigma_2 + 2 \rho_{13}\sigma_1\sigma_3 + 2 \rho_{23}\sigma_2\sigma_3$ and the stochastic integrals haven't been explicitly stated, because their expectation will be zero.

By taking the expectation of $f$ and defining $m(t) := \mathrm{E}[f(t)]$, we will get (with Fubini's theorem) $$m(t) = m(0) + \frac{1}{2}k\int_0^t m(s) ds.$$ Differentiating with respect to t and solving the resulting ODE leads then to the result.