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I have the following model for my model

$\Delta X_{t} = \mu \Delta t + \rho \Delta X_{t-1} + \sigma \sqrt{\Delta t} Z_t$

with the following initial conditions -

$\Delta X_{1} = \mu \Delta t + \sigma \sqrt{\Delta t} Z_1$

and

$X_{0} = Z_0 $

all the $Z$ variables are i.i.d Standard normal variables $\sim \mathcal{N}(0,1) $

My goal is to find the distribution of $X_T$

This is my progress so far -

$\begin{bmatrix} X_{T} \\ X_{T-1} \\ X_{T-2} \\ \vdots \\ X_2 \\ X_1 \\ X_0 \end{bmatrix} = \mu \Delta t \begin{bmatrix} 1 \\ 1 \\ 1 \\ \vdots \\ 1 \\ 1 \\ 0 \end{bmatrix} + \begin{bmatrix} 0 & 1+\rho & -\rho & \cdots & 0 & 0 & 0 \\ 0 & 0 & 1+\rho & \cdots & 0 & 0 & 0 \\ 0 & 0 & 0 & \cdots & 0 & 0 & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & 0 & 1+\rho & -\rho \\ 0 & 0 & 0 & \cdots & 0 & 0 & 1 \\ 0 & 0 & 0 & \cdots & 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} X_{T} \\ X_{T-1} \\ X_{T-2} \\ \vdots \\ X_2 \\ X_1 \\ X_0 \end{bmatrix} + \begin{bmatrix} \sigma \sqrt{\Delta t} & 0 & 0 & \cdots & 0 & 0 & 0 \\ 0 & \sigma \sqrt{\Delta t} & 0 & \cdots & 0 & 0 & 0 \\ 0 & 0 & \sigma \sqrt{\Delta t} & \cdots & 0 & 0 & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & \sigma \sqrt{\Delta t} & 0 & 0 \\ 0 & 0 & 0 & \cdots & 0 & \sigma \sqrt{\Delta t} & 0 \\ 0 & 0 & 0 & \cdots & 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} Z_{T} \\ Z_{T-1} \\ Z_{T-2} \\ \vdots \\ Z_2 \\ Z_1 \\ Z_0 \end{bmatrix} $

Condensing it I get

$\mathbf{x} = (\mu \Delta t) \mathbf{k} + \mathbf{R} \mathbf{x} + \mathbf{S} \mathbf{z}$

Simplifying the expression, I get -

$\mathbf{x} = \left(\mathbf{I} - \mathbf{R} \right)^{-1}\left( (\mu \Delta t) \mathbf{k} + \mathbf{S} \mathbf{z}\right)$

This works but is very inefficient to code up with the matrix inversion. Any way I can get the closed form solution for $ X_T$? Or at least a more efficient way to compute it?

EDIT 1: I managed to get the expected value using the following steps.

$X_T = X_0 + \Delta X_1 + \Delta X_2 + \cdots + \Delta X_T $

If we look at the expected value of any of the $\Delta X_t$ variables we see

$\mathbb{E}\left[\Delta X_t \right] = \mu \Delta t + \rho (\mathbb{E}\left[\Delta X_{t-1} \right] )$

This, in combination with the initial condition, gives me -

$\mathbb{E}\left[\Delta X_1 \right] = \mu \Delta t $

$\mathbb{E}\left[\Delta X_2 \right] = \mu \Delta t + \rho (\mu \Delta t)$

$\mathbb{E}\left[\Delta X_T \right] = \mu \Delta t + \rho (\mu \Delta t) + \cdots + \rho^{T-1}(\mu \Delta t)$

Adding them all up, I get -

$ \sum_{i=1}^{T}\mathbb{E}\left[\Delta X_1 \right] = \sum_{i=1}^{T} \mu \rho^{i-1} (T-i+1) \Delta t $

After multiplying this equation with $\rho$, subtracting from itself, and simplifying the equation, I get

$ \sum_{i=1}^{T}\mathbb{E}\left[\Delta X_1 \right] = \left(\frac{\mu \Delta t}{1- \rho} \right)\left[T - \frac{1-\rho^{T+1}}{1- \rho} \right] $

Finally, I have the expected value of $X_T$ as

$\mathbb{E}\left[\Delta X_T \right] = \mathbb{E}\left[\Delta X_0 \right] + \sum_{i=1}^{T}\mathbb{E}\left[\Delta X_1 \right] = \sum_{i=1}^{T}\mathbb{E}\left[\Delta X_1 \right]$

Giving me

$\mathbb{E}\left[\Delta X_T \right] \sim \mathcal{N}\left( \left(\frac{\mu \Delta t}{1- \rho} \right)\left[T - \frac{1-\rho^{T+1}}{1- \rho} \right] , ?? \right)$

Still trying to figure out the variance.

EDIT 2: I followed the same process for the variance.

$ Var\left( \Delta X_t \right) = \rho^{2} Var( \Delta X_{t-1} ) + \sigma^2 \Delta t$

Like with the expected values, this gives me.

$ Var\left( \Delta X_1 \right) = \sigma^2 \Delta t$

$ Var\left( \Delta X_2 \right) = \sigma^2 \Delta t + \rho^{2} (\sigma^2 \Delta t)$

$ Var\left( \Delta X_T \right) = \sigma^2 \Delta t + \rho^{2} (\sigma^2 \Delta t) + \cdots \rho^{2T-2} (\sigma^2 \Delta t)$

We also have to take into account the covariances that exist

$Cov(\Delta X_t , X_{t-1}) = \rho Var(\Delta X_{t-1}) $

This gives us

$Cov(\Delta X_2 , X_1) = \rho (\sigma^2 \Delta t) $

$Cov(\Delta X_3 , X_2) = \rho (\sigma^2 \Delta t) + \rho^3 (\sigma^2 \Delta t)$

$Cov(\Delta X_T , X_{T-1}) = \rho (\sigma^2 \Delta t) + \rho^3 (\sigma^2 \Delta t) + \cdots + \rho^{2T-3} (\sigma^2 \Delta t)$

Putting it all together we get

$Var\left( \sum_{i=1}^{T} \Delta X_1 \right) = \sigma^2 \Delta t \left( T + \sum_{i=1}^{T-1} (T-i)( 2 \rho^{2i-1} + \rho^{2i} ) \right)$

Solving for $\sum_{i=1}^{T-1} (T-i)( 2 \rho^{2i-1} + \rho^{2i} )$ we get

$Var\left( \sum_{i=1}^{T} \Delta X_1 \right) = \sigma^2 \Delta t \left( T + \frac{1}{1-\rho^2}\left( (T-1)(2\rho^{1} + \rho^{2}) - \frac{1-\rho^{2T-2}}{1-\rho^2}(2\rho^3 + \rho^4) \right) \right)$

Since $X_0$ is independent we get the variance of $X_T$ as

$Var(X_T) = 1 + \sigma^2 \Delta t \left( T + \frac{1}{1-\rho^2}\left( (T-1)(2\rho^{1} + \rho^{2}) - \frac{1-\rho^{2T-2}}{1-\rho^2}(2\rho^3 + \rho^4) \right) \right)$

Final distribution of $X_T$ is

$\mathcal{N}\left( \left(\frac{\mu \Delta t}{1- \rho} \right)\left[T - \frac{1-\rho^{T+1}}{1- \rho} \right] , 1 + \sigma^2 \Delta t \left( T + \frac{1}{1-\rho^2}\left( (T-1)(2\rho^{1} + \rho^{2}) - \frac{1-\rho^{2T-2}}{1-\rho^2}(2\rho^3 + \rho^4) \right) \right)\right)$

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  • $\begingroup$ Because $R$ is nilpotent ($R^{T+1}=0$), inverting it is easy to do via $$(I-R)^{-1}=I+R+R^2+\cdots+R^T.$$ However, it's usually inefficient to invert matrices directly: simply solve the equation $$(I-R)\mathbf{x} = (\mu\Delta t)\mathbf{k} + \mathbf{Sz}.$$ You don't need to go this far, though, because all you really need are the mean and covariance of $x_T$--you know the distribution is Normal. $\endgroup$ – whuber May 24 at 20:42

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