2
$\begingroup$

In Example 7.1 of "Introducing Monte Carlo Methods with R", the authors write

$(X,Y)\sim N\Bigg((0,0),\begin{pmatrix}1 &\rho \\ \rho & 1\end{pmatrix}\Bigg)$

Then,

  1. Given $x_t$,
  2. $Y_{t+1}\mid x_t \sim N(\rho x_t, 1-\rho^2)$
  3. $X_{t+1}\mid y_{t+1} \sim N(\rho y_{t+1}, 1-\rho^2)$

Finally, they say: The subchain $(X_t)_t$ satisfies $$ X_{t+1} \mid X_t = x_t\sim N(\rho ^2 x_t,1-\rho^4) $$.

I'm at a loss for how to show this. I did

$$ \begin{align} f(x_{t+1}\mid x_{t}) &= \int_\mathbb{R} f(x_{t+1}, y_{t+1} \mid x_t)dy_{t+1}\\ &=\frac{1}{2\pi(1-\rho^2)}\int_{\mathbb{R}}exp\Big[ \frac{-1}{2(1-\rho^2)}\big((y_{t+1} - \rho x_t)^2+ (x_{t+1}-\rho y_{t+1})^2\big) \Big] dy_{t+1} \end{align} $$ where I have assumed $X_{t+1}$ is conditionally independent of $X_t$ given $Y_{t+1}$.

But I just don't know where to go from there. Any help would be appreciated.

$\endgroup$
3
$\begingroup$

Sorry if this is unclear in our book, but when \begin{align*} Y_{t+1}\mid X_t=x_t &\sim \mathrm{N}(\rho x_t, 1-\rho^2)\\ X_{t+1}\mid Y_{t-1}=y_{t+1},X_t=x_t &\sim \mathrm{N}(\rho y_{t+1}, 1-\rho^2) \end{align*} one gets $$X_{t+1}|X_t=x_t \sim \mathrm{N}(\rho \times \rho x_t,1-\rho^2+\rho^2(1-\rho^2))$$ as $$\mathbb{E}[X_{t+1}|X_t]=\underbrace{\mathbb{E}[\mathbb{E}[X_{t+1}|Y_{t+1}]|X_t]}_{\mathbb{E}[\rho Y_{t+1}|X_t]=\rho^2 X_t}$$ and $$\text{var}(X_{t+1}|X_t)=\underbrace{\mathbb{E}[\text{var}(X_{t+1}|Y_{t+1})|X_t)}_{\mathbb{E}[1-\rho^2|X_t]=1-\rho^2}+\underbrace{\text{var}(\mathbb{E}[X_{t+1}|Y_{t+1}]|X_t)}_{\rho^2\text{var}(Y_{t+1}|X_t)=\rho^2(1-\rho^2)}$$ If one wants to solve by the integral in the question, \begin{align*} &\exp\left\{\frac{-1}{2(1-\rho^2)}\big((y - \rho x_t)^2+ (x_{t+1}-\rho y)^2\big)\right\}\\ &\quad = \exp\left\{\frac{-1}{2(1-\rho^2)}\big(y^2(1+\rho^2)-2y(\rho x_t+\rho x_{t+1})+\rho^2 x_t^2+x_{t+1}^2\big)\right\}\\ &\quad = \exp\left\{\frac{-(1+\rho^2)}{2(1-\rho^2)}\big(y^2-2\rho (x_t+x_{t+1})y/(1+\rho^2)+\rho^2(x_t+x_{t+1})^2/(1+\rho^2)^2\big)\right\}\\ &\quad\times\exp\left\{\frac{-1}{2(1-\rho^2)}\big(\rho^2 x_t^2+x_{t+1}^2-\rho^2(x_t+x_{t+1})^2/(1+\rho^2)\big)\right\}\\ \end{align*} which makes the first exponential term a perfect normal density in $y$ and the second term can be simplified into \begin{align*} &\exp\left\{\frac{-1}{2(1-\rho^2)}\big(\rho^2 x_t^2+x_{t+1}^2-\rho^2(x_t+x_{t+1})^2/(1+\rho^2)\big)\right\}\\ &\quad=\exp\left\{\frac{-1}{2(1-\rho^2)(1+\rho^2)}\big( (1+\rho^2)[\rho^2 x_t^2+x_{t+1}^2]-\rho^2(x_t+x_{t+1})^2\big)\right\}\\ &\quad=\exp\left\{\frac{-1}{2(1-\rho^4}\big( x_{t+1}^2[1+\rho^2-\rho^2]-2x_tx_{t+1} \rho^2+x_t^2\rho^2[1+\rho^2-\rho^2]\big)\right\}\\&\quad=\exp\left\{\frac{-1}{2(1-\rho^4)}\big( x_{t+1}-\rho^2 x_t\big)^2\right\}\\\end{align*} which leads to the same result (!)

$\endgroup$
  • 1
    $\begingroup$ Great answer, thanks! I understand the integration which shows the result is normally distributed with given mean and variance. However, I have one follow-up question. In the first part, you used the laws of total expectation/variance to derive the mean and variance. But that part alone does not guarantee normality (right?). Is there a theorem we can appeal to which lets us conclude X_{t+1} | X_t is normal simply based on the setup? Such a theorem, paired with the laws of total exp/var, would constitute a slick and parsimonious derivation. Either way, thanks for completing the integration. $\endgroup$ – RMurphy Mar 17 '17 at 2:39
  • $\begingroup$ Indeed, I also used the fact that $(X_{t+1},Y_{t+1})$ given $X_t$ being normal implies that $X_{t+1}$ given $X_t$ is also normal. $\endgroup$ – Xi'an Mar 17 '17 at 6:58
-1
$\begingroup$

Perhaps this view helps: \begin{align} Y_{t + 1} &= \rho x_t + \sqrt{1 - \rho^2} N_t \\ X_{t + 1} &= \rho Y_{t + 1} + \sqrt{1 - \rho^2} N_{t + 1} \\ &= \rho^2 x_t + \rho \sqrt{1 - \rho^2} N_t + \sqrt{1 - \rho^2} N_{t + 1}, \end{align} where $N_t$ and $N_{t + 1}$ are standard normal distributed. It is then easy to see that $$\mathbb{E}[X_{t + 1} | x_t] = \rho^2 x_t + 0 + 0$$ and $$\mathbb{V}[X_{t + 1} \mid x_t] = \rho^2(1 - \rho^2) + (1 - \rho^2) = (1 + \rho^2)(1 - \rho^2) = 1 - \rho^4,$$ where we have used that the variance of the sum is the sum of the variances for uncorrelated variables and the third binomial formula.

$\endgroup$
  • $\begingroup$ I am curious why this answer got downvoted? $\endgroup$ – Lucas Mar 16 '17 at 10:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.