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

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


  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.


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 (!)

  • 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

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.

  • 1
    $\begingroup$ I am curious why this answer got downvoted? $\endgroup$
    – Lucas
    Mar 16 '17 at 10:20

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