$$\begin{align} p_{ij}(k, k+n) =& \Pr\left\{X_{k+n}=j\mid X_k=i\right\}\\ =& \sum_{r=1}^R \Pr\left\{X_{k+n}=j\mid X_u=r\color{red}, X_k=i\right\}\Pr\left\{X_u=r\mid X_k=i\right\} \end{align}$$

Markov Chains; Using the law of total probability, one can obtain the following equation. I am not entirely sure how it can be obtained. Does the comma between $X_u$ and $X_k$ means intersection between them? $X_u$ is supposed to be a partition of the sample space I think. Any help is appreciated.

  • $\begingroup$ $X_u$ is presumably supposed to be a random variable. $\endgroup$ – Nate Eldredge Sep 24 '13 at 23:06

The comma should mean intersection, as you have guessed.

Remember from the definition of conditional probability,

$$\begin{align} \Pr\left[X_{k+n}=j\mid X_k=i\right] =& \frac{\Pr[X_{k+n}=j, X_k=i]}{\Pr[X_k=i]}\\ =& \frac{\sum_r\Pr[X_{k+n}=j, X_k=i, X_u=r]}{\Pr[X_k=i]}\\ =& \sum_r\frac{\Pr[X_{k+n}=j\mid X_k=i, X_u=r]\Pr[X_k=i, X_u=r]}{\Pr[X_k=i]}\\ =& \sum_r\Pr[X_{k+n}=j\mid X_k=i, X_u=r]\Pr[X_u=r\mid X_k=i] \end{align}$$

where the summation is over all possible value of $X_u$.

  • $\begingroup$ Splendid job Peterwhy. The comma does mean intersection as the values of X(u) are summed. Thanks a bunch. $\endgroup$ – qwert87 Sep 24 '13 at 23:36

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