Let's say the Bernoulli random variables $X_1,X_2,...,X_n$ follow a Markov process of order 1. Does this imply that for $t=1,...,n$ \begin{equation} \mathbb{E}(X_t)=P(X_t=1\mid X_{t-1}=x_{t-1}) \end{equation} ?


1 Answer 1


Almost, but not quite.

A Bernoulli random variable can only take values $0$ and $1$, leading to the straightforward calculation $$E(X) = 1\cdot P(X=1) + 0\cdot P(X=0) = P(X=1).$$

If we condition on an event $B$, almost nothing changes: $$E(X|B) = P(X=1|B).$$

So for a markov chain, it is true that $$E(X_t|X_1, X_2, \cdots X_{t-1}) = E(X_t|X_{t-1}) = P(X_t=1|X_{t-1} = x_{t-1})$$ However, this is not what you have written and generally speaking $E(X_t)$ is not the same as $E(X_t|X_{t-1})$.

Example, suppose $X_1 \sim \text{Bern}(1/2)$ and $X_t \sim \text{Bern}(p_{x_{t-1}})$ for $t \geq 2$, where $p_0 = 0$ and $p_1=1$. Then for any $t$ we have $$E(X_t) = \frac{1}{2}$$ but we also have $$E(X_t|X_{t-1}) = P(X_t=1|X_{t-1}=x_{t-1}) = \begin{cases} 0, & x_{t-1}=0 \\[1.2ex] 1, & x_{t-1}=1 \end{cases}$$

  • $\begingroup$ The sudden appearance of "$x_{t-1}$" in your assertion is difficult to interpret. What does it mean? It looks superfluous as well as being undefined. What would be the difference, if any, between $P(X_t=1\mid X_{t-1})$ and $P(X_t=1\mid X_{t-1}=x_{t-1})$? $\endgroup$
    – whuber
    Dec 17, 2019 at 18:00
  • 1
    $\begingroup$ @whuber, It may be a slight abuse of notation, but (as I'm sure you can intuit) $x_{t-1}$ refers to the realized value of the RV $X_{t-1}$. I am trying to show that $E(X_t|X_{t-1})$ depends explicitly on the realized value of $X_{t-1}$. Moreover, I don't see that I have written $P(X_t=1|X_{t-1})$ anywhere in this answer (although I do use this notation in the expected value). If you feel that this answer can be improved by changing the notation, constructive comments/suggestions (or edits) are welcome. $\endgroup$
    – knrumsey
    Dec 17, 2019 at 18:46

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