Given a variable $x_t \in \{0,1\}$, then we sample $x_{t+1}$ in the following way $$ x_{t+1} = x_{tmp} x_t + (1-x_{tmp})(1-x_t ), \ x_{tmp} \sim Ber(x_{tmp};p). $$ Does $x_{t+1}$ follows the distribution $Ber(x_{t+1} ; p^{x_{t}}(1-p)^{1 - x_{t}})$? If so, how to prove it? Or not, what is the distribution of $p(x_{t+1}|x_{t})$?

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    $\begingroup$ What does "$x$" mean in "$Ber(x;p)$"?? Unless you get your notation absolutely right, we have to believe your question might be ambiguous. $\endgroup$
    – whuber
    Commented Jul 30, 2021 at 14:21
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    $\begingroup$ @whuber sorry to confuse you. I have edited the question. Actually it is $Ber(x_{tmp}; p)$ $\endgroup$
    – jzin
    Commented Jul 31, 2021 at 7:06

1 Answer 1


You have that the PMF of $x_{t+1}|x_{t}$. From what you've given $x_{t+1}|x_{t}$ can only take the values 0 and 1.

Let's find the probability of $x_{t+1}=1$ which can be realized in two possible ways,

$p(x_{t+1}=1|x_{t}=1)=p(x_{tmp})=p$ and $p(x_{t+1}=1|x_{t}=0)=p(1-x_{tmp})=1-p$.

Those two cases can be merged irrelevant of the specific value of $x_{t}$ as,


Hence, as you noted the distribution is

$$x_{t+1}|x_{t}\sim Bern(x_{t+1};p^{x_{t}}(1-p)^{1-x_{t}})$$


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