It seems each $X(T)$ is Bernoulli, that is $X(T)=0$ or $1$ almost surely. As such, the variance of $X(T)$ is $p(T)(1-p(T))$ a(T)(1-a(T))$ where $p(T)=E(X(T))$. a(T)=E(X(T))$. But $p(0)=1$ a(0)=1$ and $p(T+1)=p(T)p$ a(T+1)=a(T)p$ hence $p(T)=p^T$ a(T)=p^T$ and the variance of $X(T)$ is $p^T(1-p^T)$.
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It seems each $X(T)$ is Bernoulli, that is $X(T)=0$ or $1$ almost surely. As such, the variance of $X(T)$ is $p(T)(1-p(T))$ where $p(T)=E(X(T))$. But $p(0)=1$ and $p(T+1)=p(T)p$ hence $p(T)=p^T$ and the variance of $X(T)$ is $p^T(1-p^T)$. |
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