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)$. 1 It seems each$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)\$.