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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))$$a(T)(1-a(T))$ where $p(T)=E(X(T))$$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 $p^T(1-p^T)$.

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 $p^T(1-p^T)$.

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