Suppose a_n∈ [0,1] and X_n is a sequence of i.i.d random variables with p.d.f : p(X_n=1)= p(X_n= -1)=0.5 . ∑_(n=1)^∞▒a_n X_n is convergent with probability 1, is ∑_(n=1)^∞▒a_n^2 convergent?
To determine whether the series ∑_(n=1)^∞▒a_n^2 is convergent, we can use the comparison test. In this case, we will compare the series to the series ∑_(n=1)^∞▒a_n.
The comparison test states that if 0 ≤ a_n ≤ b_n for all n and the series ∑_(n=1)^∞▒b_n converges, then the series ∑_(n=1)^∞▒a_n also converges. Conversely, if the series ∑_(n=1)^∞▒b_n diverges, then the series ∑_(n=1)^∞▒a_n also diverges.
In our case, since a_n ∈ [0, 1] for all n, we have 0 ≤ a_n^2 ≤ a_n for all n. Therefore, we can conclude that 0 ≤ ∑_(n=1)^∞▒a_n^2 ≤ ∑_(n=1)^∞▒a_n.
Now, let's analyze the series ∑_(n=1)^∞▒a_n. Given that the series ∑_(n=1)^∞▒a_nX_n is convergent with probability 1, it means that the series ∑_(n=1)^∞▒a_nX_n converges almost surely. That is, it converges with probability 1.
Since X_n is a sequence of i.i.d random variables with p.d.f p(X_n = 1) = p(X_n = -1) = 0.5, we can rewrite the series as ∑_(n=1)^∞▒a_n(1 or -1), where each term of the series is either a_n or -a_n with probability 0.5.
Given that a_n is bounded by [0,1] and each term can be either a_n or -a_n, we can say that the series ∑_(n=1)^∞▒a_n converges absolutely, and therefore it converges.
Using the comparison test, we have shown that 0 ≤ ∑_(n=1)^∞▒a_n^2 ≤ ∑_(n=1)^∞▒a_n. Since ∑_(n=1)^∞▒a_n converges, it implies that ∑_(n=1)^∞▒a_n^2 also converges.
Therefore, the series ∑_(n=1)^∞▒a_n^2 is convergent.