Suppose that $(X_n)_{n\ge1}$ is a sequence of independent random variables with $E[|X_n|] < \infty$ for all $n$ and $E(X_n) = \mu$. Prove that

$$\sum_{n=1}^{\infty}\frac{1}{2^n}X_n = \mu \; a.s$$

I am stuck with this question and not sure how to go about it. I have proven that the sum converges absolutely almost surely but am not sure if this is useful towards my goal.

  • $\begingroup$ Hint 1: $\sum_{n=0}^{\infty} \frac{1}{2^n} = 2$. Hint 2: Looks like there's a typo in your question. Probably the sum should be for $n>0$ $\endgroup$ – Callus Oct 8 '14 at 3:34
  • $\begingroup$ Whoops sorry - I changed it now $\endgroup$ – John Oct 8 '14 at 4:09

Your claim does not hold without further assumptions.

To see this, first note that we can replace $X_n$ by $X_n - \mu$ and thus assume $\mu = 0$ without loss of generality (this uses $\sum_{n=1}^\infty 2^{-n}= 1$).

Let us assume that your claim is true (for any such sequence).

Now, let $f := \sum_{n=1}^\infty 2^{-n} X_n$. By assumption that your theorem is true, we see $f \equiv 0$ almost surely.

Also, set $Y_1 := 2 X_1$ and $Y_n := X_n$ for $n \geq 2$. Note that the sequence $(Y_n)_n$ also satisfies the requirements of your supposed theorem. Hence, the random variable $g := \sum_{n=1}^\infty 2^{-n} Y_n$ is also $0$ almost surely.

But $g = f + X_1$, which implies $X_1 \equiv 0$ almost surely. After renormalization, this means $X_1 \equiv \mu$ a.s. But it is clear that one can find examples of sequences $(X_n)_n$ satisfying your assumptions such that $X_1 \equiv \mu$ does not hold almost surely.

  • 1
    $\begingroup$ +1. Your example is more effective than another natural one which is $X_n=\pm2^n$, since $|X_n|$ may be uniformly bounded. $\endgroup$ – Did Oct 8 '14 at 5:51

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