Let $(Y_n)$ and $(Z_n)$ be two independent sequences of iid Bernoulli-distributed random variables with parameter $p$ and $q$, respectively. For $n=1,2,3,...$, define $$ X_n = 3Y_{n+1} + 4Y_{n+1}Z_{n-3} + Y_{n+3}Z_{n-1}$$ Question: Show that the variance of $S_n = \sum_{i=1}^n X_i $ is converging for $n$ going to infinity, i.e. $ \lim_{n \rightarrow \infty} var(S_n) < \infty$.

My "Ansatz": First, I observed that $X_3$ is dependent $X_1$ and that $X_5$ is dependent on $X_3$, and so on. The same behaviour for $X_n$ for $n$ being even, i.e. $X_4$ being dependent on $X_2$. So I can decompose $S_n$ into $S_n^* = \sum_{i=2,4,6,...}X_i$ and $S_n^{**} = \sum_{i=1,3,5,...}X_i$. For simplicity assume that $n$ is some number number which can be divided by 4 (e.g. 48), s.t. $S_n^*$ and $S_n^{**}$ are of the same size. Then one can state that $$ var(S_n) = var(S_n^{*}+S_n^{**}) = 2*var(S_n^{*})$$

  • $\begingroup$ Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. $\endgroup$ Jun 24 '18 at 8:41
  • $\begingroup$ @TonyHellmuth I just stated my Ansatz in the question. Thanks $\endgroup$
    – Selos
    Jun 24 '18 at 8:59
  • $\begingroup$ Cool! This might be of some use :) $\endgroup$ Jun 24 '18 at 9:23

I'm not sure how $X_2$ is defined unless we index the $Z_i$ with negative indices, so I started the sum from $i=4$. Also, the variance is clearly unbounded for

$$ \lim_{n\to\infty}\sum_{i=4}^n X_i, $$

so I'm assuming instead you mean the standard normalization

$$ \lim_{n\to\infty}\frac{1}{\sqrt{n}}\sum_{i=4}^nX_i. $$

I'm going to compute what you called $\text{Var}(S^*_n)$, but with the $1/\sqrt{n}$ normalization above.

Writing out the first few terms of the sequence, it's easy to see that, for even $n\geqslant6$,

$$ \tag{1}\label{eqn:1}\sum_{i=4,6,\dots}^{n}X_i=Y_5(3+4Z_1)+Y_{n+3}Z_{n-1}+\sum_{i=6,8,\dots,}^{n}Y_{i+1}(3+5Z_{i-3}) $$

Since each summand in the sum above is independent, in light of (\ref{eqn:1}), we have that

\begin{align} \text{Var}\bigg(\sum_{i=4,6,\dots}^nX_i\bigg) &=\text{Var}\bigg(Y_5(3+4Z_1)+Y_{n+3}Z_{n-1}+\sum_{i=6,8,\dots,}^{n}Y_{i+1}(3+5Z_{i-3})\bigg)\\ &=\underbrace{\text{Var}(Y_5(3+4Z_1))}_{\displaystyle\equiv\alpha}+\underbrace{\text{Var}(Y_{n+3}Z_{n-1})}_{\displaystyle\equiv\beta}+\sum_{i=6,8,\dots,}^{n}\underbrace{\text{Var}(Y_{i+1}(3+5Z_{i-3}))}_{\displaystyle\equiv\gamma}\\ &=\alpha+\beta+\big(\frac{n-4}{2}\big)\gamma \end{align}

Note that the constants $\alpha,\beta$ and $\gamma$ are independent of $n$. Our result is then

\begin{align} \text{Var}\bigg(\frac{1}{\sqrt{n}}\sum_{i=4,6,\dots}^nX_i\bigg) &=\frac{1}{n}\bigg(\alpha+\beta+\big(\frac{n-4}{2}\big)\gamma\bigg)\\ &=\frac{\alpha+\beta}{n}+\bigg(\frac{1}{2}-\frac{2}{n}\bigg)\gamma \end{align}

which is clearly finite as $n\to\infty$. My calculations give that

$$ \gamma=25q(1-q)p+p(1-p)(3+5q)^2. $$

If I made any algebraic mistakes let me know!

  • $\begingroup$ Great, thanks for your help! One question remains: How did you derived the second-last variance of the sum term? I couldn't simplify so much.. $\endgroup$
    – Selos
    Jun 24 '18 at 17:36
  • $\begingroup$ @QuantFinance You mean the term with the $(n/2-1)$ in front? $\endgroup$
    – David M.
    Jun 24 '18 at 17:43
  • $\begingroup$ Yes,exactly this term. $\endgroup$
    – Selos
    Jun 24 '18 at 17:52
  • $\begingroup$ @QuantFinance Glad you asked--I made a mistake. See updated answer. $\endgroup$
    – David M.
    Jun 24 '18 at 18:10
  • $\begingroup$ Perfect, thanks a lot! One additional think I noticed: You say, that Var[ Y(3+4Z) ] = Var[ 3Y ] + Var[ 4YZ ]. This seems not correct to me, since this is only possible if 3Y is independent of 4YZ, correct? $\endgroup$
    – Selos
    Jun 24 '18 at 19:47

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