1
$\begingroup$

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 $$ var(S_n) = var(S_n^{*}+S_n^{**}) = 2*var(S_n^{*})$$

$\endgroup$
  • $\begingroup$ Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. $\endgroup$ – Tony Hellmuth 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 :) $\endgroup$ – Tony Hellmuth Jun 24 '18 at 9:23
1
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.