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Let $X_1,X_2....X_{2n}$ be random variables such that $V(X_i)=4;(i=1,2...2n)$ and $Cov(X_i,X_j) =3, (1 \leq i\ne j\leq2n)$ . Then

$V(X_1-X_2+X_3-X4......+X_{2n-1}-X_{2n})$

I am not able to figure out the covariance terms. According to formula sum of dependent positive random variables we have $\begin{align}\operatorname{Var}\left(\sum_{i=1}^nX_i\right) &= \sum_{i=1}^n\operatorname{Var}(X_i)+{2\sum_{i<j }} \operatorname{Cov(X_i,X_j)}\end{align} \\$

In my case if we take odd terms and even terms together they will give us positive co variance consecutive terms gives negative covariance. I am confused how to merge my result or is there standard result for alternating signs ?

$V(X_1-X_2+X_3-X4......+X_{2n-1}-X_{2n})=8n-()$?

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$\newcommand{\Var}{\operatorname{Var}}\newcommand{\Cov}{\operatorname{Cov}}$You want to find: \begin{align} \Var\left(\sum_{k=1}^{2n}(-1)^{k-1}X_k\right) \end{align} By setting $Y_k=(-1)^{k-1}X_k$ we get according to your formula: \begin{align} \Var\left(\sum_{k=1}^{2n}(-1)^{k-1}X_k\right)=\sum_{k=1}^{2n}\Var((-1)^{k-1}X_k)+2\sum_{i<j}\Cov((-1)^{i-1}X_i,(-1)^{j-1}X_j) \end{align} We have $\Var(aX)=a^2\Var(X)$ for constant $a$. Now figure out what is $\Cov(aX,bY)$ in terms of $\Cov(X,Y)$. Can you do that?


Edit. Avoiding a large amount of comments. One has $$\sum_{i<j}\Cov((-1)^{i-1}X_i,(-1)^{j-1}X_j)=3\sum_{j=1}^{2n}(-1)^j\sum_{i=1}^{j-1}(-1)^i$$ Moreover $\sum_{i=1}^N(-1)^i=\frac{1}{2}((-1)^N-1)$ so: $$\sum_{j=1}^{2n}(-1)^j\sum_{i=1}^{j-1}(-1)^i=\frac{1}{2}\sum_{j=1}^{2n}((-1)^{2j-1}-(-1)^j)=-n-\frac{1}{4}((-1)^{2n}-1)=-n$$ putting everything together: \begin{align} \Var\left(\sum_{k=1}^{2n}(-1)^{k-1}X_k\right)=8n+2\cdot 3\cdot (-n)=2n \end{align}

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  • $\begingroup$ $2\sum_{i<j}(-1)^{i+j} $Cov$ (X_i,X_j)$? $(i+j=2n)$ it wll be overall positive? $\endgroup$ – Daman deep Jan 30 '18 at 8:41
  • $\begingroup$ @Damn1o1 Yep! That is it! $\endgroup$ – Shashi Jan 30 '18 at 8:42
  • $\begingroup$ So that will give me $8n-6n=2n$. Thanks you ve been very helpful. I am learning these things. $\endgroup$ – Daman deep Jan 30 '18 at 8:44
  • $\begingroup$ @Damn1o1 I'm not sure about the positivity. Be careful with $\sum_{i<j}$ it means: $$\sum_{i<j}=\sum_{i<j\leq 2n} = \sum_{j=1}^{2n}\sum_{i=1}^{j-1}$$ hence $$\sum_{j=1}^{2n} (-1)^{j}\sum_{i=1}^{j-1}(-1)^i\Cov(X_i,X_j)=3\sum_{j=1}^{2n}(-1)^j\sum_{i=1}^{j-1}(-1)^i$$ now figure out what $\sum_{i=1}^{j-1}(-1)^i$ is. $\endgroup$ – Shashi Jan 30 '18 at 8:58
  • $\begingroup$ Moreover let $b_j:=\sum_{i=1}^{j-1}(-1)^i$. Then we have $b_1=0$ , $b_2=-1$ , $b_3=0$, $b_4=-1$ etc etc. So you can easily find a closed form for $b_j$. $\endgroup$ – Shashi Jan 30 '18 at 9:00

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