correlation between $\sum_{i=1}^{98}X_i$ and $\sum_{i=3}^{100}X_i$

Question

Let $X_1,...,X_{100}$ be iid $N(0,1)$ random variables. The correlation between $\sum\limits_{i=1}^{98}X_i$ and $\sum\limits_{i=3}^{100}X_i$ is equal to

(A) $0$ (B) $\dfrac{96}{98}$ (C) $\dfrac{98}{100}$ (D) 1

My Steps: $96$ of these $98$ variables of each series have the same value . So, B should be the correct option.

Did I solve this correctly ? Please help me confirm my solution.

Answer 1

Since the $X_i$ are i.i.d. and have unit variance, we have $$\DeclareMathOperator{Cor}{Cov} \Cor(X_i,X_j) = \begin{cases} 1 & \text{if } i = j, \\ 0 & \text{if } i \neq j. \end{cases}$$ Therefore $$ \begin{align*} \Cor\left(\sum_{i=1}^{98} X_i, \sum_{j=3}^{100} X_j\right) &= \sum_{i=1}^{98} \sum_{j=3}^{100} \Cor(X_i,X_j) \\ &= \sum_{i=1}^{98} \sum_{j=3}^{100} [i = j] \\ &= |\{1,\ldots,98\} \cap \{3,\ldots,100\}| \\ &= |\{3,\ldots,98\}| \\ &= 96. \end{align*} $$ In order to calculate the actual correlation coefficient, we need to normalize by the standard deviations: $$ \DeclareMathOperator{Var}{Var} \rho\left(\sum_{i=1}^{98} X_i, \sum_{j=3}^{100} X_j\right) = \frac{\Cor\left(\sum_{i=1}^{98} X_i, \sum_{j=3}^{100} X_j\right)}{\sqrt{\Var\left(\sum_{i=1}^{98} X_i\right)}\sqrt{\Var\left(\sum_{i=3}^{100} X_i\right)}} = \frac{96}{98}, $$ since the variance of both $\sum_{i=1}^{98} X_i$ and $\sum_{i=3}^{100} X_i$ is $98$ due to additivity of variance.

Answer 2

(B) is correct and here is one formal demonstration.

I'll write $C$ below for covariance and $V$ for variance. Also, let $S=\sum_{i=1}^{98}X_i$ and $T=\sum_{i=3}^{100} X_i$. Due to independence, we have $$ V(S)=\sum_{i=1}^{98}V(X_i)=98,\quad V(T)=\sum_{i=3}^{100}V(X_i)=98. $$ Moreover, using $C(X_i,X_j)=0$ for $i\neq j$, we have $$ C(S,T)=C\left(\sum_{i=1}^{98}X_i,\sum_{j=3}^{100} X_j\right)=\sum_{i=1}^{98}\sum_{j=3}^{100}C(X_i,X_j)=\sum_{i=3}^{98}C(X_i,X_i)=96. $$ It follows that $\frac{C(S,T)}{\sqrt{V(S)V(T)}}=\frac{96}{98}\cdot$