Number of variables, $n=10$
Variance of each variable, $\operatorname{Var}\left(X_i\right)=1$
Correlation between different variables, $\operatorname{Corr}\left(X_i, X_j\right)=1 / 4$ (for $i \neq j$ )
The variance of a sum of variables is the sum of all their variances plus the sum of all their covariances:
$$
\operatorname{Var}\left(\sum_{i=1}^n X_i\right)=\sum_{i=1}^n \operatorname{Var}\left(X_i\right)+\sum_{i \neq j} \operatorname{Cov}\left(X_i, X_j\right)
$$
There are $n=10$ variables, and each has a variance of 1 .
$$
\sum_{i=1}^{10} \operatorname{Var}\left(X_i\right)=\sum_{i=1}^{10} 1=10 \times 1=10
$$
We are given the correlation, so we first find the covariance for any pair $i \neq j$ :
$$
\begin{gathered}
\operatorname{Cov}\left(X_i, X_j\right)=\operatorname{Corr}\left(X_i, X_j\right) \times \sqrt{\operatorname{Var}\left(X_i\right) \times \operatorname{Var}\left(X_j\right)} \\
\operatorname{Cov}\left(X_i, X_j\right)=\frac{1}{4} \times \sqrt{1 \times 1}=\frac{1}{4}
\end{gathered}
$$
We need to sum the covariance $\frac{1}{4}$ for all pairs where $i \neq j$.
- There are $n$ choices for $i$.
- For each $i$, there are $n-1$ choices for $j$ (since $j$ cannot be $i$).
- This gives a total of $n(n-1)$ pairs.
Total number of pairs $=n(n-1)=10(10-1)=10(9)=90$ Now, multiply this by the covariance for each pair:
$$
\begin{gathered}
\sum_{i \neq j} \operatorname{Cov}\left(X_i, X_j\right)=90 \times \frac{1}{4}=\frac{90}{4}=22.5 \\
\operatorname{Var}\left(\sum_{i=1}^{10} X_i\right)=(\text { Sum of Variances })+(\text { Sum of Covariances }) \\
\operatorname{Var}\left(\sum_{i=1}^{10} X_i\right)=10+22.5=32.5
\end{gathered}
$$
