I've got the following random variable for which I must find the expected value and variance:
$X_h =\sum_{i=1}^{15} X_i$
Where $X_i$ is a random variable of the set $s = \{0, 1, 3\}$, corresponding to the probability density function $f(x) =
\left\{
\begin{array}{ll}
0.6 & \mbox{if } x = 3 \\
0.2 & \mbox{otherwise}
\end{array}
\right.$.
I got the expected value by simply stating that:
$E(X_h) = E(\sum_{i=1}^{15} X_i) = \sum_{i=1}^{15} E(X_i) = \sum_{i=1}^{15} (\sum_s x\cdot f_x(x)) =\sum_{i=1}^{15} (0\cdot 0.2 + 1 \cdot 0.2 + 3 \cdot 0.6) = 30$
I did the following for the variance (which appears to be incorrect):
$Var(X_h) = E(X_h^2)-E(X_h)^2 = E((\sum_{i=1}^{15} X_i)\cdot(\sum_{i=1}^{15} X_i)) - E(X_h)^2$
$=E(\sum_{i=1}^{15}\sum_{j=1}^{15} X_i\cdot X_j) - E(X_h)^2$
$=\sum_{i=1}^{15}\sum_{j=1}^{15} E(X_i\cdot X_j) - E(X_h)^2$
$= \sum_{i=1}^{15}\sum_{j=1}^{15} (\sum_s X_i\cdot X_j \cdot f_x(X_i)) - E(X_h)^2$
$= \sum_{i=1}^{15}\sum_{j=1}^{15} (\sum_s X_i^2 \cdot f_x(X_i)) - E(X_h)^2$
$= \sum_{i=1}^{15}\sum_{j=1}^{15} (3^2 \cdot 0.6+1^2\cdot 0.2 + 0^2\cdot 0.2) - E(X_h)^2$
$= 15^2(3^2 \cdot 0.6+1^2\cdot 0.2 + 0^2\cdot 0.2) - 30^2$
$= 360$
Now my solution paper claims the following:
$Var(X_h) = \sum_{i=1}^{15} Var(X_i) = 15 \cdot 1.6 = 24$
I've been using this answer to compute the variance here but I think there may be a flaw somewhere. Is correct that even if the variables are independent, that calculating variance formally should yield the same result? What did I do wrong?