I am trying to calculate mean $E[Y]$ and variance $Var[Y]$ of sums, when rolling a dice of 6 outcomes $(1,2,3,4,5,6)$, 10 times. I get the mean as below.
Mean of sums for 10 tosses
$ Y = 10X, \\ \\ E[Y] = E[10X] = 10E[X] \\ \\ E[X] = 1 \frac{1}{6} + 2 \frac{1}{6} + 3 \frac{1}{6} + 4 \frac{1}{6} + 5 \frac{1}{6} + 6 \frac{1}{6} = \frac{21}{6} = 3.5 \\ \\ E[Y] = 10E[X] = 10(3.5) = 35 \\ \\ $
Variance for 10 tosses
Here I thought as follows.
$
Var[Y] = Var[10X] = 10^2Var[X] = 100Var[X] \\
Var[X] = \frac{(1-3.5)^2}{6} + \frac{(2-3.5)^2}{6} + \frac{(3-3.5)^2}{6} + \frac{(4-3.5)^2}{6} + \frac{(5-3.5)^2}{6} + \frac{(6-3.5)^2}{6} \\
= 2.91 \\
Var[Y] = 100(2.91) = 291
$
However as per here, Variance is
$ Var[Y] = 10Var[X] = 10(2.91) = 29.1 $
because each throw is an independent event. I also verified statistically this being correct.
Questions:
1. If each throw is an independent event, then mean would have been $E[X_1 + X_2 + \cdot + X_10]$, Isn't it? I could not imagine how $E[X_1] = \frac {10}{6}$ and so on?
2. I am confused when to apply scaling $E[aX]$ or $E[X_1 + X_2 + \cdot + X_10]$. Because, this affects variance as above, while both assumption for mean might result in same value. Kindly clarify.
The error here is right at the beginning: it's actually not true that $Y = 10X$. That would be equivalent to rolling one die and multiplying the result by $10$, which is quite a different process than rolling ten dice and taking their sum. (Proof: The probability that [the sum of ten dice is $10$] is $6^{-10}$, but the probability that [a single die rolled multiplied by $10$ is $10$] is $1/6$.)
What you really want here is $Y = \sum_{i=1}^{10} X_i$, where the $X_i$ represent independent die rolls. From here, it follows that $$\operatorname{Var}(Y) = \operatorname{Var} \left( \sum_{i=1}^{10} X_i \right) = \sum_{i=1}^{10} \operatorname{Var}(X_i) = \sum_{i=1}^{10} 2.91 = 29.1.$$
EDIT: Oops, I forgot to answer your questions at the end as well!