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Let be a dice with the sides $\{0,0,6,0,3,3\}$ which will be rolled $n$ times since $n\in\mathbb{N}$. Now I'm modelling the following experiment: Let $X_i$ be the random variable of the $i$-th throw and $S:=\sum_{i=1}^n X_i$ the sum of the diced values. The exercise is to determine the variance of $S$.

I've calculated the mean of $S$: Since the random variables are independent, then

$$E(X_1) = 0\cdot P(X_1=0)+3\cdot P(X_1=3)+6\cdot P(X_1=6) = 0 + \frac{3}{3}+\frac{6}{6} = 1+1=2,$$ so

$$E(S) = E(X_1+\ldots+X_n) = n\cdot E(X_1) = 2n.$$

Then, the second moment of $X_1$ is

$$E(X_1^2) = 0 + \frac{1}{3}\cdot 9+\frac{1}{6}\cdot 6^2 = 3+6=9,$$

and the variance of $X_1$ is $Var(X_1) = E(X_1^2)-E(X_1)^2 = 9 - 4 = 5.$

For the sum $S$, I determine by the aid of the independece of $X_i$:

$\begin{align*} Var(S) &= Var(X_1+\ldots + X_n) = n\cdot Var(X_1) = 5n.\\ \end{align*}$

Are my calculations and reasons correctly?

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  • $\begingroup$ Isn't $P(X_1=3)=1/2$? $\endgroup$ Sep 3 '19 at 10:43
  • $\begingroup$ I've corrected the numbers of the dice. It has to be $\{0,0,6,0,3,3\}$. $\endgroup$
    – MathCracky
    Sep 3 '19 at 10:46
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All is fine but $$ \operatorname{Var}(X_1+\ldots + X_n) = \operatorname{Var}(n\cdot X_1). $$

You have to take the variance summandwise in case of independence, i.e. $$ \operatorname{Var}(S)= \operatorname{Var}(X_1+\ldots + X_n) =n\cdot \operatorname{Var}(X_1). $$

Illustration: Let $X=\text{weight of a pack of sugar in kg}$. In that case $X_1+\dots +X_n$ is the sum of the weights of onethousand packages in kg whereas $1000 X$ is the weight of one package, measured in grams.

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  • $\begingroup$ Could you please elaborate Why first one is incorrect ?? $\endgroup$
    – tourism
    Sep 3 '19 at 10:58
  • $\begingroup$ @tourism For showing that you only need a counterexample. $\endgroup$
    – drhab
    Sep 3 '19 at 11:37
  • $\begingroup$ Because you deal with a sum and not with changing units. $\endgroup$ Sep 3 '19 at 12:43
  • $\begingroup$ I understood... Got confused with $Var(X_1+X_2+...X_n) = Var(n \times X_1)$ which is not true $\endgroup$
    – tourism
    Sep 3 '19 at 20:32
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You took a nice route when you went for analyzing $S$ as the sum $X_1+\cdots+X_n$.

However an essential mistake is made by calculation of variance.

We do not have: $\mathsf{Var}(X_1+\cdots+X_n)=\mathsf{Var}(nX_1)$.

What we do have is:$$\mathsf{Var}(X_1+\cdots+X_n)=\mathsf{Var}(X_1)+\cdots+\mathsf{Var}(X_n)$$which can be exploited nicely.

This on base of independence of the $X_i$.

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  • $\begingroup$ Thanks for the answer! I've corrected the formula of the variance and I hope that the result should be now correctly. $\endgroup$
    – MathCracky
    Sep 3 '19 at 11:01
  • $\begingroup$ You are welcome. Everything seems fine to me now. $\endgroup$
    – drhab
    Sep 3 '19 at 11:05

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