Variance of the sum of n-independent random variables 
*

*Let $X_{1}, X_{2},\ldots,X_{n}$ be $n$-independent non-identical random variables.

*Here I define $a = 5$ and $E\left(X_{i}\right) = a/n,\quad 1 \leq i \leq n$.

*Then I simulate the Summation process of random variables. The result shows that expectation value approaches $5$, while the variance of the summation of $X_{i}$ approaches zero.


I am struggling to show that the variance approaches zero.
Do you have any ideas ?.
 A: Usually you have mutually independent variables $\{Y_1, Y_2,  Y_3, ...\}$ and you define for all positive integers $n$: 
$$L_n = \frac{Y_1+Y_2+...+Y_n}{n}$$ 
Then
\begin{align} 
E[L_n]&=\frac{E[Y_1]+E[Y_2]+...+E[Y_n]}{n}\\
Var(L_n)&=\frac{Var(Y_1)+Var(Y_2)+...+Var(Y_n)}{n^2}
\end{align}
If $E[Y_i]=a$ and $Var(Y_i)\leq \sigma_{max}^2$ for all $i$ then the above equations imply
\begin{align}
E[L_n]&=a \quad \forall n \in \{1, 2, 3, ...\}\\
Var(L_n)&\leq \frac{\sigma_{max}^2}{n} \quad \forall n \in \{1, 2, 3, ...\}
\end{align}
So indeed the mean $E[L_n]$ does not change but the variance $Var(L_n)$ goes to zero as $n\rightarrow\infty$. 

It looks like you are defining $X_i=Y_i/n$ for some reason, in which case $E[X_i]=E[Y_i]/n=a/n$ and your results are consistent with the above:
\begin{align}
X_1+...+X_n &= L_n\\
E[X_1+...+X_n]&=E[L_n]=a\\
Var(X_1+...+X_n)&=Var(L_n)\leq \frac{\sigma_{max}^2}{n}
\end{align}
However the notation $X_i=Y_i/n$ is a bit awkward since then $X_i$ depends on both $i$ and $n$. So it would be difficult/impossible to change the value of $n$ during a simulation run.

Useful variance formulas are these: Let $X, Y, X_1, ..., X_n$ be independent random variables and let $c,d$ be given real numbers.  Then
\begin{align}
Var(aX+b) &= a^2Var(X)\\
Var(X+Y) &= Var(X)+Var(Y)\\
Var\left(\sum_{i=1}^n X_i\right) &= \sum_{i=1}^n Var(X_i)
\end{align}
Using these formulas you can compute many things:  Let $\{Y_i\}_{i=1}^{\infty}$ be independent random variables with $E[Y_i]=a$ and $Var(Y_i)=\sigma^2$ for all $i$, let $b_1, ..., b_n$ be real numbers, and define
\begin{align}
Z &= \sum_{i=1}^n b_i Y_i\\
L_n &= \frac{1}{n}\sum_{i=1}^n Y_i\\
G_n &=\frac{1}{\sqrt{n}}\sum_{i=1}^n(Y_i-a)
\end{align}
Use the above formulas to compute $E[Z], E[L_n], E[G_n]$ and $Var(Z), Var(L_n), Var(G_n)$.
