# 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 ?.

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 $$Var(Z), Var(L_n), Var(G_n)$$.

• Thank you very much! If I substitute E(X1)+E(X2)+...+E(Xn) = a for the condition E(Xi)=a/n, will the variance of Ln be zero when n approaches infinity? – Michael.Andy May 23 at 16:12
• I think you are referring to the second part of my answer. Indeed if you have $X_i=Y_i/n$ then you get $Var(X_1+...+X_n)\leq \sigma_{max}^2/n$. Now if $\sigma_{max}$ is a fixed number (independent of $n$) then we can say $\lim_{n\rightarrow\infty} \sigma_{max}^2/n=0$. However, the notation $X_i=Y_i/n$ is awkward. Are you re-starting a new simulation every time you change $n$? – Michael May 23 at 16:15
• Yes, I change n and re-start a new simulation every time. But actually form the point of the original problem(Using Monte Carlo method to get the pdf of the sum of random variables), it is expected that when n becomes larger, the summation of random variable's pdf will be more accurate. But follow the previous process, the variance approaches zero which makes me confused. – Michael.Andy May 23 at 16:41
• I'm not sure what your model is, or what you expect to happen. The "Law of Large Numbers" is a standard theorem and indeed the variance of $L_n$ should go to zero as described above. On the other hand another theorem in that genre is the "Central Limit Theorem" which says that, under a different scaling $G_n = \frac{1}{\sqrt{n}}\sum_{i=1}^n (Y_i-a)$, the distribution of $G_n$ converges to a Gaussian distribution. What you said about PDFs in your last comment suggests you expect something like this, in which case you need to divide by $\sqrt{n}$, not $n$, and need to subtract the mean. – Michael May 23 at 20:59