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$1.$ Prove: If $X_1,X_2,X_3,\ldots,X_n$ are independent random variables then: $$D\left(\sum_{i=1}^n X_i\right)=\sum_{i=1}^n D(X_i)$$

Proof: Because of independence we have:

$$D(\sum_{i=1}^n X_i)=E(\sum_{i=1}^n (X_i-EX_i))^2= \sum_{i=1}^n E(X_i-EX_i)^2+ \sum_{i \neq j} E((X_i-EX_i)(X_j-EX_j))= \sum_{i=1}^n DX_i +**\sum_{i\neq j}(EX_i-EX_i)(EX_j-EX_j)**=\sum_{i=1}^n DX_i$$

What I dont understand is how the expressions within the stars are derived as well as how this independence comes into play here.

$2.$ Find : $EX(X-1)$ if $X=G(p)$- geometric distribution of $p$.

Answer(which I completely don't understand): $$\sum_{k=2}^\infty k(k-1)q^{k-1}p$$

Background of my understanding: I know how to get EX , and I;m aware that the answer here is close to the second derivative of EX.. That is all..

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    $\begingroup$ You want to show that $E[(X_i-E(X_i))(X_j-E(X_j))]=E[X_i-E(X_i)]E[X_j-E(X_j)]$. (Note the typo between the stars.) This is a consequence of independence. By the way, what you call dispersion is usually called variance in English. That will help in a search. $\endgroup$
    – André Nicolas
    Commented May 24, 2015 at 23:10
  • $\begingroup$ now I am confused as to why the sum over that is equal to zero ? $\endgroup$
    – Bozo Vulicevic
    Commented May 24, 2015 at 23:14
  • $\begingroup$ Note that $E(X_i)$ is a constant, which temporarily I will call $\mu_i$. Then $E[X_i-\mu_i]=E[X_i]-\mu_i=\mu_i-\mu_i=0$. Informally, if $\mu_i$ is the average value of $X_i$, then the average value of $X_i-\mu_i$ is 0$. $\endgroup$
    – André Nicolas
    Commented May 24, 2015 at 23:17
  • $\begingroup$ "Dispersion" is a broader term than variance. Variance and standard deviation are measures of dispersion, mean absolute deviation from the mean is a measure of dispersion, range is a measure of dispersion, interquartile range is a measure of dispersion, entropy of a discrete categorical distribution is a measure of dispersion, and there are others. $\endgroup$
    – Michael Hardy
    Commented May 24, 2015 at 23:20
  • $\begingroup$ @MichaelHardy: But in this instance "dispersion" is used to mean variance. Most other measures of dispersion do not have the additivity property. $\endgroup$
    – André Nicolas
    Commented May 24, 2015 at 23:22

1 Answer 1

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$\newcommand{\E}{\operatorname{E}}$If $U$, $V$ are undependent and $\E U$ and $\E V$ both exist then $\E(UV)=(\E U)(\E V)$.

Therefore $$ \E\Big((X_i-\E X_i)(X_j-\E X_j\Big) = \E(X_i-\E X_i) \E(X_j-\E X_j) $$ and $$ \E(X_i-\E X_i) = \E X_i - \E(\E X_i)) = \E X_i - \E X_i=0 $$ and similarly for $j$.

\begin{align} & \sum_{k=2}^\infty k(k-1)q^{k-1}p = qp \sum_{k=2}^\infty k(k-1)q^{k-2} = qp \sum_{k=2}^\infty \frac{d^2}{dq^2} q^k \\[10pt] = {} & qp\frac{d^2}{dq^2}\sum_{k=2}^\infty q^k \qquad (\text{This step is problematic.}) \\[10pt] = {} & qp \frac{d^2}{dq^2} \frac{q^2}{1-q}. \end{align} Now do the differentiation.

The derivative of the sum is the sum of the derivatives when the number of terms is finite. When it is infinite, then that isn't always true. But it is true of power series in the interior of the intervals of convergence, and that is applied here.

There are also discrete ways of solving this problem.

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  • $\begingroup$ Id like to see the second answer I must admit. $\endgroup$
    – Bozo Vulicevic
    Commented May 24, 2015 at 23:37

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