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Expectation of Sums of Random Variables

Example (Expectation of a Binomial Random Variable): Let X be a binomial random variable with parameters n and p. Recalling that such a random variable represents the number of successes in nindependent trials when each trial has probability p of being a success, we have that $X=X_1+X_2+\cdots+X_n$ where
$X_i=\left\{ \begin{array}{ll} 1&\mbox{ if the }i\mbox{th trial is a success} \\ \\ 0&\mbox{ if the }i\mbox{th trial is a failure} \end{array}\right .$
Hence, X_i is a Bernoulli random variable having expectation E[X_i]=1(p)+0(1-p). Thus
$E[X]=E[X_1]+E[X_2]+\cdots+E[X_n]=np\qquad\rule[0.02em]{1.0mm}{1.5mm}$

Example (Mean of a Negative Binomial Random Variable): If independent trials, having a constant probability p of being successes, are performed, determine the expected number of trials required to amass a total of r successes.
Solution:: If X denotes the number of trials needed to amass a total of r successes, then X is a negative binomial random variable. It can be represented by $X=X_1+X_2+\cdots+X_r$ where
X₁ is the number of trials required to obtain the first success,
X₂ the number of additional trials until the second success is obtained,
X₂ the number of additional trials until the third success is obtained, and so on.
That is, X_i represents the number of additional trials required, after the (i-1)st success, until a total of i successes are amassed. A little thought reveals that each of the random variables X_i is a geometric random variable with parameter p. Hence, $E[X_i]=1/p,i=1,2,\ldots,r$ ; and thus
$E[X]=E[X_1]+\cdots+E[X_r]=\displaystyle\frac{r}{p}\qquad\rule[0.02em]{1.0mm}{1.5mm}$

Example (Mean of a Hypergeometric Random Variable): If n balls are randomly selected from an urn containing N balls of which m are white, find the expected number of white balls selected.
Solution:: Let X denote the number of white balls selected, and represent X as $X=X_1+\cdots+X_m$ where
$X_i=\left\{ \begin{array}{ll} 1&\mbox{ if the }i\mbox{th white ball is selected} \\ \\ 0&\mbox{ otherwise } \end{array}\right .$
Now,
$\begin{array}{rcl} E[X_i]&=&P\{X_i=1\} \\ \\ &=&P\{i\mbox{th white ball is sel... ...n-1}}{\displaystyle{N\choose n}} \\ \\ &=&\displaystyle\frac{n}{N} \end{array}$
Hence, $E[X]=E[X_1]+\cdots+E[X_m]=\displaystyle\frac{mn}{N}$ .
We could also have obtained the above result by using the alternative representation $X=Y_1+\cdots+Y_n$ where
$Y_i=\left\{ \begin{array}{ll} 1&\mbox{ if the }i\mbox{th white ball is selected} \\ \\ 0&\mbox{ otherwise } \end{array}\right .$
Since the ith ball selected is equally likely to be any of the N balls, it follows that $E[Y_i]=\displaystyle\frac{m}{N}$ so $E[X]=E[Y_1]+\cdots+E[Y_n]=\displaystyle\frac{nm}{N}\qquad\rule[0.02em]{1.0mm}{1.5mm}$