Suppose the population size is 10 and the independent measurements pertaining to a particular interest(for example number of siblings) are as following, 1,1,2,2,3,4,4,4,5,5,5.

So the frequency is as following,

number of siblings , frequency

1 , 2

2 , 2

3 , 1

4 , 3

5 , 3

Suppose i collect a sample of size of 3, the sample mean is a random variable denoted as $\bar X_3$, where $\bar X_3 = \frac{X_1+X_2+X_3}{3}$, $X_1,X_2,X_3$ are identical independent random variables.

The population mean $\mu$, is $E[X_i]$, which is also equivalent to $E[\bar X_n]$.

I have some questions that i need help on.

Are my thoughts 1) and 2) below correct?

1) the random variables $X_1, X_2, X_3$ each takes the possible value of 1,2,3,4 or 5.

The actual value that $X_i$ takes is the number of siblings that the $i$-th person in my sample has.

2) To calculate the population mean $\mu$, i will need the probabilities associated to the frequency. So $E[X_i] = \frac{2}{11}\cdot 1 + \frac{2}{11}\cdot 2 + \frac{1}{11}\cdot 3 + \frac{3}{11}\cdot 4 + \frac{3}{11}\cdot 5$

and this agrees to the definition of population mean, $\frac{x_1+x_2+...+x_N}{N}$, where $N$ is the population size.

  • $\begingroup$ In your example the value 1 occurs once, not twice. As a check, add up the frequencies you listed: $2+2+1+3+3$ which comes to 11 not 10 as it should. As to your questions: (1) is correct. And (2) would be, if the arithmetic is done correctly. You are asking, in effect, is $EX = \sum_i P(X=i) i = \sum_i i f_i /N$ if $N=\sum_i f_i$ and $f_i$ is the frequency of $i$ in the population. $\endgroup$ Aug 12 '17 at 14:21
  • $\begingroup$ oops, i missed out a '1', edited. i found the following description online, "$x_1,...,x_n$ can be viewed as a realizations of identical, independent random variables $X_1,...,X_n$" Any idea what this means? $\endgroup$ Aug 12 '17 at 14:47

First, it would help to get the numbers right. Second, this might be easier to understand with more carefully chosen notation.

Numerical computations. There are two ways to find the population mean: $\mu = (1+1+2+2+3+4+4+4+5+5+5)/11 = 3.2727\;$ and $\mu = [2(1) + 2(2) + 3(1) + 4(3) + 5(3)]/(2+2+1+3+3) = 36/11 = 3.2727.$

Formulas. If the individual numbers of siblings in the population are denoted $s_1 = 1,\,s_2=1, \dots, s_{11} = 5.$ then the population size is $N = 11$ and $\mu = \frac{\sum_{i=1}^N s_i}{N}.$
Also, if the $k=5$ values are $v_1 =1, v_2 = 2, \dots, v_5 = 5$ siblings with corresponding frequencies $f_1 = 2, f_2 = 2, \dots, f_5 = 3,$ then $\mu = \frac{\sum_{j=1}^k f_jv_j}{\sum_{j=1}^k f_j} = \frac{\sum_{j=1}^k f_jv_j}{N} = \sum_{j=1}^k r_iv_i,$ where $r_j = f_j/N$ are called relative frequencies.

Theoretical statements. If you are sampling $n = 3$ subjects from this population at random and with replacement, you can say that the $n$ values $X_i$ you observe are independent and identically distributed random variables. The expected number of siblings on $i$th observation $E(X_i) = \mu$.

  • $\begingroup$ The part that i don't fully understand is the realization of observations as identical, independent random variables. Is it right that the i.i.d random variables each takes possible values of 1,2,3,4 or 5 in my example? And why do we let the observations be i.i.d random variables? What is the intuition and purpose for it? I tried to find it in the textbook but i can't. :( $\endgroup$ Aug 12 '17 at 18:07
  • $\begingroup$ Yes, each of the $X_i$ takes one of those possible values. It seems you are just at the beginning of random variables. Different texts have various developments. I think it's best to get that info from the text as you move along. $\endgroup$
    – BruceET
    Aug 12 '17 at 18:47
  • $\begingroup$ I've completed an introductory course to probability. Doing some self-learning on statistics right now before taking regression analysis during the next semester. Is there any text on Statistics you would recommend? Right now im reading "Mathematical statistics and data analysis" by Rice J.A, it's the same book my school used for probability course. $\endgroup$ Aug 12 '17 at 20:24
  • $\begingroup$ Rice is a fine modern text. I have taught several classes from it. A more traditional book that also does justice to both theory and practice (but does not have much about computer applications), is the book by Bain and Englehardt, now available in a relatively inexpensive paperback edition. A more elementary book is by Wackerly, Mendenhall, & Shafer; new editions are expensive, but for your purposes an older edition is just as good. $\endgroup$
    – BruceET
    Aug 13 '17 at 0:23
  • $\begingroup$ I'm starting on the book Mathematical statistics with applications by Wackerly, Mendenhall, Shafer. Thanks for the suggestion! $\endgroup$ Aug 13 '17 at 9:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.