Suppose that the possible values of Y in a population are b1

Suppose that the possible values of Y in a population are b₁, b₂, … , b_k and the number of individuals with value b_i in the population is n_i (there are n₁ individuals with value b_1, n₂ individuals with value b₂, …, n_k individuals with value b_k). Now, you draw one individual at random from this population and note the value of the variable Y for that person. This yields a random variable. Show that the expected value of this random variable is the same as the population mean of Y.

Solution

population mean is given as

Ybar=(n1b1+n2b2+n3b3+n4b4+......+nkbk)/(n1+n2+n3+n4+....+nk)

let X be the random variable denoting the value of Y of the person drawn at random.

so X can take the values b1 or b2 or b3 or.........or bk with probailities n1/(n1+n2+n3+n4+....+nk) or n2/(n1+n2+n3+n4+....+nk) or n2/(n1+n2+n3+n4+....+nk) or........or nk/(n1+n2+n3+n4+....+nk).

so the probability distribution of X is

X:                      b1          b2                b3           .................         bk

P[X=x]:     n1/(n1+n2+n3+....+nk)    n2/(n1+n2+n3+....+nk) n2/(n1+n2+n3+....+nk) ... nk/(n1+n2+n3+n4+....+nk)

hence E[X]=b1n1/(n1+n2+n3+n4+....+nk)+b2n2/(n1+n2+n3+n4+....+nk)+b3n3/(n1+n2+n3+n4+....+nk)+.......+bknk/(n1+n2+n3+n4+....+nk)=(n1b1+n2b2+n3b3+n4b4+......+nkbk)/(n1+n2+n3+n4+....+nk)=Ybar [proved]