let the random variable X have a mean of m and variance n^2. define Y=a+bx where a and b are constants satisfying -∞<a<∞ and b>0. select a and b so that Y has a mean of zero and variance equal to one.

Question

statistical theory

Answer 1

Suppose the set of data for X is { Xsub1 Xsub2 Xsub3 ... XsubN }, then m=S(XsubJ)/N where S represents the sum of XsubJ for J=1 to N. But X=(Y-a)/b and XsubJ=(YsubJ-a)/b.

Therefore m=S(XsubJ)/N=S((YsubJ-a)/b)/N=S(YsubJ)/N-Na/bN=m'-a/b where m' is the mean of Y. We know that m'=0. m'=m+a/b=0, m=-a/b, and a=-bm.

S((XsubJ-m)^2)/N is the variance=n^2. So n^2=S(XsubJ-m)^2/N=S(((YsubJ-a)/b-m)^2)/N=

S((YsubJ/b-a/b+a/b)^2/N=S((YsubJ)^2)/Nb^2. Variance of Y=1=S((YsubJ)^2)/N.

So n^2=1/b^2 and b^2=1/n^2, b=1/n>0. Therefore, since a=-bm, then a=-m/n.

ANSWER: a=-m/n and b=1/n.