SECTION 6.3 Parameters and Statistics 367
E(X 1 +X 2 +···+Xk) =E(X 1 ) +E(X 2 ) +···+E(Xk).If the random variablesXandY are independent, then we may write
the density functionfXY(x,y) as a product: fXY(x,y) = fX(x)fY(y),
from which it follows immediately that
E(XY) = E(X)E(Y), whereXandY are independent.In particular, this shows the following very important result. Assume
that we are to taken independent samples from a given population
having meanμ. IfXdenotes the average of these samples, thenXis a
itself a random variable and
X =
X 1 +X 2 +···+Xn
n,
whereX 1 , X 2 , ...,Xnare independent random variables from this pop-
ulation. We have, therefore, that
E(X) =
E(X 1 ) +E(X 2 ) +···+E(Xn)
n
= μ.We now turn our attention to variance. However, a couple of pre-
liminary observations are in order. First of all, letX be a continuous
random variable, let a be a real constant, and set Y = X+a. We
wish first to compare the density functions fY and fX. Perhaps it’s
already obvious thatfY(x) =fX(x−a), but a formal proof might be
instructive. We have
∫t
−∞fY(x)dx=P(Y ≤t) =P(X+a≤t) =P(X≤t−a) =∫t−a
−∞ fX(x)dx.But a simple change of variable shows that