On Wed, 16 Jul 2003 10:57:30 -0700, Jon and Mary Miller
<[EMAIL PROTECTED]> wrote:
> Horst Kraemer wrote:
>
> >On Wed, 16 Jul 2003 16:38:40 +0800, "ZHANG Yan" <[EMAIL PROTECTED]>
> >wrote:
> >
> >
> >
> >>Hea, I am asking for a question about the expected value.
> >>
> >>Suppose the postive continuous random variable X, Y, Z1, Z2,Z3,Z4. Moreover,
> >>Y, Z1,Z2,Z3,Z4 are independent. The pdf of them are given. They satisfy the
> >>following relationship.
> >>
> >> __
> >> | Y if Y<Z
> >>X =
> >> | Z1 if 2Z>Y>Z
> >> | Z2 if 3Z>Y>2Z
> >> | Z3 if 4Z>Y>3Z
> >> | Z4 other
> >> ---
> >>
> >>I need to find the expected value of X. What should I do?Thanks in advance!
> >>
> >>
> >
> >I assume that Z is some constant. and y(t) is the pdf of Y.
> >
> >Using
> >
> > Z
> >p0 = Pr {Y<Z} = int y(t) dt
> > -oo
> >
> > 2Z
> >p1 = Pr {Z<Y<2Z} = int y(t) dt
> > Z
> >
> > 3Z
> >p2 = Pr {2Z<Y<3Z} = int y(t) dt
> > 2Z
> >
> > 4Z
> >p3 = Pr {3Z<Y<4Z} = int y(t) dt
> > 3Z
> >
> > oo
> >p4 = Pr {4Z<Y} = int y(t) dt
> > 4Z
> >
> >(Sum p_i = 1)
> >
> >Then E(X) = p0*E(Y)+p1*E(Z1)+p2*E(Z2)+p3*E(Z3)+p4*E(Z4)
> >
> This should be p0*E(Y|Y<=Z) + the rest. The difference being that
> E(Y|Y<=Z) = int(-oo,Z)(t*y(t) dt) whereas E(Y) = int(-oo,+oo)(t*y(t) dt).
Yes, of course, but E(Y|Y<=Z) = int(-oo,Z)(t*y(t) dt)/p0, therefore
you may write
E(X) = p0`*E(Y|><=Z) + p1*E(Z1)+p2*E(Z2)+p3*E(Z3)+p4*E(Z4)
or
E(X) = int(-oo,Z)(t*y(t) dt + p1*E(Z1)+p2*E(Z2)+p3*E(Z3)+p4*E(Z4)
Regards
Horst
.
.
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