Horst Kraemer wrote:
>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,
>
Skitt's Law (among other appelations): Every spelling (in this case,
formula) correction contains its own error.
Well, to OP, you've now got the right formula.
This is why mathematicians avoid calculations.
> 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)
>
>
>
Jon Miller
.
.
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