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).
>It is immaterial whether Y,Z1,Z2,Z3,Z4 are independent or not. If for
>instance Y=Z1=Z2=Z3=Z4 then E(X)=E(Y).
>
>
Jon Miller
.
.
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