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# S2 Continous Random Variable Q watch

1. f(x) = 6x(1-x) 0<x<1
= 0 elsewhere

use integration to show the following:
i) E[X] = 1/2..........done
ii) V[X] = 1/20

hence state the value of:
a)E[2X-1]
b) V[5X-3]

calculate tha value of E[X^-1] and hence state the value of E[1/4X^-1]

2. v(X) = integral of (x^2)f(x) - [e(x)]^2
=> integral of 6x^3-6x^4 = [(6/4)x^4 - (6/5)x^5] (between 1 and 0)
=6/4 - 6/5 = 6/20
and then minus the expectation squared to get
6/20 - (1/4)=6/20 - 5/20 = 1/20
3. Then there's a couple of results for the next part:-
E(a + bX) = a + bE(X)
So E[2X-1] = -1+2(1/2) = 0
And Var(a +bX) = (b^2)Var(X)
b) V[5X-3] = 25(1/20) = 5/4
4. I can't think how to find E(x^-1) Sorry (and sorry for posting 3 separate posts!)

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Updated: November 10, 2004
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