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Math Help - Question about mean and variance

  1. #1
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    Question about mean and variance

    Question : If X_1 has mean 4 and variance 9 while X_2 has mean -2 and variance 5, and the two are independent, determine
    i) E(2X_1 + X_2 -5)
    ii) Var(2X_1 + X_2 -5)

    My answers are
    i) E(2X_1 + X_2 -5) = 2[E(X_1)] + [E(X_2) -5] = 8 + [-2 -5] = 1


    ii) Var(2X_1 + X_2 -5) = 2[Var(X_1)] + [Var(X_2) -5] = 2(9) + [5 -5] = 18 + 0 = 18


    Is my answer correct and is my method for solving the answer crorect or no
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    Scalars a and b acting on a random variable X as:

    aX+b

    and act on a random variables variance as:

    a^2\sigma^2_x;

    Why? Think about it this way. If you had the ages of 10 people, and you multiplied their ages by 2 and added 10 years, the 10 years isn't going to affect the variation because they have all been adjusted evenly by 10 years. What will affect their ages is multiplying by the 2, and the variance is adjust by the square of the scalar.
    Last edited by ANDS!; December 10th 2009 at 02:33 PM. Reason: Less confusing.
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  3. #3
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    Quote Originally Posted by zorro View Post
    Question : If X_1 has mean 4 and variance 9 while X_2 has mean -2 and variance 5, and the two are independent, determine
    i) E(2X_1 + X_2 -5)
    ii) Var(2X_1 + X_2 -5)

    My answers are
    i) E(2X_1 + X_2 -5) = 2[E(X_1)] + [E(X_2) -5] = 8 + [-2 -5] = 1


    ii) Var(2X_1 + X_2 -5) = 2[Var(X_1)] + [Var(X_2) -5] = 2(9) + [5 -5] = 18 + 0 = 18


    Is my answer correct and is my method for solving the answer crorect or no
    E(aX_1 + bX_2) = a E(X_1) + b E(X_2) and E(cX + d) = c E(X) + d and so your answer to (i) is correct.

    However, if X_1 and X_2 are independent then Var(aX_1 + bX_2) = a^2 Var(X_1) + b^2 Var(X_2) and Var(cX + d) = c^2 Var(X). Therefore your answer to (ii) is wrong.
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    Is this correct?

    Quote Originally Posted by mr fantastic View Post
    E(aX_1 + bX_2) = a E(X_1) + b E(X_2) and E(cX + d) = c E(X) + d and so your answer to (i) is correct.

    However, if X_1 and X_2 are independent then Var(aX_1 + bX_2) = a^2 Var(X_1) + b^2 Var(X_2) and Var(cX + d) = c^2 Var(X). Therefore your answer to (ii) is wrong.

    Mr fantastic is this correct

    Var(2X_1 + X_2 - 5) = 2^2 [Var(X_1)] + Var(X_2) - 5 = 4 * 9 + 5 -5 = 36 .......Is this correct
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    Quote Originally Posted by zorro View Post
    Mr fantastic is this correct

    Var(2X_1 + X_2 - 5) = 2^2 [Var(X_1)] + Var(X_2) - 5 = 4 * 9 + 5 -5 = 36 .......Is this correct
    Not quite.

    Var(X_2 - 5) = Var(X_2) (using Var(cX + d) = c^2 Var(X) and noting that c = 1 and d = -5). Therefore Var(2 X_1 + X_2 - 5) = 2^2 Var(X_1) + Var(X_2 - 5) = 2^2 Var(X_1) + Var(X_2) = .....
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    Is this correct?

    Quote Originally Posted by mr fantastic View Post
    Not quite.

    Var(X_2 - 5) = Var(X_2) (using Var(cX + d) = c^2 Var(X) and noting that c = 1 and d = -5). Therefore Var(2 X_1 + X_2 - 5) = 2^2 Var(X_1) + Var(X_2 - 5) = 2^2 Var(X_1) + Var(X_2) = .....


    Is the answer 41?
    Last edited by mr fantastic; December 10th 2009 at 05:55 PM. Reason: Edited the quote (I had edited my post whch this quote comes from).
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  7. #7
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    Quote Originally Posted by zorro View Post
    Is the answer 41?
    Yes.
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  8. #8
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    Thanks everyone

    Quote Originally Posted by mr fantastic View Post
    Yes.

    Thanks every one who has helped me
    Cheers
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