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anonymous

  • 4 years ago

Can somebody give me advice on how to find variance and the expected value; when one considers the function f(Y) = 1/9 is when 0 <= y <= 9 and the probability is 0 everywhere else when y is not in that interval(0-9). please note that this is a uniform distribution function and Y is continous. Im struggling with the integrals. How can you find an expected value for Y, when all values in the interval [0,9] have the same probability? thanks for your help!

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  1. dumbcow
    • 4 years ago
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    \[E(X) = \int\limits_{0}^{9}p(y)*y = \int\limits_{0}^{9}\frac{y}{9} dy\]

  2. dumbcow
    • 4 years ago
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    Variance is E(X^2) - E(X)^2

  3. anonymous
    • 4 years ago
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    why isn't this oh, what dumbcow said

  4. anonymous
    • 4 years ago
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    But when I put in 9 for Y: 1/9*9 = 1. But I think the answer is 4.5. Note that I am having trouble with integrals, even basic ones. Is the expected value equal to the average value in this case?

  5. dumbcow
    • 4 years ago
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    yes it is same as average value since it is uniformly distributed. E(X) = 4.5

  6. anonymous
    • 4 years ago
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    ah I see, thank you!

  7. dumbcow
    • 4 years ago
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    no problem :)

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