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    Expectation of normal distribution

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    I know this problem is very easy, finding E(x) from the distribution function. I tried to do it by integration by parts, one way I took x to be my first function and the whole exp term to be my 2nd function, but it didn't work, then I split the exponential term to 2 terms combined one with x as my first function and took the second as the 2nd function but it didn't work either. Can someone help me?

    Please provide a detailed proof/answer. Justify all your claims.

    See attached file for full problem description.

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    https://brainmass.com/statistics/normal-distribution/expectation-normal-distribution-78092

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    Please see the attached file.

    .f(x) = __1__ exp-( x - μ )^2
    √2πσ^2 2σ^2

    α
    E(x) = ∫ x __1__ exp-(x - μ)^2 dx
    -α √2πσ^2 2σ^2

    let ( x - μ ) = t ; dx = dt ; dx = √2σ dt
    √2σ √2σ

    x = √2σ t + μ
    Substituting in the above integral we get
    α
    ∫ (√2σ t + μ) { __1__ exp(- t^2 ) }√2σdt
    -α √2πσ^2

    α
    ∫ (√2σ t + μ) { __1__ ...

    Solution Summary

    The expert examines expectations of normal distributions.

    $2.19

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