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    I'm stuck on the following question:
    Name:  d3fee9591bb448dca7ac9ddb3d724404.png
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    I've done 3a) and I got
    E((X-a)^2) = σ^2 + μ^2 + a^2 -2aμ

    For part b I don't understand what the question wants me to do, I've looked online and seen some things on integration for the minimum which I have no idea how to do, I failed the previous probability module and I think this one follows on and expects me to know this.

    Any help or pointers would be appreciated
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    (Original post by Sayless)
    I'm stuck on the following question:
    Name:  d3fee9591bb448dca7ac9ddb3d724404.png
Views: 32
Size:  16.7 KB
    I've done 3a) and I got
    E((X-a)^2) = σ^2 + μ^2 + a^2 -2aμ

    For part b I don't understand what the question wants me to do, I've looked online and seen some things on integration for the minimum which I have no idea how to do, I failed the previous probability module and I think this one follows on and expects me to know this.

    Any help or pointers would be appreciated
    For part a i get that \mathbb{E}((X-a)^{2})=\mu^{2}+\sigma^{2}-2a\mu+a^{2}

    For part b. As \mu and \sigma are constants taking the derivative with respect to a gives \frac{\text{d}(\mathbb{E}((X-a)^{2}))}{ \text{d}a}=-2\mu+2a.

    For \mathbb{E}((X-a)^{2}) to be minimised, \frac{\text{d}(\mathbb{E}((X-a)^{2}))}{ \text{d}a}=0

    I have not done this really in much detail so this could be wrong.
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    (Original post by Cryptokyo)
    For part a i get that \mathbb{E}((X-a)^{2})=\mu^{2}+\sigma^{2}-2a\mu+a^{2}

    For part b. As \mu and \sigma are constants taking the derivative with respect to a gives \frac{\text{d}(\mathbb{E}(X))}{ \text{d}a}=-2\mu+2a.

    For \mathbb{E}(X) to be minimised, \frac{\text{d}(\mathbb{E}(X))}{ \text{d}a}=0

    I have not done this really in much detail so this could be wrong.
    We got the same answer for part 'a' it looks like. The derivative method makes sense, using that method I got a = μ, by value I thought it meant any real number, I overlooked the fact that it could equal μ but that is also a real number.

    Then for part C all I would have to do is plug in a = μ into the answer to part a, that would give me:
    E((X-a)^2) = σ^2

    Is that right?
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    (Original post by Sayless)
    We got the same answer for part 'a' it looks like. The derivative method makes sense, using that method I got a = μ, by value I thought it meant any real number, I overlooked the fact that it could equal μ but that is also a real number.

    Then for part C all I would have to do is plug in a = μ into the answer to part a, that would give me:
    E((X-a)^2) = σ^2

    Is that right?
    Yes that is what I got for part c .
 
 
 
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