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Math Help - Expected squared euclidean distance

  1. #1
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    Expected squared euclidean distance

    Hi,

    can someone help me with this problem?

    Let X, Y be two independent d-dimensional random variables with Gaussian distribution N(\mu_{X},\sigma^2 \mathbb{I}), N(\mu_{Y},\sigma^2 \mathbb{I}), where \mathbb{I} is the identity matrix in \mathbb{R}^d, \sigma \in \mathbb{R} with \sigma > 0, and \mu_X , \mu_Y \in \mathbb{R}^d.

    Compute the expected squared distance E(||X-Y||^2).

    Hint: Use that ||X-Y||^2 = ||X||^2 + ||Y||^2 - 2<X,Y>
    This is what I've got so far:

    E(||X-Y||^2) =

    E(||X||^2 + ||Y||^2 - 2<X,Y>) =

    E(||X||^2) + E(||Y||^2) - 2E(<X,Y>) =

    E(\sum_{i = 1}^d X_i^2) + E(\sum_{i = 1}^d Y_i^2) - 2E(\sum_{i=1}^d X_i Y_i) =

    E(\sum_{i = 1}^d X_i^2) + E(\sum_{i = 1}^d Y_i^2) - 2 \sum_{i = 1}^d E(X_i Y_i) =

    E(\sum_{i = 1}^d X_i^2) + E(\sum_{i = 1}^d Y_i^2) - 2 \sum_{i = 1}^d E(X_i) E(Y_i) =

    E(\sum_{i = 1}^d X_i^2) + E(\sum_{i = 1}^d Y_i^2) - 2 \sum_{i = 1}^d \mu_{X_i} \mu_{Y_i} =

    E(\sum_{i = 1}^d X_i^2) + E(\sum_{i = 1}^d Y_i^2) - 2 <\mu_X , \mu_Y>

    Now, what could be done further (especially regarding the 2 first addends)?

    Thank you very much!
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  2. #2
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    Quote Originally Posted by Kyle_Katarn View Post
    Hi,

    can someone help me with this problem?

    This is what I've got so far:

    E(||X-Y||^2) =

    E(||X||^2 + ||Y||^2 - 2<X,Y>) =

    E(||X||^2) + E(||Y||^2) - 2E(<X,Y>) =

    E(\sum_{i = 1}^d X_i^2) + E(\sum_{i = 1}^d Y_i^2) - 2E(\sum_{i=1}^d X_i Y_i)

    E(\sum_{i = 1}^d X_i^2) + E(\sum_{i = 1}^d Y_i^2) - 2 \sum_{i = 1}^d E(X_i Y_i) =

    E(\sum_{i = 1}^d X_i^2) + E(\sum_{i = 1}^d Y_i^2) - 2 \sum_{i = 1}^d E(X_i) E(Y_i) =

    E(\sum_{i = 1}^d X_i^2) + E(\sum_{i = 1}^d Y_i^2) - 2 \sum_{i = 1}^d \mu_{X_i} \mu_{Y_i} =

    E(\sum_{i = 1}^d X_i^2) + E(\sum_{i = 1}^d Y_i^2) - 2 <\mu_X , \mu_Y>

    Now, what could be done further (especially regarding the 2 first addends)?

    Thank you very much!
    E\left(\sum_{i = 1}^d X_i^2\right)=\sum_{i=1}^d E(X_i^2)=\sum_{i=1}^d \left[\sigma_{X_i}^2 + \overline{X}_i^2\right]<br />
=d\sigma^2 + \sum_{i=1}^d \overline{X}_i^2<br />
=d\sigma^2 + \sum_{i=1}^d (\mu_X)_i^2 <br />
=d\sigma^2 + \Vert \mu_X \Vert^2<br />

    .
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  3. #3
    MHF Contributor

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    Paris, France
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    Quote Originally Posted by Kyle_Katarn View Post
    Hi,

    can someone help me with this problem?

    This is what I've got so far:

    E(||X-Y||^2) =

    E(||X||^2 + ||Y||^2 - 2<X,Y>) =

    E(||X||^2) + E(||Y||^2) - 2E(<X,Y>) =

    E(\sum_{i = 1}^d X_i^2) + E(\sum_{i = 1}^d Y_i^2) - 2E(\sum_{i=1}^d X_i Y_i) =
    (...)
    E(\sum_{i = 1}^d X_i^2) + E(\sum_{i = 1}^d Y_i^2) - 2 <\mu_X , \mu_Y>

    Now, what could be done further (especially regarding the 2 first addends)?

    Thank you very much!
    E[\|X\|^2]=E\left[\sum_{i=1}^d X_i^2\right]=\sum_{i=1}^d E[X_i^2] and E[X_i^2]={\rm Var}(X_i)+E[X_i]^2=\sigma^2+(\mu_X^{(i)})^2, so that E[\|X\|^2]=d\sigma^2+\|\mu_X\|^2.

    By the way: if you know it, you can use the fact that X-Y is a d-dimensional r.v. with distribution \mathcal{N}(\mu_X-\mu_Y,2\sigma^2 I_d), and apply the above computation to X-Y instead of just X. This way you'll be able to get the result quicker (almost instantly), so you can check yours.
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  4. #4
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    Oh man, I totally forgot the law Var(X) = E(X^2) - (E(X))^2! Thanks so much to both of you!
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