Expected squared euclidean distance

**Kyle_Katarn** · Jan 18th 2009, 04:46 AM

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!

**Constatine11** · Jan 18th 2009, 05:53 AM

Originally Posted by Kyle_Katarn

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!

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

.

**Laurent** · Jan 18th 2009, 06:01 AM

Originally Posted by Kyle_Katarn

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.

**Kyle_Katarn** · Jan 18th 2009, 08:23 AM

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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