Let $X_1,X_2,...,X_n$ be realizations of a random variable $X_0$. Let $Y_1,Y_2,...,Y_n$ be realizations of a random variable $Y_0 \neq X_0$. The distribution of $X_0$ and $Y_0$ is $N(0,\sigma^2_x)$ and $N(0,\sigma^2_y)$ and they are independent of each other. How can I find the pdf of the euclidean distance /difference between the two initial conditions i.e., $Z = {||X_0 - Y_0||}_2$

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    $\begingroup$ The question can't be answered without specifying the joint distribution (or equivalent information) $\endgroup$
    – Glen_b
    Jul 28, 2016 at 0:27
  • $\begingroup$ +1 to @Glen_b 's comment . Plus, the way I read your question, the nonlinear dynamical map, $f$, is totally irrelevant to answering your question - if that is not the case, then why do $f$ and iterateis beyond $X_0$ and $Y_0$ matter? $\endgroup$ Jul 28, 2016 at 0:30
  • $\begingroup$ @Glen_b: Due to the deterministic property, $P({x_{j+1} | x_j}) $ = 1 same for $y$ and $P_x(x_0) = N(0,\sigma_x^2\mathbf{I}) $; same for $y$ for $j =1:n$ $\endgroup$
    – SKM
    Jul 28, 2016 at 0:31
  • $\begingroup$ Perhaps you have not correctly written out the problem you really want to solve. $\endgroup$ Jul 28, 2016 at 0:33
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    $\begingroup$ Just for the record, my comment above has a typo, and was intended to be $X_0 - Y_0$ is $N(0,\sigma_x^2 + \sigma_y^2)$,, as I think you already understood to be the case. $\endgroup$ Jul 28, 2016 at 1:29

1 Answer 1


As you stated in a comment, $X_0$ and $Y_0$ are independent, therefore the column vector $V = [X_0,Y_0]$ is Bivariate Normal with mean being the zero vector, and covariance matrix being the diagonal matrix with $\sigma^2_x$ and $\sigma^2_y$ on the diagonal.

$X_0 -Y_0 = BV$, where B is the row vector [1 -1]. Applying the properties of an affine transformation of a Multivariate Normal https://en.wikipedia.org/wiki/Multivariate_normal_distribution#Affine_transformation , results in $X_0 - Y_0$ being $N(0,\sigma^2_x$ + $\sigma^2_y)$.

If you want to just understand more simply what is going on with the variance, then look at the properties of Variance provided at https://en.wikipedia.org/wiki/Variance#Properties . In particular, if $X$ and $Y$ are random variables and $a$ and $b$ are constants, then $$\operatorname{Var}(aX + bY) = a^2\operatorname{Var}(x) + b^2\operatorname{Var}(Y) + 2ab\operatorname{Cov}(X,Y).$$In this case, use $a = 1, b = -1$.

Finally, note that $Z = \operatorname{abs}(X_0 - Y_0)$, and therefore has the Half-Normal distribution http://en.wikipedia.org/wiki/Half-normal_distribution with parameter $\sigma^2_x$ + $\sigma^2_y$.


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