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., $Z = {||X_0 - Y_0||}_2$

  • 1
    $\begingroup$ The question can't be answered without specifying the joint distribution (or equivalent information) $\endgroup$ – Glen_b -Reinstate Monica Jul 28 '16 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? $\endgroup$ – Mark L. Stone Jul 28 '16 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 '16 at 0:31
  • $\begingroup$ Perhaps you have not correctly written out the problem you really want to solve. $\endgroup$ – Mark L. Stone Jul 28 '16 at 0:33
  • 1
    $\begingroup$ 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. $\endgroup$ – Mark L. Stone Jul 28 '16 at 1:29

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 $\sigma^2_x$ + $\sigma^2_y$.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.