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What is the PDF of $Z=\sqrt{(X-x_0)^2+(Y-y_0)^2}$ when $X$ and $Y$ are i.i.d. normal random variables with zero mean, i.e., $X \sim N(0,\sigma^2)$ and $Y\sim N(0,\sigma^2)$?

Silverfish
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Bahar
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    No guarantees, just my thoughts about it: If $X \tilde \: N(0,\sigma^2)$, then I would say $(X - x_0) \tilde \: N(-x_0, \sigma^2)$. Which means we are looking at the root of the squared sum of two normally distributed random variables. [As Wikipedia states](http://en.wikipedia.org/wiki/Chi-squared_distribution) _In probability theory and statistics, the chi-squared distribution (...) with $k$ degrees of freedom is the distribution of a sum of the squares of $k$ independent standard normal random variables._ Therefore, I would expect $Z^2$ to be $\chi^2$-distributed, and $Z$ $\chi$-distributed? – der_herr_g Oct 11 '14 at 15:24
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    Since the joint pdf of $X$ and $Y$ has circular symmetry about the origin, use a _rotation of axes_ to bring the point $(x_0,y_0)$ to lie at $\left(\sqrt{x_0^2+y_0^2},0\right) = (c,0)$ on the $x$-axis. This transforms the problem to finding the pdf of $Z = \sqrt{(X-c)^2+Y^2}$ which at least reduces the clutter a little bit. – Dilip Sarwate Oct 11 '14 at 21:52
  • for distribution of d^2: check here: https://www.google.ca/url?sa=t&rct=j&q=&esrc=s&source=web&cd=1&cad=rja&uact=8&ved=0CB8QFjAA&url=http%3A%2F%2Fwww.researchgate.net%2Fpublication%2F242097952_On_the_Distribution_of_the_Distance_Between_Two_Multivariate_Normally_Distributed_Points%2Flinks%2F00b49523a86fbbe2f1000000&ei=KAuHVMPLJIiuyATh1YDIBw&usg=AFQjCNF1kCKYERX2n_fRqXv2D4YeBmfHDA&sig2=MR2MaaBDOVpoEgRNLo1VOw&bvm=bv.81449611,d.aWw for distribution of d and only for a specific case check here: http://link.springer.com/article/10.1007%2Fs00418-014-1192-3 –  Dec 09 '14 at 15:02
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    A [thread on the distribution of the squared Euclidean distance](http://stats.stackexchange.com/questions/9220) contains closely related material and references. – whuber Dec 09 '14 at 16:34
  • A non-central chi distribution? https://en.wikipedia.org/wiki/Noncentral_chi_distribution – shabbychef Dec 06 '15 at 04:40

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