Distribution of distances from the origin of randomly generated points

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Robin04
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Homework Statement
We randomly generate points in 4 dimensional Euclidean space. The expecte value ##\mu## of the coordinates is 0 and the standard deviation is ##\sigma = 2.5##. Their distribution is normal.
What's the distribution of the distance of these points from the origin?
Relevant Equations
Density of the normal distribution: ##\rho (x)=\frac{1}{\sqrt{2 \pi \sigma}}e^{-\frac{(x-\mu)^2}{2\sigma^2}}##
I'm not really sure how to do this. Maybe somehow I should transform the density function. Can you give me a hint?
 
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How do you write the radial distance from the origin in terms of the 4 coordinates?
 
marcusl said:
How do you write the radial distance from the origin in terms of the 4 coordinates?
##d=\sqrt{x_1^2+x_2^2+x_3^2+x_4^2}##
 
Right. You know that each variable is IID random with the same normal distribution. What distribution applies? (You may have seen the more usual case for 2 dimensions.)
 
marcusl said:
Right. You know that each variable is IID random with the same normal distribution. What distribution applies? (You may have seen the more usual case for 2 dimensions.)
I'm not sure what you mean by that. I think I haven't seen the problem for 2 dims.
 
Hmm, we at PF are supposed to guide you to the answer without giving it outright, but I'm not sure what to do here...
The distribution you're looking for is related to chi-squared. Have you seen that?
 
marcusl said:
Hmm, we at PF are supposed to guide you to the answer without giving it outright, but I'm not sure what to do here...
The distribution you're looking for is related to chi-squared. Have you seen that?
Oh yes, we learned an equation for that: ##\rho_n (x) = \frac{1}{\Gamma (n/2) 2^{n/2}}x^{\frac{n}{2}-1}e^{-\frac{x}{2}}##
In my case n would be 4. And I also have to transform this as I need the square root, right?
 
Well, this is a starting point, it's the distribution for [itex]d^2[/itex]. You want a related distribution that has the name "chi" in it.
 
marcusl said:
That's it--a chi distribution.
Thank you very much! Sorry for my clumsiness, I need to clean my head about this topic :D
 
Just to repeat, the sum of squares of normals is Chi-squared -distributed.
 
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Oh, but where do ##\sigma## and ##\mu## come into the game?
 
WWGD said:
Just to repeat, the sum of squares of normals is Chi-squared -distributed.
But the problem is looking for the square-root of the sum of squares, which has a Chi distribution.
 
marcusl said:
But the problem is looking for the square-root of the sum of squares, which has a Chi distribution.
You're right, I should have completed it. I was trying to do it step-by-step, but I did not finish--my bad:
Start with indepent ID Normals. The sum of their squares is Chi-square-distributed. The square root in previous step is Chi-distributed.
 
Robin04 said:
Oh, but where do ##\sigma## and ##\mu## come into the game?
Take a look at the first equation in the Wikipedia article and you'll see that the pdf is for normalized, zero-mean random variables (just as is the pdf you quoted for Chi-squared in post #7). You need to "unnormalize" the variables and you'll find sigma shows up in the pdf expression.
 
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