Probability Density Function of |X|^(1/2)+|Y|^(1/2)+|Z|^(1/2) with N(0,1)

[forumfann]

Could anyone help me figure out the the probability density function (pdf) of |X|^(1/2)+|Y|^(1/2)+|Z|^(1/2) if X, Y and Z are distributed normally with mean 0 and variance 1, N(0,1) ?

Thanks in advance.

 

[statdad]

Are your variables independent? If so, first work out the distribution of $$|X|^{1/2}$$. The other distributions will be identical and you can use standard procedures to find the distribution of their sum.

 

[forumfann]

Yes, the variables are independent. But what are the standard procedures? It there a easier way to get the pdf if one has more random variables than three?

 

[flatmaster]

If X, Y, and Z, are distributed normally, just make them a normal (gausian) PDF of a random variable.

X=Ae^(ax). You're lucky that you can use little x,y, and z for your random variables to match up with their distributions.

 

[statdad]

"X=Ae^(ax). " That isn't the form of a normal distribution.

Okay, suppose $$X \sim n(0,1)$$. Think this way.

1) You should be able to write down the distribution of $$|X|$$ - it's a pretty
standard result, and if you're working on this problem I'm guessing you know this.
2) Use a standard transformation (if [tex]W = |X|, find the distribution of square root of W). This gives the distribution of $$|X|^{1/2}$$.<br /> 3) Since $$X, Y$$ and $$Z$$ are i.i.d, the same is true for <br /> <br /> $$|X|^{1/2} + |Y|^{1/2} + |Z|^{1/2}$$<br /> <br /> so the distribution of their sum should be relatively easy to obtain.[/tex]