Distribution of ratio std Normal and sqrt chi squared

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WWGD
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Hi all,
I am trying to understand two things from a paper
The ratio of a standard normal by the square root of a a chi squared divided by its df ( degrees of freedom) is a t distribution. So
1) What is the dist of square root of Chi squared? I know a normal squared is a chi squared, but a chi squared may not necessarily come about ad the square of a normal
2) Why does the ratio of a standard normal by the square root of a chi squared a t distribution? What result is this?
Only somewhat related result can think of is that ratio of independent standard normals ( of course, nonzero denominator) is a Cauchy. Edit: I wanted to double check the claim that the square root of a chi squared is a chi squared because this does not seem true about the square root of a square normal, which seems should be normal.
Thanks.
 
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andrewkirk
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The distribution of a random variable that is the square root of a chi-squared distribution with k degrees of freedom is the chi distribution with k degrees of freedom. Details are here.
: I wanted to double check the claim that the square root of a chi squared is a chi squared
the claim is incorrect. It's not another chi-squared, it's a chi.
2) Why does the ratio of a standard normal by the square root of a chi squared a t distribution?
It depends how one defines a t-distribution - as a ratio of a standard normal to a chi, or by its pdf or cdf. When I learned it, it was defined as the ratio, so a standard normal divided by a chi is a t-distribution by definition. One then has to do the work to derive the pdf and cdf from that definition.

Alternatively, if one defines a t-distribution by its pdf or cdf, one has to do the work to prove that a RV that is the ratio of a std normal to a chi has that pdf or cdf.
 
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WWGD
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The distribution of a random variable that is the square root of a chi-squared distribution with k degrees of freedom is the chi distribution with k degrees of freedom. Details are here.
the claim is incorrect. It's not another chi-squared, it's a chi.
It depends how one defines a t-distribution - as a ratio of a standard normal to a chi, or by its pdf or cdf. When I learned it, it was defined as the ratio, so a standard normal divided by a chi is a t-distribution by definition. One then has to do the work to derive the pdf and cdf from that definition.

Alternatively, if one defines a t-distribution by its pdf or cdf, one has to do the work to prove that a RV that is the ratio of a std normal to a chi has that pdf or cdf.
Thanks, so a chi can coincide with a normal? I know there are issues with the square root not being an injection ( or not bijection once we choose one of the roots). I guess this is why the square root of a squared normal is not a normal? Or do we just consider the composition to be an absolute value?
 
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WWGD
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I understand it comes down to a change of variable and using the Jacobian but I am writing on my phone since my PC is down.
 
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andrewkirk
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I know there are issues with the square root not being an injection ( or not bijection once we choose one of the roots). I guess this is why the square root of a squared normal is not a normal?
Yes, that's the reason. See the blue line in this graph, which is the chi distribution with one degree of freedom, which is the +ve sqrt of a squared std normal RV. It looks just like half a bell curve, and it is. If you insert ##k=1## in the formula given for the pdf you get double the pdf of a standard normal.
 

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