# Distribution of ratio std Normal and sqrt chi squared

• I
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## Main Question or Discussion Point

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.

• 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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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.

andrewkirk
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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