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We can relate the student-t and beta distributions as such:

If $X$ has a Student's t-distribution with degree of freedom $\nu$ then one can obtain a Beta distribution:

$$\frac{\nu}{\nu + X^2} \sim \text{B}(1/2, \nu/2)$$

If our goal is to compute the CDF of the student-t at a point $x$, how would we do this using the beta CDF?

For example, I can compute the CDF of the beta distribution given by the incomplete beta function

$$I_{x=0.4}(a=1/2, b=1/2) = 0.43590578$$

I guess in this case $b=1/2 \implies \nu = 1$. But for the incomplete beta, we need $0\leq x \leq 1$. So how would we evaluate the CDF of the student-t if, say if $x=-1$ and $\nu=2.4$ ?

Using the transformation above, I didn't get matching results

$$z = \frac{\nu}{\nu + X^2} = \frac{2.4}{2.4 + (-1)(-1)} = 0.705882352941$$ $$\text{student-t CDF}(\nu=2.4, x=-1) = 0.203620621107$$ $$\neq$$

$$I_{z=0.705882352941}(a=1/2, b=\nu/2=1.2)=0.8826576230306$$

PyRsquared
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  • The answer from which you obtained this quotation addresses your question in the most explicit way possible: by providing (what is intended to be) simple code. Please notice how it treats negative and positive XXX differently. The point, of course, is that the Student t distribution is symmetric about 0,0,0, which means F(−x)=1−F(x)F(−x)=1−F(x)F(-x)=1-F(x) for any x.x.x. – whuber Mar 02 '19 at 17:28

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