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I'm asked to derive the pdf of the t-distribution following way.

$$V = \frac{Z}{\sqrt{U/n}}$$ where $Z$ is the standard normal and $U$ is the chi-squared distribution with degree of freedom n.

Letting $Y = \sqrt{U/n}$, we can find the pdf of $Y$ to be

$$F_Y(y) = P(\sqrt{U/n} \leq y) = P(U/n \leq y^2) = P(U \leq ny^2) = F_U(ny^2)$$ $$f_Y(y) = 2yn f_U(ny^2)$$

And then using the formula for finding pdf of ratio of random variables, we get

$$f_V(v) = \int^{\infty}_{0} y f_Z(vy) f_Y(y) dy$$

But I cannot get the correct answer trying to integrate the above integral.

Am I right up to this point?

Thanks for your help.

Phil
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  • You forgot the Jacobian term $y$ in the integral. It should be $\int_0^\infty yf_Z(vy)f_Y(y)~dy$. – Shubham Johri Dec 04 '19 at 05:23
  • substitute in the PDF for the chi squared distribution and the standard gaussian, and you will get an expression of the form:

    $C\int_0^{\infty}y^{n}e^{-y^2(v^2+n^2)}dy$. For $n$ even this is $\frac{C}{2}\mathbb{E}[N^n]$ where $N$ is normal with mean 0 and variance $\frac{1}{2(v^2+n^2)}$.substitute in the PDF for the chi squared distribution and the standard gaussian, and you will get an expression of the form:

    $C\int_0^{\infty}y^{n}e^{-y^2(v^2+n^2)}dy$. For $n$ even this is $\frac{C}{2}\mathbb{E}[N^n]$ where $N$ is normal with mean 0 and variance $\frac{1}{2(v^2+n^2)}$.

     – fGDu94 Dec 04 '19 at 07:17
  • https://math.stackexchange.com/questions/474733/derivation-of-the-density-function-of-student-t-distribution-from-this-big-integ, https://math.stackexchange.com/questions/472165/derive-student-t-distribution-using-transformation-theorem – StubbornAtom Dec 04 '19 at 14:19
  • Possible duplicate of Derivation of the density function of student t-distribution from this big integral. – NCh Dec 04 '19 at 23:54

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