How to prove that when X is from Normal Distribution and Y is from Chi-square Distribution with parameter f and X,Y are independent then X/sqrt(Y/f) is from t distribution with parameter t?
I got this far that I have to solve a quite hard integral and then gave up. Has anyone solved it?
Unfortunately this can't be shown by considering moment generating functions - often one of the easiest ways to demonstrate relationships between distributions - as the $t$ distribution has no MGF.
But so long as you can spot the kernel of the integral using standard PDFs (a very handy trick!) it is fairly easy to show that that $T=\frac{X}{\sqrt(Y/f)}$ has the required PDF:
$f_{T}(t)=\frac{1}{\sqrt{f\pi}}\cdot\frac{\Gamma(\frac{f+1}{2})}{\Gamma(\frac{f}{2})}\cdot(1+\frac{t^{2}}{f})^{-\frac{f+1}{2}}$ on support $(-\infty,\infty)$
Proof: $Y\sim\chi_{f}^{2}$ has $f_{Y}(y)=\frac{y^{\frac{f}{2}-1}e^{-\frac{y}{2}}}{2^{\frac{f}{2}}\Gamma(\frac{f}{2})}$ on support $[0,\infty)$ and $X\sim N(0,\,1)$ has $f_{X}(x)=\frac{1}{\sqrt{2\pi}}e^{-\frac{x^{2}}{2}}$ on support $(-\infty,\infty)$. By independence, their joint PDF is $f_{Y,X}(y,x)=\frac{y^{\frac{f}{2}-1}e^{-\frac{y}{2}}}{2^{\frac{f}{2}}\Gamma(\frac{f}{2})}\cdot\frac{1}{\sqrt{2\pi}}e^{-\frac{z^{2}}{2}}$for $y\geq0$, $x\in\mathbb{R}$. Define the transformed random variables $W=\frac{Y}{2}$ (so $W\geq0$) and $T=\frac{X}{\sqrt{Y/f}}$ (so $T$ can take all real values and we aim to show $T\sim t_{f}$). Then writing $Y$ and $X$ in terms of $W$ and $T$ gives $Y=2W$ and $X=T\sqrt{\frac{2W}{f}}$ so $\frac{\partial y}{\partial w}=2$, $\frac{\partial y}{\partial t}=0$ and $\frac{\partial x}{\partial t}=\sqrt{\frac{2w}{f}}$. The Jacobian is $J(w,t)=\begin{vmatrix}\frac{\partial y}{\partial w} & \frac{\partial y}{\partial t}\\ \frac{\partial x}{\partial w} & \frac{\partial x}{\partial t} \end{vmatrix}=2\sqrt{\frac{2w}{f}}$. The joint PDF of $W$ and $T$, on its support $w\geq0$, is:
$\begin{align*} f_{W,T}(w,t) & =f_{Y,X}(y(w,t),x(w,t))\mid J(w,t)\mid\\ & =\frac{(2w)^{\frac{f}{2}-1}e^{-w}}{2^{\frac{f}{2}}\Gamma(\frac{f}{2})}\cdot\frac{1}{\sqrt{2\pi}}e^{-\frac{t^{2}(2w)}{2f}}\cdot2\sqrt{\frac{2w}{f}}\\ & =\frac{\sqrt{2w}\cdot w^{\frac{f}{2}-1}}{\Gamma(\frac{f}{2})\sqrt{2f\pi}}\cdot\frac{2^{\frac{f}{2}-1}\cdot2}{2^{\frac{f}{2}}}\cdot e^{-w-\frac{t^{2}w}{f}}\\ & =\frac{1}{\Gamma(\frac{f}{2})\sqrt{f\pi}}\cdot w^{\frac{f}{2}-\frac{1}{2}}e^{-(1+\frac{t^{2}}{f})w} \end{align*}$
The marginal PDF of $T$ is then obtained by $f_{T}(t)=\int_{0}^{\infty}f_{W,T}(w,t)\,\mathrm{d}w$. The kernel of the integrand is $w^{\frac{f}{2}-1}e^{-(1+\frac{t^{2}}{f})w}$, recognisably the kernel of a Gamma PDF. Using $\int_{0}^{\infty}t^{n-1}e^{-\lambda n}\mathrm{d}t=\Gamma(n)\lambda^{-n}$ with $\lambda=1+\frac{t^{2}}{f}$ and $n=\frac{f}{2}+\frac{1}{2}$ gives $f_{T}(t)=\frac{1}{\sqrt{f\pi}}\cdot\frac{\Gamma(\frac{f+1}{2})}{\Gamma(\frac{f}{2})}\cdot(1+\frac{t^{2}}{f})^{-\frac{f+1}{2}}$ as required.
Some texts put this the other way round: they use what you wrote as the definition of the $t$ distribution and then use this method to obtain the PDF!