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First I want to prove that the sum $Y_1+...+Y_n$ where $Y_i=X_i^2$ and $X$ is standard normally distributed has density $f_n(x)=c_n x^{n/2-1}e^{-x/2}1_{x>0}$

I do not want to derive it, I would like to prove it by induction. For $n=1$ and $n=2$ I derived the densities, so this case is done. For the step $n\rightarrow n+1$ I think there is not much to do: $f_{n+1}(x)=c_{n+1}x^{\frac{n+1}{2}-1}e^{x/2}=c*c_n x^{n/2+n/2-1}e^{-x/2}=c c_n x^{n/2-1}e^{-x/2}=c f_n$ so this should also be done.

Now I would like to use my formula for $f_n$ to derive the density (=student t-distribution) from the random variable $T=\frac{X_0}{\sqrt{(X_1^2+...+X_n^2)/n}}$, it should be $f_T(t)=k_n(\frac{1}{1+t^2/n})^{(n+1)/2}$ but I do not see how this can be done. I think we need to start somehow like $F_T(t)=P(T\le t)=P(\frac{X_0}{\sqrt{(X_1^2+...+X_n^2)/n}}\le t)=...$

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You are missing the independence of the $X_i$ in your formulation. Also, how are you getting $f_{n+1}(x)$? You need to convolve $f_n(x)$ with $f_1(x)$, but you seem to be multiplying densities? –  Dilip Sarwate Jun 2 '13 at 23:29
    
You mean $f_{n+1}=\int_0^x f_{Y_{n+1}}(x-y)f_{Y_1+...+Y_n}(y) dy$ I do not know how to go on with this. –  Alexander Jun 2 '13 at 23:50
    
@Alexander: When you write the convolution integral the product of the exponential terms becomes independent of the integral variable, and only a monomial remains in the integral. Computing the integral is straightforward then. –  S.B. Jun 3 '13 at 0:06
    
@Alexander What S.B. said is nearly but not quite right. Specifically, the exponential terms do become independent of the variable of integration, but you don't get just a monomial to integrate. For the details of the calculation for general gamma random variables, see this answer of mine. –  Dilip Sarwate Jun 3 '13 at 1:34
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1 Answer 1

For induction you need to show that $f_{n+1} = f_1 * f_n$ where $*$ is the convolution. As for the student-$t$ distribution write: $$P(\frac{X_0}{\sqrt{\frac{1}{n}\sum_{i=1}^nX_i ^2}}\leq t)=\mathbb{E}\left[\Phi\left(t\sqrt{\frac{1}{n}\sum_{i=1}^nX_i ^2}\right)\right],$$ where $\Phi$ is the normal CDF. Take the derivative with respect to $t$ and you'll have $$f(t)=\mathbb{E}\left[\phi\left(t\sqrt{\frac{1}{n}\sum_{i=1}^nX_i ^2}\right)\sqrt{\frac{1}{n}\sum_{i=1}^nX_i ^2}\right]\\ =\frac{1}{\left(2\pi\right)^{n/2}}\int_\mathbb{R^n}\phi\left(\sqrt{t^2+n}\sqrt{\frac{1}{n}\sum_{i=1}^nX_i ^2}\right)\sqrt{\frac{1}{n}\sum_{i=1}^nX_i ^2}\mathrm{d}X_1\ldots\mathrm{d}X_n\\=\frac{1}{\left(2\pi\right)^{n/2}}\times\frac{1}{\left(1+t^2/n\right)^\frac{n+1}{2}}\int_\mathbb{R^n}\phi\left(\sqrt{\sum_{i=1}^n\hat{X}_i ^2}\right)\sqrt{\sum_{i=1}^n\hat{X}_i ^2}\mathrm{d}\hat{X}_1\ldots\mathrm{d}\hat{X}_n,$$ with $\phi(z)=\exp(-z^2/2)/\sqrt{2\pi}$ being the normal PDF. The last integral is independent of $t$ so you got what you needed. (The value of the integral is the mean of a $\chi$-distributed random variable of ordern $n$.)

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