Question

We define a process $\chi_k^n=\sum _{i=1}^k x_i^n$ where x_i are iid gaussian processes. I try to find the distribution of $\chi_k^n$. If k=1 then we get $f(x^n=y)=\frac1n y^{\frac{1-n}{n}}\exp(-y^{2/n}/2)$ and then try to do convolution but then to say something about the integral $$C_{n}\int_{-\infty}^{\infty}\frac1n (l^2-h^2)^{\frac{1-n}{n}}\exp(-(l+h)^{2/n}/2)\exp(-(l+h)^{2/n}/2)\exp(-(l-h)^{2/n}/2)dh $$ What can I do n>2????????

Can I say somthing about the density, formula or upper bound?

Thanks.

Answer 1

Firstly you forgot to multiply the density $f(x^n=y)$ by $1/\sqrt{2\pi}$. I think if you obtained the density of the random variable $X_{1,2}=X_1^n+X_2^n$ by the convolution method, the problem no more posed , because for $X_1^n+X_2^n+X_3^n=X^n_{1,2}+X_3^n=X^n_{1,2,3}$, and you have the density of $X^n_{1,2}$, the density of $X^n_{3}$, you can calculate there convolution, i.e the density of $X^n_{1,2,3}$. If the calculation is very difficult with the convolution (I think) you can use the characteristic function. You calculate the function characteristic of the variable $X_1^n$ that one noted $\psi_{X_1^n}(t)$. As the two variables $X_1^n$ and $X_2^n$ are i.i.d, then $\psi_{X_1^n+X_2^n}(t)=\psi_{X_1^n}(t)\cdot \psi_{X_2^n}(t)=(\psi_{X_1^n}(t))^2$ and so on for variable $X_1^n+X_2^n+...+X_k^n$ we will have $\psi_{X_1^n+X_2^n+\ldots +X_k^n}(t)=(\psi_{X_1^n}(t))^k$. Just well calculate $\psi_{X_1^n}(t)$.