$X_1$, $X_2$, . . . are iid logistic random variables with mean $0$
and variance $2$, and
$$T_n=\frac{X_1+···+X_n}{\sqrt{X_1^2+···+X_n^2}}$$Consider the
sequence $T_1$, $T_2$, . . . and give the pmf or pdf of the limiting
distribution.
I have that for a logistic random variable $X$
$$\mathsf{Var}(X)=\frac{\pi^2\beta^2}{3}$$
and so $\beta=\sqrt{\frac{6}{\pi^2}}$ giving that $X\sim Logistic\left(\mu=0,\beta=\sqrt{\frac{6}{\pi^2}}\right)$ with pdf
$$f_X(x)=\frac{1}{\sqrt{\frac{6}{\pi^2}}}\frac{e^\frac{-x}{\sqrt{\frac{6}{\pi^2}}}}{\left(1+e^\frac{-x}{\sqrt{\frac{6}{\pi^2}}}\right)^2}, -\infty\lt x\lt\infty$$
I think it may be useful to note that by symmetry of $X$ around $0$, we have $$\lim_{n\rightarrow\infty}\sum_{i=1}^n X_i=0$$
My intuition tells me this will converge to a degenerate random variable for which $$P(T=0)=1$$
but I'm not sure how to go about formalizing this.
Best Answer
After reviewing @Did's helpful comment:
We have that by CLT $$\sqrt{n}\left(\frac{\sum{X_n}}{n}-0\right)\stackrel{d}{\rightarrow}\mathsf{N}(0,2)$$
Hence
$$\frac{1}{\sqrt{n}}(X_1+...+X_n) \stackrel{d}{\rightarrow}\mathsf{N}(0,2)$$
We also have that
$$\mathsf E(X^2)=\mathsf{Var}(X)+\mathsf E(X)^2=2$$
Thus by the law of large numbers
$$V_n\stackrel{p}{\rightarrow}2$$
and hence
$$\sqrt{V_n}\stackrel{p}{\rightarrow}\sqrt{2}$$
By Skutsky's Theorem we get that
$$\frac{U_n}{\sqrt{V_n}}\stackrel{p}{\rightarrow}\frac{1}{\sqrt{2}}\mathsf N(0,2)=\mathsf N(0,1)$$
Thus the pdf of the limiting distribution is
$$f_T(t)=\frac{1}{\sqrt{2\pi}}\mathsf{exp}\left(-\frac{t^2}{2}\right), -\infty\lt t\lt\infty$$