Distribution of sum of squared differences between two standard normal variables

Question

The following question should be answered:

Is the distribution of $\, T_n = (X_1 - X_2)^2 + ... + (X_{n-1}-X_n)^2$, (with $X_i \sim N(0,1) \,i.i.d.)$, $\chi^2$-distributed?

For clarity, I made the following assumptions (see comments below):

• $T_n$ has only independent differences, meaning $T_n = (X_1 - X_2)^2 + (X_3 - X_4)^2 + (X_5 - X_6)^2 + ... + (X_{n-1} - X_n)^2$.
• $P(X_i = X_j) = 0$

The additive property of $\chi^2$-distributed variables, if needed, is known (this was to prove in the first part of the question).

Following this post I started with the first part, and proved that the difference of $X_1-X_2 \,$ is the same as $X_1 + aX_2$ with $\, a=-1$ and therefore N(0,2) distributed (using the characteristic function). The same follows for all other differences of $T_n$.

I struggle with the next steps, proving the distribution of $(X_1-X_2)^2$ (and all other squared differences) and finally the distribution of $T_n$. In our class it was stated, that the distribution is not really $\chi_2$-distributed, but these are stacked $\chi_2$-squares (?).

Can somebody help please?

Thanks so much!

Answer 1

If $\ X_i\ $ are independent standard normal variates then it's automatically true that $$ P(X_i=X_j)=\cases{1&if $\ i=j$\\ 0&if $\ i\ne j\ $,} $$ so your assumption that $\ P(X_i=X_j)=0\ $ for $\ i\ne j\ $ is redundant.

If $$ Q_r=T_{2r}=\sum_{i=1}^r(X_{2i-1}-X_{2i})^2\ , $$ then $\ Q_r\ $ is the sum of the squares of $\ r\ $ independent random variables $\ Y_i=X_{2i-1}-X_{2i}\sim N(0,2)\ $. Therefore $\ \frac{Q_r}{2}\ $ is the sum of the squares of $\ r\ $ independent standard normal variates, $\ \frac{Y_i}{\sqrt{2}}\ $. Thus, while $\ Q_r\ $ isn't, strictly speaking, $\ \chi^2\ $ distributed, $\ \frac{Q_r}{2}\ $ follows a chi-square distribution with $\ r\ $ degrees of freedom.