# Distribution of sum of squared differences between two standard normal variables

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?

• $$T_n$$ 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 (?).

Thanks so much!

• If $X_i=X_1$ for all $i$ then $T_n=0$. Did you forget independence? Dec 18, 2022 at 11:36
• Even with independence of the $X_i$, you do not get a $\chi^2$ distribution. Find $E[T_n]$ and $Var(T_n)$. I suspect these are $2n-2$ and $12n-16$ for $n>1$, when the variance of a $\chi^2$ distribution is double its mean Dec 18, 2022 at 11:49
• With independence of the $X_i$, you will have $X_{i}-X_{i+1} \sim N(0,2)$ and so $(X_{i}-X_{i+1})^2 \sim 2 \chi^2_1$ but $(X_{i}-X_{i+1})^2$ is not independent of $(X_{i-1}-X_i)^2$ Dec 18, 2022 at 11:55
• Thank you for your comments!! Hmmm, I would interprete the sum $T_n$ in this way: $T_n = (X_1 - X_2)^2 + (X_3 - X_4)^2 + ... (X_{n−1}−{X_n})^2$, so each difference is independent from each other. I don't think, that there should be dependent differences. And yes, it is also not clear, if $P(X_i = X_j) = 0$ or not. I would guess so, or at least for 1 difference. I think, the question, that should be answered here, is, what the distribution of the sum of squared N(0,2) would look like. I hope these assumptions are helpful? I added these assumptions in my original post. Dec 18, 2022 at 12:21
• Oh, sorry, I overlooked it. Yes, all $X_i$ are i.i.d. I added it to my original post. Dec 18, 2022 at 15:09

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.