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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!

m09s19
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    If $X_i=X_1$ for all $i$ then $T_n=0$. Did you forget independence? – Kavi Rama Murthy Dec 18 '22 at 11:36
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    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 – Henry Dec 18 '22 at 11:49
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    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 $(X_{i-1}-X_i)^2$ – Henry Dec 18 '22 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. – m09s19 Dec 18 '22 at 12:21
  • Oh, sorry, I overlooked it. Yes, all $X_i$ are i.i.d. I added it to my original post. – m09s19 Dec 18 '22 at 15:09
  • Why do you assume $\ (X_1-X_2)^2+\dots+(X_{n-1}-X_n)^2\ $ means $\ (X_1-X_2)^2+(X_3-X_4)^2+\dots+(X_{n-1}-X_n)^2\ $ (where $\ n\ $ is forced to be even) rather than $\ (X_1-X_2)^2+$$,(X_2-X_3)^2+$$,\dots$$\ +(X_{n-1}-X_n)^2\ $, which seems to me to be a more natural interpretation? – lonza leggiera Dec 19 '22 at 01:20

1 Answers1

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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.

lonza leggiera
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