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Is there a quick way to calculate this?

$$\operatorname{Var}\left[\frac{1}{n-1} \sum_{i=1}^{n-1} (X_{i+1} - X_{i})^2\right]$$

$X_{i+1}$ and $X_{i}$ ~ $N(\mu, \sigma^2)$

I know that for $\operatorname{Var}[S^2]$ where $S^2$ is the sample variance we can use chi-square distribution. Is it possible to count it that way in this case?

vitamin d
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Mia
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2 Answers2

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The difference $Y_i \sim X_{i+1}-X_i$ has a normal distribution with mean $0$ and variance $2\sigma^2$. Then let $Y_i^2 \sim (X_{i+1}-X_i)^2$. But $Y_i^2$ is not standard normal such that we can use $Y_i^2 \sim \chi^2(1)$. We have to adjust the variance.

Note that for a random variable A which has variance $Var(A)$, $Var(cA) = c^2Var(A)=1\implies c = \frac{1}{\sqrt{Var(A)}}$ for $Var(cA)=1$.

In our case $c = \frac{1}{\sigma\sqrt{2}}$ if $Var(A) = 2\sigma^2$.

In other words, if $c = \frac{1}{\sigma\sqrt{2}}$, then $Var(cY_i)=Var(\frac{Y_i}{\sigma\sqrt{2}}) = 1$. Then $\left(\frac{Y_i}{\sigma\sqrt{2}}\right)^2$ has a $\chi^2(1)$ distribution. Further, $\sum\limits_{i=1}^{n-1} \left(\frac{Y_i}{\sigma\sqrt{2}}\right)^2$ has a $\chi^2(n-1)$ distribution.

Therefore $\sum_{i=1}^{n-1}\left(\frac{X_{i+1} - X_{i}}{\sigma\sqrt{2}}\right)^2$ has a $\chi^2(n-1)$ distribution.

Now variance \begin{align*} Var\left[\frac{1}{n-1}\sum_{i=1}^{n-1}\left(X_{i+1} - X_{i}\right)^2\right] &= Var\left[\frac{(\sigma\sqrt{2})^2}{n-1}\sum_{i=1}^{n-1}\left(\frac{X_{i+1} - X_{i}}{\sigma\sqrt{2}}\right)^2\right] \\ &= \frac{4\sigma^4}{(n-1)^2}Var\left[\sum_{i=1}^{n-1}\frac{ (X_{i+1} - X_{i})^2}{2\sigma^2}\right] \\ &= \frac{4\sigma^4}{(n-1)^2}Var(\chi^2(n-1)) \\ &= \frac{4\sigma^4}{(n-1)^2}2(n-1) \\ &= \frac{8\sigma^4}{(n-1)} \\ \end{align*}

Therefore:

$$\operatorname{Var}\left[\frac{1}{n-1} \sum_{i=1}^{n-1} (X_{i+1} - X_{i})^2\right] = \frac{8\sigma^4}{(n-1)}$$

EDIT: Corrected calculation error in $\sigma^2$ as per Tommik's answer.

Rahul Madhavan
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  • I forgot to add that we have a sample with a normal distribution $\mathbb{X} = (X_{1} , \cdots , X_{n}) $. So is this solution correct in this case too? Are the variables independent? – Mia Apr 19 '21 at 20:06
  • Yes it works only when ${X_1,\dots,X_n}$ are i.i.d with the same distribution $\sim \mathcal N(\mu,\sigma^2)$. If on the other hand they are dependent on each other, then we need the distribution of $X_i-X_{i-1}$ – Rahul Madhavan Apr 20 '21 at 07:35
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Your solution can be nice only if we assume uncorrelation between $(X_{i+1};X_i)$ but anyway there is an evident error in you calculations

$$\frac{X_{i+1}-X_i}{\sigma\sqrt{2}}\sim N(0;1)$$

thus the requested variance is

$$\frac{1}{(n-1)^2}\mathbb{V}\left[2\sigma^2 \sum_{i=1}^{n-1}\left(\frac{Y_i}{\sigma\sqrt{2}}\right) ^2 \right]=$$

$$=\frac{4\sigma^4}{(n-1)^2}\cdot\mathbb{V}\left[\chi_{(n-1)}^2\right]=\frac{4\sigma^4}{(n-1)^2}\cdot2(n-1)=\frac{8\sigma^4}{n-1}$$

tommik
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