I was provided a question which went as follows:
If $Y_1,Y_2,...,Y_n$ is a random sample from $N\left(\mu,\sigma^2\right)$, prove that $\bar{Y}$ is independent of $\sum_{i=1}^{n-1}\left(Y_i-Y_{i+1}\right)^2$
Now, we have $$Y_i\sim N\left(\mu,\sigma^2\right)\\\implies Y_i-Y_{i+1}\sim N\left(0,2\sigma^2\right)\\\implies \frac{\left(Y_i-Y_{i+1}\right)^2}{2\sigma^2}\sim\chi^2_1\\\implies\sum_{i=1}^{n-1}\frac{\left(Y_i-Y_{i+1}\right)^2}{2\sigma^2}\sim\chi^2_{n-1}$$
Further, I found that $E\{\sum_{i=1}^{n-1}\left(Y_i-Y_{i+1}\right)^2\}=2(n-1)\sigma^2$.
Also, we have that $\bar{Y}\sim N\left(\mu,\frac{\sigma^2}{n}\right)$.
Now, to prove that these two are independent, I tried to prove that the covariance between $\bar Y$ and $\sum_{i=1}^{n-1}\left(Y_i-Y_{i+1}\right)^2$ is $0$. Thus, I get that, $$COV\left(\bar Y,\sum_{i=1}^{n-1}\left(Y_i-Y_{i+1}\right)^2\right)=E\left(\bar Y\sum_{i=1}^{n-1}\left(Y_i-Y_{i+1}\right)^2\right)-E\left(\bar Y\right)E\left(\sum_{i=1}^{n-1}\left(Y_i-Y_{i+1}\right)^2\right)\\=E\left(\bar Y\sum_{i=1}^{n-1}\left(Y_i-Y_{i+1}\right)^2\right)-2(n-1)\mu\sigma^2$$.
However, I am stuck here as to how to proceed to prove that $$E\left(\bar Y\sum_{i=1}^{n-1}\left(Y_i-Y_{i+1}\right)^2\right)=2(n-1)\mu\sigma^2$$
A little help would be very great. Also, are there any alternative methods for this problem?
First of all, showing $\overline Y$ and $\sum_{i=1}^{n-1}(Y_i-Y_{i+1})^2$ are uncorrelated does not prove their independence. This is because $(\overline Y,\sum_{i=1}^{n-1}(Y_i-Y_{i+1})^2)$ is not jointly normal.
On the other hand, $(\overline Y,Y_i-Y_{i+1})$ is jointly normal for every $i=1,\ldots,n-1$. Therefore, showing $\overline Y$ and $Y_i-Y_{i+1}$ are uncorrelated would prove the independence of $\overline Y$ and $Y_i-Y_{i+1}$ for every $i=1,\ldots,n-1$. This would imply $\overline Y$ is independent of the vector $(Y_i-Y_{i+1})_{1\le i\le n-1}$, and hence independent of $\sum_{i=1}^{n-1}(Y_i-Y_{i+1})^2$.
To show they are uncorrelated, use the bilinearity of covariance:
$$\operatorname{Cov}(\overline Y,Y_i-Y_{i+1})=\operatorname{Cov}(\overline Y,Y_i)-\operatorname{Cov}(\overline Y,Y_{i+1})$$
and $$\operatorname{Cov}(\overline Y,Y_i)=\operatorname{Cov}\left(\frac1n\sum_{j=1}^n Y_j,Y_i\right)=\frac1n\sum_{j=1}^n \operatorname{Cov}(Y_j,Y_i)$$