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I haven't worked with time-series in a while now and stumbled upon them in a different setting. Given $X_t\sim\mathcal{N}(0,\sigma^2)$ for $t=1,\ldots,n$ and the process $Y_t$ for $t=1,\ldots,n-1$ defined by: $$ Y_t = X_{t+1} - X_t$$. Imagine I only observe $Y_t$ and now want to estimate $\sigma^2$ from these observations. It seems $Var(Y)=2\sigma^2$, but I have problems deriving this theoretically. I want to be able to show that the properties of a consistent estimator for the variance of $X$ also work for $Y$, especially the Mean Absolute Deviation $MAD$. $Y_t$ can be interpreted as a $MA(1)$ process with known autocovariance function, but I have problems deriving this result.

Any help or hint how to assess this would be greatly appreciated.

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I assume that $X_t$ are iid normal distributed with mean $0$ and variance $\sigma^2$. $Var(Y_t) = Var(X_{t+1}-X_t) = Var(X_{t+1})+Var(X_t)-2Cov(X_{t+1},X_{t}) \\= Var(X_{t+1})+Var(X_t) = \sigma^2 + \sigma^2 - 0 = 2\sigma^2$.

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  • $\begingroup$ Hey, thank you very much for the reply. I considered the same, but was wondering the following: Is the sample variance still an unbiased estimator then, despite the dependence of the $Y_t$ ? $\endgroup$ – Momo May 20 '16 at 18:17
  • $\begingroup$ Sure. Let's have a little diffrent look at your question: Assuming $X_t \stackrel{iid}{\sim} N(0,\sigma^2)$. This means $Y_t \stackrel{iid}{\sim} N(a,b)$ with $a=E(X_t-X_{t-1})=0$ and $b=Var(X_t-X_{t-1})=\sigma^2+\sigma^2$. Now it should be obvious. BTW: With the sample variance you get an estimator for the variance of $Y_t$, not $X_t$ $\endgroup$ – dan May 21 '16 at 11:57
  • $\begingroup$ But for $Y_t$ constructed the described value, the assumption that $Y_t$ is iid does not hold, as for each $t\in{2,\ldots,n-2}$ the conditional properties $\mathbb{P}(Y_t\leq x|Y_{t-1}) \neq \mathbb{P}(Y_t\leq x)$ $\mathbb{P}(Y_t\leq x|Y_{t+1}) \neq \mathbb{P}(Y_t\leq x)$. I'm confused on wether this has an effect on the unbiasednes of the estimator of variance of $Y_t$. If we had independent realizations of $Y_t$, I would not have an issue of that. One thing I thought about was deriving estimator properties from a partition of the $Y_t$ like in $\endgroup$ – Momo May 21 '16 at 12:27
  • $\begingroup$ W Stadje (1984) A note on sample means and variances of dependent normal variables, Series Statistics, 15:2, 205-218, DOI: 10.1080/02331888408801759 $\endgroup$ – Momo May 21 '16 at 12:30
  • $\begingroup$ Right, my inattention, the $iid$ shouldn't be there, because the autocorrelation function $\rho(k) = .5$ with $k=|1|$ and $0$ when $|k|>=2$. Let $\gamma(k):=E(Y_t\cdot Y_{t+k})$ be the autocovariance function ($Y_t$ is weakly stationary with mean $0$) and let $(Y_1,\dots,Y_N)$ be your sampled vector. Then $\hat{\gamma}(k)=\frac{1}{N}\sum_{t=1}^{N-k}(Y_t-\bar{Y})(Y_{t+k}-\bar{Y})$ with $k\geq 0$. The bias of this estimator converges to $0$ when $N$ goes to infinity: According to my notes you have $E(\hat{\gamma}(k))=\gamma(k)-\frac{k}{n}\gamma(k)$. For $k=0$ you are unbiased. $\endgroup$ – dan May 21 '16 at 13:54

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