1
$\begingroup$

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.

Improve this question
$\endgroup$

1 Answer 1

Reset to default
3
$\begingroup$

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

Improve this answer
$\endgroup$
5
  • $\begingroup$ 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
    Commented May 20, 2016 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
    Commented May 21, 2016 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
    Commented May 21, 2016 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
    Commented May 21, 2016 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
    Commented May 21, 2016 at 13:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.