2
$\begingroup$

I have some time series data, $X_t$, and I need to fit a heavy tail distribution to the first difference i.e. $Y_t=X_t-X_{t-1}$. Prior to fit this distribution, I need to test the iid (identical independent distribution) assumption. So far, I check the ACF (autocorrelation function) plot of $Y_t$ to see if there is any serial autocorrelation. This shows no significant autocorrelation. I also applied Ljung–Box test and McLeod Li test to $Y_t$ in order to check independency and conditional heteroscedascity (ARCH), respectively. Neither of these two tests was significant. Hence, there is no evidence of dependency and presence of conditional heteroscedascity in $Y_t$. Moreover, from the plot of $Y_t$, I can conclude that the unconditional variance of $Y_t$ is constant over time. Finally, I performed runs test over $Y_t$ to check for randomness. Based on this test, I couldn’t reject the randomness of $Y_t$.

Question: Under these conditions, can I claim that the $Y_t$'s are iid? If not, what other test should I perform in order to show the iid assumption?

$\endgroup$
2
$\begingroup$

The $Y_t$ are very unlikely to be independent random variables since $Y_t = X_t-X_{t-1}$ and $Y_{t+1} = X_{t+1} - X_t$ both are functions of $X_t$. So you cannot assume independence: the burden of proof is on you to persuade other people by reasoned argument that $Y_t$ and $Y_{t+1}$ are independent. If you want to use some statistical method such as hypothesis testing to provide support (not proof) for your thesis, then you need to set it up so that is the null hypothesis that $Y_t$ and $Y_{t+1}$ are dependent random variables and you need to be able to reject the null definitively. The way you have it, your null hypothesis is that the random variables are independent. Remember that failing to reject your null hypothesis is by no means a persuasive "proof" that your null hypothesis is true. A failure to reject the null is not the same as a whole-hearted embrace of the null.

I will give a proof that the $Y_t$'s are not independent by proving that they are correlated random variables. Let $R_X(t) = \text{cov}(X_{\tau}, X_{t+\tau})$ be the autocovariance function of the $X_t$ stochastic process or time series. Then, the autocovariance function of the $Y_t$ process is $$\begin{align*} R_Y(t)& = \text{cov}(Y_\tau, Y_{t+\tau})\\ &=\text{cov}(X_\tau-X_{\tau-1},X_{t+\tau}-X_{t+\tau-1})\\ &= \text{cov}(X_\tau, X_{t+\tau}) - \text{cov}(X_\tau, X_{t+\tau-1}) - \text{cov}(X_{\tau-1}, X_{t+\tau}) + \text{cov}(X_{\tau-1},X_{t+\tau-1})\\ &= R_X(t) - R_X(t-1) - R_X(t+1) + R_X(t)\\ &= 2R_X(t) - R_X(t-1) - R_X(t+1) \end{align*}$$ that is, the $R_Y$ sequence is the convolution of the $R_X$ sequence and the autocorrelation function of the transformation sequence $h = (1,-1)$ which is $R_h = (-1,2,-1)$. Thus, even if the $X_t$ are an iid sequence so that $R_X(t) = 0$ for all $t\neq 0$, it cannot be that $R_Y(t)=0$ for all $t\neq 0$. At the very least, $R_Y(\pm 1) = -R_X(0) \neq 0$. So, while the $Y_t$'s can be identically distributed, they are not independent. Call them id if you wish, but not iid.


If your tests are not revealing a large correlation at lag $1$, that is fine. You should not be trying to reject the hypothesis that $Y_t$ and $Y_{t-1}$ are independent, but rather to reject the hypothesis that $Y_{t}$ and $Y_{t-1}$ are correlated, and to reject this hypothesis is reasonable only if the correlation at lag $1$ is very close to $0$. "Not large" is not good enough: it should be negligibly small.

$\endgroup$
6
  • $\begingroup$ Thanks for the reply. If your argument is correct, then the ACF plot would have shown a high autocorrelation at lag 1. But I haven't seen such a thing. Also, you have assumed that $cov(X_t,X_{t−1})=...=0$. I don't have that assumption. $\endgroup$
    – Stat
    Nov 20 '12 at 4:03
  • $\begingroup$ I think you mean "accept" instead of "reject" in "You should not be trying to reject the hypothesis that $Y_t$ and $Y_{t−1}$ are independent..." Is that correct? Also, as I said in my question, the ACF plot shows that $Y_t$ are not serially autocorrelated at (5% significant level) or as you said "small enough". But, I didn't claim that they are independent by just looking at the ACF plot. I did 3 more tests, especially Ljung–Box test (en.wikipedia.org/wiki/Ljung%E2%80%93Box_test) with the null hypothesis of independence in a given time series. $\endgroup$
    – Stat
    Nov 21 '12 at 5:39
  • 1
    $\begingroup$ Although your calculations look correct to me, independence of increments $X_{t+1}-X_{t}$ and $X_t-X_{t-1}$ is a common assumption in the construction of many sorts of stochastic processes. This assumption is common in finance for instance, where flexible distributions are typically used for modelling log-returns (see). $\endgroup$
    – user10525
    Nov 21 '12 at 11:24
  • $\begingroup$ What Dilip Sarwate said is correct statistically. But I think, even though I observe small autocorrelation ( i.e. not statistically significant) , it is still valid to consider $Y_t$’s as independent increments. This has been mentioned in the paper "Dependence of increment in time series via large deviations", "The classic way is estimating the autocorrelation function of time series increments. These estimates are rather small and mutually independent in the case of independent increments". $\endgroup$
    – Stat
    Nov 21 '12 at 18:21
  • $\begingroup$ @Procrastinator I am aware of processes with independent increments but if the $Y_t$'s are assumed to be iid so that $\{X_t\}$ is such a process, then $$\begin{align*}X_t&=Y_t+X_{t-1}\\&=Y_t+Y_{t-1}+X_{t-2}\\&= \cdots\\&= \sum_{i=1}^t Y_t\end{align*}$$ has properties $E[X_t]=t\mu_y$ and $\text{var}(X_t)=t\sigma_Y^2$. The OP has made no such claim about the $\{X_t\}$ process, and so unless the observations ${x_t}$ very strongly support the hypothesis that $\text{var}(X_t)$ is increasing linearly with $t$, I would not be inclined to believe any test for independence of the increments. $\endgroup$ Nov 21 '12 at 22:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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