IID assumption for $Y_t=X_t-X_{t-1}$

Question

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?

Answer 1

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.

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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.