How do you express the variogram $\gamma(u)$ in terms of correlation for a stationary process?

Question

The Analysis of Longitudinal Data textbook by Diggle et al. (2002) mentioned twice (p48 f. and then on p82) that given the following definition of the variogram, \begin{equation} \gamma(u) = \frac{1}{2}\mathrm{E}[\{Y(t) - Y(t-u)\}^2], \quad u \geq 0 \end{equation} for some stationary stochastic process $\{Y(t)\}$ and $\mathrm{Var}(Y(t)) = \sigma^2$, we can rewrite the variogram in terms of some correlation function $\rho(Y(t), Y(t-u)) := \rho(u)$: \begin{equation} \gamma(u) = \sigma^2 \{1 - \rho(u)\}. \end{equation}

How is the second equation derived from the first?

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My answer (following @whuber's answer):

First, expand the definition of the variogram,\begin{align} \gamma(u) &= \frac{1}{2}\mathrm{E}[\{Y(t) - Y(t-u)\}^2] \\ &= \frac{1}{2}\mathrm{E}[Y^2(t)] - \mathrm{E}[Y(t)Y(t-u)] + \frac{1}{2}\mathrm{E}[Y^2(t-u)]. \end{align}

Then, I use $\mathrm{Var}(X) = \mathrm{E}[X^2] - \mathrm{E}[X]^2 \implies \mathrm{E}[X^2] = \mathrm{Var}(X)+\mathrm{E}[X]^2$ to continue, writing $\mathrm{E}[Y(t)] =: \mu(t)$,\begin{align} \gamma(u) &= \frac{1}{2}\sigma^2 + \frac{1}{2}\mu^2(t) - \mathrm{E}[Y(t)Y(t-u)] + \frac{1}{2}\sigma^2 + \frac{1}{2}\mu^2(t-u) \\ &= \sigma^2 - \mathrm{E}[Y(t)Y(t-u)] + \frac{1}{2}\mu^2(t) + \frac{1}{2}\mu^2(t-u). \end{align}

This gives the $\sigma^2$ part. To try to work in the covariance, \begin{align} \mathrm{Cov}\{Y(t), Y(t-u)\} &= \mathrm{E}[\{Y(t)-\mu(t)\}\{Y(t-u)-\mu(t-u)\}] \\ &= \mathrm{E}[Y(t)Y(t-u)] - \mu(t)\mu(t-u) \\ &= \sigma^2 \rho(u). \end{align}

Fit this expression into that last expression for the variogram, \begin{align} \gamma(u) &= \sigma^2 - \mathrm{E}[Y(t)Y(t-u)] + \mu(t)\mu(t-u) - \mu(t)\mu(t-u) + \frac{1}{2}\mu^2(t) + \frac{1}{2}\mu^2(t-u) \\ &= \sigma^2 - \sigma^2\rho(u) - \mu(t)\mu(t-u) + \frac{1}{2}\mu^2(t) + \frac{1}{2}\mu^2(t-u) \\ &= \sigma^2 - \sigma^2\rho(u) + \frac{1}{2}\{\mu(t)-\mu(t-u)\}^2. \end{align}

Now, since $\{Y(t)\}$ is a stationary process, $\mu(t) = \mu(t-u)$, \begin{align} \gamma(u) &= \sigma^2 - \sigma^2\rho(u) + \frac{1}{2}\{\mu(t)-\mu(t-u)\}^2 \\ &= \sigma^2 - \sigma^2\rho(u) + 0 \\ &= \sigma^2\{1 - \rho(u)\}. \end{align}

Answer 1

Stationarity is a bit of a distraction because this is a simple property of any two random variables $Y(t)$ and $Y(t-u)$ having common (finite) variances $\sigma^2$ and zero expectations.

Fix $t$ and $u.$ Because $E[Y(t)]=E[Y(t-u)]=0$ and $E\left[Y^2(t)\right] = E\left[Y^2(t-u)\right] = \sigma^2,$

$$\operatorname{Cov}(Y(t), Y(t-u)) = E[Y(t)Y(t-u)] = \sigma^2 \rho(Y(t),Y(t-u)).$$

This is an immediate consequence of the definitions of covariance, correlation, and variance.

The linearity property of expectation implies

$$\begin{aligned} 2\gamma(t,u) &= E\left[(Y(t)-Y(t-u))^2\right]\\ &= E\left[Y^2(t)\right] + E\left[Y^2(t-u)\right] - 2 E\left[Y(t)Y(t-u)\right]\\ &= 2\sigma^2 - 2\operatorname{Cov}(Y(t), Y(t-u))\\ &= 2\sigma^2 - 2\sigma^2\rho(Y(t),Y(t-u)) \end{aligned}$$

Stationarity (or just second-order stationarity) implies the correlation $\rho(t,u) = \rho(u)$ depends only on $u,$ permitting us to conclude

$$\gamma(u) = \gamma(t,u) = \sigma^2(1-\rho^2(u)).$$