# Why does differencing White Noise induce autocorrelation of $-0.5$?

I am curious about the following problem. Let's have a variable given by white noise,

$$y_t \sim \operatorname{NID}(0,1).$$

Let's say we difference it,

$$\Delta y_t = y_t - y_{t-1}.$$

And now, if we measure the ACF of $$\Delta y_t$$, we get that the first lag is autocorrelated by $$-0.5$$.

The question is: Why $$-0.5$$, specifically?

For notational simplicity, let $$X_t \equiv \Delta Y_t.$$

By definition, $$\text{Corr}(X_t,X_{t-1}) = \frac{\text{Cov}(X_t,X_{t-1})}{\sqrt{\text{Var}(X_t)}\sqrt{\text{Var}(X_{t-1})}}.$$

Here, \begin{aligned} \text{Var}(X_t) &= \text{Var}(Y_t-Y_{t-1}) \\ &= \text{Var}(Y_t) - 2\text{Cov}(Y_t,Y_{t-1}) + \text{Var}(Y_{t-1}) \\ &= 1-0+1 \\ &= 2. \end{aligned} and the same for $$\text{Var}(X_{t-1})$$. Thus
\begin{aligned} \sqrt{\text{Var}(X_t)}\sqrt{\text{Var}(X_{t-1})} &= \sqrt{\text{Var}(X_t)}\sqrt{\text{Var}(X_{t})} \\ &= \text{Var}(X_t). \end{aligned}

Also, \begin{aligned} \text{Cov}(X_t,X_{t-1}) &= \text{Cov}(Y_t-Y_{t-1},Y_{t-1}-Y_{t-2}) \\ &= \text{Cov}(Y_t,Y_{t-1}) - \text{Cov}(Y_t,Y_{t-2}) - \text{Cov}(Y_{t-1},Y_{t-1}) + \text{Cov}(Y_{t-1},Y_{t-2}) \\ &= 0 - 0 - \text{Var}(Y_{t-1}) + 0 \\ &= 0 - 0 - \text{Var}(Y_{t}) + 0 \\ &= 0-0-1+0 \\ &= -1. \end{aligned}

Thus, $$\text{Corr}(X_t,X_{t-1}) = \frac{-1}{2} = -0.5.$$