This problem was given in an introductory course to Time Series:
Let $(X_t)$ be a weakly stationary process with mean $\mu$ and autocovariance function $\gamma(k)=\text{Cov}(X_t,X_{t+k})$. Let us now define another series, $\delta_t$, which is given below: $$\delta_t = \sum_{i=a}^{b}c_iX_{t+i}$$ Show that any $k$th order difference (denoted as $\nabla^kX_t$) can be written in the form of $\delta_t$.
My main issue is the approach I used to solve the problem. It seemed evident to me that induction was the most direct way to prove this statement. The base case can be easily proven by a first-order differencing, as seen here: $$\nabla Y_t = Y_t - Y_{t-1}$$ Which can obviously be written in the form of $\delta_t$. But the induction step is kind of stumping me. I've done the following: $$\nabla^{k+1}Y_{t} = \nabla(\nabla^{k}Y_t) = \nabla(\sum_{i=a}^{b}c_iX_{t+i}) = \sum_{i=a}^{b}c_i\nabla X_{t+i}$$ I am not certain as to how to continue from this step. We could expand the first-order difference out, but I don't know how that would help me. How would you guys approach this problem?
Let $B$ be the backshift operator: $B X_t = X_{t-1}$. Then, notice that we can write the first-order difference operator as $$\nabla X_t= X_t - X_{t-1} = (I - B)X_t$$ By applying the first-order difference operator $k$ times, we get $$\nabla^k = (I-B)^k = \sum_{r=0}^k \binom{k}{r}(-1)^r B^r$$ Thus, $$\nabla^k X_t = \sum_{r=0}^k \binom{k}{r}(-1)^r B^r X_t = \sum_{r=0}^k \binom{k}{r}(-1)^r X_{t-r}$$ Which is of the form you want once you read off the values of $a, b,$ and $c_i$.