# Find moments of a discrete process using $\mathcal{Z}$-Transform method

For a linear process relating a variable $$Y_i$$ to random, independent variables $$X_i$$ using the equation:

$$\begin{equation} Y_i = aY_{i-1} + (1-a)X_i \end{equation}$$

which has the solution:

$$\begin{equation} Y_n = a^nY_0 + (1-a)\sum_{i=0}^{n-1} a^i X_{n-i} \end{equation}$$

it is relatively straightforward to investigate how the moments of the input ($$X_i$$) are transformed into the m$$th$$ moments of the output ($$Y_i)$$ using the expectation of the expansion of $$(Y_i-\mu_{Y_i})^m$$. Using this method (and assuming $$Y_0=0$$, $$|a|<1$$ and $$n\rightarrow\infty$$) it can be shown that:

$$\begin{equation} \mathbf{E}\left[Y_n\right] = \mathbf{E}\left[X_n\right] \quad \mathbf{var}\left[Y_n\right] = \frac{\left(1-a\right)^2}{1-a^2}\mathbf{var}\left[X_n\right] = \frac{1-a}{1+a}\mathbf{var}\left[X_n\right] \quad \mathbf{skew}\left[Y_n\right] = \frac{\left(1-a\right)^3}{1-a^3}\sqrt{\left(\frac{1+a}{1-a}\right)^3}\mathbf{skew}\left[X_n\right] \end{equation}$$

and so on. I have calculated these expressions for up to the 5th moment and have verified them numerically.

However, if I use a more complicated version of this simple system (for example, cascading through several such recursions) the expectations models become very cumbersome. I am therefore looking to obtain these moment relationships between output, input and the linear process parameter $$a$$ using the $$\mathcal{Z}$$-transform of the impulse-response function for the system. I have determined that this would be:

$$\begin{equation} H\left(z\right) = \frac{z\left(1-a\right)}{z-a} \end{equation}$$

which arises from:

$$\begin{equation} Y(z) -az^{-1}Y(z) = (1-a)X(z) \quad \mbox{so} \quad Y(z) = \frac{z(1-a)}{z-a}X(z) \quad \mbox{and so} \quad H(z)=\frac{Y(z)}{X(z)} \end{equation}$$

and that, for example the first moments from this would be:

\begin{align} \mu_0^{'} &= H(z)_{z=1} = 1\\ \mu_1^{'} &= \left[-z\frac{d}{dz}H(z)\right]_{z=1} = \frac{a}{1-a}\\ \mu_2^{'} & = \left[-z\left(-z\frac{d}{dz}H(z)\right)\right]_{z=1} = \frac{a(1+a)}{(a-1)^2} \end{align}

and that the second central moment should be given by:

$$\begin{equation} \mu_2 = \mu_2^{'}-(\mu_1^{'})^2 = \frac{a}{(1-a)^2} \end{equation}$$

Clearly, this is not the same as above - where am I going wrong? And how would I apply this to the higher moments up to the 5th or 6th?

• Could you elaborate how you derived $\mu^{'}_{1}$? I somehow keep getting $0$ as a result.
– Max
Jan 10 at 9:07
• $-z\frac{d}{dz}\left[\frac{z(1-a)}{z-a}\right] = z\frac{a(1-a)}{(a-z)^2}$ so at $z=1$ this gives the result above. Jan 10 at 9:35
• Found my mistake, thanks.
– Max
Jan 10 at 9:52