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Chande Momentum Oscillator
This study calculates and displays the Chande Momentum Oscillator of the data specified by the Input Data Input, as well as three horizontal lines determined by the user.
Let \(X\) be a random variable denoting the Input Data, and let \(X_t\) be the value of the Input Data at Index \(t\). The Up Change and Down Change of the Input Data at Index \(t\) are denoted as \(U(X)\) and \(D(X)\), respectively, and we compute them for \(t > 0\) as follows.
\(\displaystyle{U_t(X) =\left\{ \begin{matrix} X_t - X_{t - 1} & X_t \geq X_{t - 1} \\ 0 & X_t < X_{t - 1} \end{matrix}\right .}\)\(\displaystyle{D_t(X) =\left\{ \begin{matrix} 0 & X_t \geq X_{t - 1} \\ X_{t - 1} - X_t & X_t < X_{t - 1} \end{matrix}\right .}\)
Let the Input CMO Length be denoted as \(n_{CMO}\). Then we denote the Chande Momentum Oscillator at Index \(t\) for the given Inputs as \(CMO_t(X,n_{CMO})\), and we compute it for \(t \geq n_{CMO}\) in terms of Moving Summations as follows.
\(\displaystyle{CMO_t(X,n_{CMO}) =\left\{ \begin{matrix} 100 \cdot \frac{\mathrm{sum}_t(U(X),n_{CMO}) - \mathrm{sum}_t(D(X),n_{CMO})}{\mathrm{sum}_t(U(X),n_{CMO}) + \mathrm{sum}_t(D(X),n_{CMO})} & \mathrm{sum}_t(U(X),n_{CMO}) + \mathrm{sum}_t(D(X),n_{CMO}) \neq 0 \\ CMO_{n - 1}(X,n_{CMO}) & \mathrm{sum}_t(U(X),n_{CMO}) + \mathrm{sum}_t(D(X),n_{CMO}) = 0 \end{matrix}\right .}\)Finally, the Line 1, Line 2, and Line 3 Inputs determine the levels of three horizontal lines that are displayed with \(CMO_t(X,n_{CMO})\).
Inputs
Spreadsheet
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*Last modified Monday, 26th September, 2022.