If the square of a time series is stationary, is the original time series stationary?Downsampling stationary...
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If the square of a time series is stationary, is the original time series stationary?
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If the square of a time series is stationary, is the original time series stationary?
Downsampling stationary time series data, effect on varianceAutocorrelation vs Non-stationaryForecasting Time Series: Stationary vs Non-StationaryCross-correlation of two non-stationary time series?Compare stationary time seriesDetermining if a time series is covariance stationary or a random walkMean Square Convergence of a Linear Process Defined in Terms of a Stationary Time SeriesMaking a time series stationary by demeaningStationary vs. Trend-Stationary Time Series: Auto.Arima difference parameterDetect stationary section in an overall non-stationary time series
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I found a solution that stated that if the square of a time series is stationary, so is the original time series, and vice-versa.
However I don't seem able to prove it, anyone has an idea if this is true, and if it is how to derive it?
time-series self-study stationarity
asked 6 hours ago
VictorVictor
586
$endgroup$
add a comment |
$begingroup$
I found a solution that stated that if the square of a time series is stationary, so is the original time series, and vice-versa.
However I don't seem able to prove it, anyone has an idea if this is true, and if it is how to derive it?
time-series self-study stationarity
asked 6 hours ago
VictorVictor
586
$endgroup$
$begingroup$
Have you tried starting from the definition of stationarity?
$endgroup$
– Matt P
5 hours ago
add a comment |
$begingroup$
I found a solution that stated that if the square of a time series is stationary, so is the original time series, and vice-versa.
However I don't seem able to prove it, anyone has an idea if this is true, and if it is how to derive it?
time-series self-study stationarity
asked 6 hours ago
VictorVictor
586
$endgroup$
I found a solution that stated that if the square of a time series is stationary, so is the original time series, and vice-versa.
However I don't seem able to prove it, anyone has an idea if this is true, and if it is how to derive it?
time-series self-study stationarity
time-series self-study stationarity
asked 6 hours ago
VictorVictor
586
asked 6 hours ago
VictorVictor
586
asked 6 hours ago
VictorVictor
586
asked 6 hours ago
VictorVictor
586
asked 6 hours ago
VictorVictor
586
586
$begingroup$
Have you tried starting from the definition of stationarity?
$endgroup$
– Matt P
5 hours ago
add a comment |
$begingroup$
Have you tried starting from the definition of stationarity?
$endgroup$
– Matt P
5 hours ago
$begingroup$
Have you tried starting from the definition of stationarity?
$endgroup$
– Matt P
5 hours ago
$begingroup$
Have you tried starting from the definition of stationarity?
$endgroup$
– Matt P
5 hours ago
add a comment |
1 Answer
1
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That conjecture is false. A simple counter-example is the deterministic time-series $X_t = (-1)^t$. This time series is not even mean stationary, but its square is strictly stationary.
answered 5 hours ago
BenBen
30.2k234132
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
That conjecture is false. A simple counter-example is the deterministic time-series $X_t = (-1)^t$. This time series is not even mean stationary, but its square is strictly stationary.
answered 5 hours ago
BenBen
30.2k234132
$endgroup$
add a comment |
$begingroup$
That conjecture is false. A simple counter-example is the deterministic time-series $X_t = (-1)^t$. This time series is not even mean stationary, but its square is strictly stationary.
answered 5 hours ago
BenBen
30.2k234132
$endgroup$
add a comment |
$begingroup$
That conjecture is false. A simple counter-example is the deterministic time-series $X_t = (-1)^t$. This time series is not even mean stationary, but its square is strictly stationary.
answered 5 hours ago
BenBen
30.2k234132
$endgroup$
That conjecture is false. A simple counter-example is the deterministic time-series $X_t = (-1)^t$. This time series is not even mean stationary, but its square is strictly stationary.
answered 5 hours ago
BenBen
30.2k234132
answered 5 hours ago
BenBen
30.2k234132
answered 5 hours ago
BenBen
30.2k234132
answered 5 hours ago
BenBen
30.2k234132
30.2k234132
add a comment |
add a comment |
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$begingroup$
Have you tried starting from the definition of stationarity?
$endgroup$
– Matt P
5 hours ago