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| bio | website | |
|---|---|---|
| location | ||
| age | 33 | |
| visits | member for | 4 months |
| seen | 4 mins ago | |
| stats | profile views | 26 |
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23m |
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AR(1) Process first term I don't think the documentation is clear about removing the mean... the shift operator E in the Details uses 1 instead of an $a_0$-term. In the Applications part there is a single example (daily exchange rates of the euro to the dollar) where it removes the mean and then add it back for forecasting... I don't think an example is clear enough for defining the function... |
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1h |
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AR(1) Process first term You said "You cannot enter (as far as I know) a nonzero-mean input into ARProcess"... so, this must be a bug, don't you think? Imagine two different processes: 1) $Y_t=10+Y_{t-1}+\epsilon_t$ and 2) $Y_t=0+Y_{t-1}+\epsilon_t$; they are clearly different but if you try to estimate both processes with EstimatedProcess[data, ARProcess[1]] you will get exactly the same result, which is not consistent with one of the processes... so I'm getting wrong results if I use Mathematica-ARProcess... |
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2h |
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AR(1) Process first term So what is the point in using Mean[ARProcess[]] if the result will be always zero? I mean, if the ARProcess[] function cannot deal with an $\alpha$-term, its much better to develop an AR function by yourself instead of using Mathematica's ARProcess[]... |
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2h |
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AR(1) Process first term I explicitly wrote that $\epsilon_t \sim N(0,2)$, i.e., $\epsilon_t$ has $0$-mean (and not 10!). The AR(1) process has mean equal to $25$ and can accept an $\alpha$-term (in this case, 10) per definition. The problem is exactly that: the ARProcess[] doesn't accept this $\alpha$-term. BTW, the $\alpha$-term has no influence on the $\epsilon_t$ mean... |
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4h |
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Evaluation in Manipulate When you said "Plot does not work as expected" I thought you we're interested in showing how $a$ and $b$ would influence your plot... |
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12h |
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AR(1) Process first term Ahah... That's exactly what I'm trying to figure out! I think it's not possible and, as such, it could be a bug in Mathematica... |
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12h |
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AR(1) Process first term Don't worry @Fred... I think it's somehow a bug in Mathematica... |
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13h |
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AR(1) Process first term @Fred it does't work... what you've suggested works like an AR(2) Process of the form $Y_t=0+10 Y_{t-1} + .6 Y_{t-2} + \epsilon_t$, which is not stationary and, as such, the mean cannot be computed. |
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14h |
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Evaluation in Manipulate Even better IMO: Manipulate[Plot[a*x+b,{x,0,1},AxesOrigin->{0,0},PlotRange->{{0,1},{-2,2}}], {a,-1,1},{b,-1,1}] |
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14h |
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Evaluation in Manipulate You could also try Manipulate[Plot[a*x+b,{x,0,1},AxesOrigin->{0, 0}],{a,-1,1},{b,-1,1}] in order to better see the results... |
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14h |
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Evaluation in Manipulate Did you try Manipulate[Plot[a*x + b, {x, 0, 1}], {a, 0, 1}, {b, 0, 1}] ? |
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18h |
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Im trying to find the eigenvectors of an 11*11 matrix but can't get it to recognise my data Try this: mat={you data}; and then Eigenvalues[mat] |
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1d |
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Smooth Kernel Distribution I'm not sure if his e-books are still available... but the R- and Matlab-codes use in e-books and classes are available at this link. |
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1d |
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Smooth Kernel Distribution I had classes with Prof. Härdle... his e-book is really amazing... |
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May 15 |
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Evaluating the different calculations You could also have two different Mathematica versions installed in your machine. Put MMA9 to run 2-3 days and use, for instance, MMA8 for simpler calculations... |
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May 15 |
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Evaluating the different calculations What have you done so far? Instead of a simple question, you could put any code to show what exactly you're trying to achieve... |
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May 13 |
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Computing Correlations and p-values OK, let me correct myself: the sampling distribution for Pearson's correlation does assume Normality, while the measure itself does not assume Normality. That's why on should use Spearman correlation instead of Pearson. |
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May 13 |
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Computing Correlations and p-values These assumptions must also be true for the Pearson correlation test... |
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May 13 |
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Computing Correlations and p-values Or even SpearmanRankTest[], why not? |
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May 13 |
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Safe values of $\mu$ and $\sigma$ when randomly sampling from a Log-Normal Distribution? What definition of "outlier" are you using? A simple "visual test" usually does not work here. Try to generate your data, compute the IQR (interquartile range) and then you can effectively say if the observation is an outler... |
