# Rod Lm

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 23m comment AR(1) Process first termI 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... 1h comment AR(1) Process first termYou 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... 2h comment AR(1) Process first termSo 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[]... 2h comment AR(1) Process first termI 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... 4h comment Evaluation in ManipulateWhen you said "Plot does not work as expected" I thought you we're interested in showing how $a$ and $b$ would influence your plot... 12h comment AR(1) Process first termAhah... 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... 12h comment AR(1) Process first termDon't worry @Fred... I think it's somehow a bug in Mathematica... 13h comment 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. 14h comment Evaluation in ManipulateEven better IMO: Manipulate[Plot[a*x+b,{x,0,1},AxesOrigin->{0,0},PlotRange->{{0,1},{-2,2}}], {a,-1,1},{b,-1,1}] 14h comment Evaluation in ManipulateYou 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... 14h comment Evaluation in ManipulateDid you try Manipulate[Plot[a*x + b, {x, 0, 1}], {a, 0, 1}, {b, 0, 1}] ? 18h comment Im trying to find the eigenvectors of an 11*11 matrix but can't get it to recognise my dataTry this: mat={you data}; and then Eigenvalues[mat] 1d comment Smooth Kernel DistributionI'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. 1d comment Smooth Kernel DistributionI had classes with Prof. Härdle... his e-book is really amazing... May15 comment Evaluating the different calculationsYou 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... May15 comment Evaluating the different calculationsWhat have you done so far? Instead of a simple question, you could put any code to show what exactly you're trying to achieve... May13 comment Computing Correlations and p-valuesOK, 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. May13 comment Computing Correlations and p-valuesThese assumptions must also be true for the Pearson correlation test... May13 comment Computing Correlations and p-valuesOr even SpearmanRankTest[], why not? May13 comment 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...