# Rod Lm

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 1h revised Forecast Future Stock Prices - Brownian Motion - Againadded 677 characters in body 2h answered Forecast Future Stock Prices - Brownian Motion - Again 2h comment Forecast Future Stock Prices - Brownian Motion - Again@Milan Ivica I don't think you need to define a new BrownianMotion function once you can use GeometricBrownianMotionProcess from Mathematica... Try, for instance, example[trend_] := RandomFunction[ GeometricBrownianMotionProcess[trend, .4, 1], {0, 10, .01}]["Path"] and then use ListLinePlot[Table[example[trend],{trend,{0,.3,.5}}]]... 1d comment CorrelationTest small bug?Could I assume the same for the Pearson correlation test? 1d accepted CorrelationTest small bug? 1d asked CorrelationTest small bug? 2d comment Why is there no PositionFunction in Mathematica?I can speak for myself, as I have several notebooks where I use Position. So, I believe Position is used more often than you think by the vast majority of users (not only unexperienced ones)... 2d comment Why is there no PositionFunction in Mathematica?BTW, congrats for the 60K ! Is there any way to "suggest" this function in the next version of Mathematica? 2d comment Why is there no PositionFunction in Mathematica?I belong to the team of Position[]-users... :-( May20 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... May20 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... May20 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[]... May20 revised AR(1) Process first termadded 26 characters in body May20 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... May20 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... May20 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... May20 comment AR(1) Process first termDon't worry @Fred... I think it's somehow a bug in Mathematica... May20 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. May20 revised AR(1) Process first termadded 7 characters in body May20 answered Evaluation in Manipulate