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| bio | website | |
|---|---|---|
| location | ||
| age | 33 | |
| visits | member for | 4 months |
| seen | 4 mins ago | |
| stats | profile views | 33 |
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1h |
revised |
Forecast Future Stock Prices - Brownian Motion - Again added 677 characters in body |
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2h |
answered | Forecast Future Stock Prices - Brownian Motion - Again |
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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}}]]... |
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1d |
comment |
CorrelationTest small bug? Could I assume the same for the Pearson correlation test? |
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1d |
accepted | CorrelationTest small bug? |
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1d |
asked | CorrelationTest small bug? |
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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)... |
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2d |
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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? |
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2d |
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Why is there no PositionFunction in Mathematica? I belong to the team of Position[]-users... :-( |
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May 20 |
comment |
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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May 20 |
comment |
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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May 20 |
comment |
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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May 20 |
revised |
AR(1) Process first term added 26 characters in body |
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May 20 |
comment |
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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May 20 |
comment |
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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May 20 |
comment |
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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May 20 |
comment |
AR(1) Process first term Don't worry @Fred... I think it's somehow a bug in Mathematica... |
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May 20 |
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. |
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May 20 |
revised |
AR(1) Process first term added 7 characters in body |
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May 20 |
answered | Evaluation in Manipulate |
