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1

I don't think you can get a solution by just throwing the problem at Solve. If you make a minor change in the argument you give to Solve (substituting 95/100 for 0get.95), you will get a message you may find more meaningful. Solve[ CDF[ChiSquareDistribution[n - 1], (n - 1)/k1] - CDF[ChiSquareDistribution[n - 1], (n - 1)/k2] == 95/100 && ...


1

You can check the documentation about the option NominalVariables http://reference.wolfram.com/language/ref/NominalVariables.html NominalVariables is an option for machine learning functions such as LinearModelFit or Classify that specifies which variables should be treated as having discrete values specified by names. So LogitModelFit[{{300, 0, ...


1

Tweak Solution Since Seth has provided 2 answers, I thought I might also put up another answer. My motivation for separating this from my original answer is that ... my original answer is a self-contained mathematical solution in transformations of random variables, essentially striving to side-step Mathematica's use of Boole which was not working, ...


0

I spent some additional time on this problem and believe I MAY have solved it in a pretty general way with Mathematica. But I'm not sure and I still think there's more help needed on the problem. I'm going to use wolfies' substitution of the conventional x and y for my earlier used f and k. Basically, what we are after is the expected value of x over a ...


2

This is a very nice problem. If I may dispense with the notation for random variables as $f$ and $k$, and refer to them instead as $X$ and $Y$ ... The Problem Let $X \sim Uniform(0,1)$ and $Y \sim Uniform(2,5)$ be independent random variables, with joint pdf $f(x,y) = \frac13$: f = 1/3; domain[f] = {{x, 0, 1}, {y, 2, 5}}; We seek a closed-form ...


2

I figured out a way to answer a related simpler question, which is Expectation[ f \[Conditioned] f + \[Alpha]*k > p], {f, k} \[Distributed] UniformDistribution[{{a, b}, {k1, k2}}] Mathematica has trouble with this computation, but if you realize that it is the centroid of a polygon, you can use a feature of Mathematica to get a (long) answer ...


4

You need to square your argument: ZTest[data, 20^2, 540., "TestStatistic"] (* 2.19469 *)


0

LaplaceTransform attempts a symbolic evaluation of the transform, which Mathematica fails to do in this case. To get an answer, write the transform explicitly as an NIntegrate multiple integral. It still takes a while, but reducing the accuracy/precision demands will speed things up: chat = NIntegrate[ GumExpExp Exp[-Ab[s] x - Ar[s] y], {x, 0, ...


8

With the data beeing data = RandomVariate[NormalDistribution[0, 1], 200]; the range of the box specified to be one sigma (approx. 68.3 %tile range) by sigma=Erf[1/Sqrt[2]] and a limit for the fences defined to be 10 % fencesLimit = 0.1 we can plot a BoxWhiskerChart using: BoxWhiskerChart[data, "Median", Method -> "BoxRange" -> (Quantile[#, ...


1

p1 = Histogram[data, {0, 9, 1}, "Count", ChartLegends -> {"Experimental Result"}, ChartStyle -> "Pastel"]; p2 = DiscretePlot[ Length@data*PDF[PoissonDistribution[2.2766917293233084`], x], {x, 0, 10}, PlotStyle -> {Red, Medium}, PlotLegends -> {"Theoretical Poisson"}]; Show[{p1, p2}, AxesLabel -> {"Time (s)", "Counts"}] ...


0

You can also depict the expected value nicely as a function of Mu and Sigma using Plot3D. px = Simplify[PDF[LogNormalDistribution[Mu, Sigma], x], x > 0] E^(-((Mu - Log[x])^2/(2 Sigma^2)))/(Sqrt[2 \[Pi]] Sigma x) Plot3D[NIntegrate[Log2[1 + x] px, {x, 0, \[Infinity]}], {Mu, -5, 10}, {Sigma, .2, 2}, AxesLabel -> {Mu, Sigma, EY}] PS: Reminds me ...


2

I don't think there is anything unusual here that can't be found in the documentation. Normalization parameters in the x and y dimensions (b, and a, respectively) can be added into the model, as can the x-axis shift (parameter c). Parameter estimates are taken from a look at the original data. nlm = NonlinearModelFit[data, a Erfc[b (x + c)], {{a, 8 ...


2

I am a little unsure what the aim is here. Looking at the data only the first third of the plot could be fitted to Erfc like function. I post this, perhaps, as motivation. d = Transpose@data; ListPlot[ds = SortBy[d, #[[1]] &], Joined -> True] Extracting the part of plot I referred to: fd = ds[[1 ;; 35]] fdlp = ListPlot[fd] Now as a rather ...


0

Using the function positionDuplicates from this answer by @Szabolcs positionDuplicates[list_] := GatherBy[Range@Length[list], list[[#]] &]; dispatchF = Thread[DeleteDuplicates[#] -> positionDuplicates[#]] &; positionBins = Union @@@ (BinLists[#, #2] /. dispatchF[#1]) &; data = RandomInteger[5, 17] (* {5, 4, 1, 3, 0, 3, 0, 0, 5, 5, 0, 0, ...


1

I suspect there may not be a closed form expression (I have not looked at it hard enough). If the aim is to not analytical but numerical, I post the following for illustration (apologies if not the intent of question): f[x_, y_] := NIntegrate[ Log2[t + 1] Exp[-(x - Log[t])^2/(2 y^2)]/(Sqrt[2 Pi] t y), {t, 0, Infinity}] rv[a_, b_, n_] := ...



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