RE:MORE: Statistics`ContinousDistributions` (integrating over a UniformDistribution)

*To*: mathgroup at smc.vnet.net*Subject*: [mg9294] RE:[mg9252] MORE: Statistics`ContinousDistributions` (integrating over a UniformDistribution)*From*: tgarza at mail.internet.com.mx*Date*: Mon, 27 Oct 1997 02:47:05 -0500*Sender*: owner-wri-mathgroup at wolfram.com

Mark wrote ><<Statistics`ContinuousDistributions` > >p[x_] :=3D PDF[UniformDistribution[0,Pi],x] > >MeanDist[f_] :=3D Integrate[ f p[x], {x, 0, Pi}] >MeanDist[ Sin[x] ] Doesn't give 2/Pi like I'd expect. (and I >don't even like forcing the limits to {0,Pi}... much better to have >{-infinity,infinity} ............... >Why can't Mathematica deal with it? Mark, There is nothing wrong here with MM3.It's just a matter of proper = definitions. To start with, there is a problem with the definition of your PDF. The = complete definition must include the domain where the function is zero, = which is x < 0 and x > 2 Pi. The function has three different = definitions (i.e., one to the left of zero, one between zero and Pi, and = one to the right of Pi) and the integral has to take this into account. = Thus, there are three domains of integration, one of them being the one = that goes between 0 and Pi, and you can't help it. That is why the = integral does not exist the way you try to evaluate it, proceeding as = though the function were defined in the same way all over the real line. Secondly, you must realize that if X is a random variable with a uniform = PDF between 0 and Pi, and EX is its expected value (or Mean, if you = wish), then Sin(X) is a random variable with a distribution function of = its own, and ESin(X) is certainly not equal to Sin(E(X)). The = expectation operator E is linear and does not "distribute" over = nonlinear functions such as Sin. If you want to find the Mean of Sin(X), then you must first find the CDF = of Sin(X), which can be done by considering that P(Sin(X) <=3D y) =3D = P(X <=3D arcsin(y)), with 0 <=3D arcsin(y) <=3D Pi, and this is equal to = 2 arcsin(y)/Pi for 0 <=3D y <=3D 1. The value of this function at y =3D = 1 is 1, as it should. Then, the PDF is obtained by differentiation of the CDF: D[2 ArcSin[x], x]=20 2/(Pi Sqrt(1 - x^2)), in the range 0 <=3D x <=3D 1, and zero elsewhere. Then, the Mean of = Sin(X) is=20 NIntegrate[2x/(Pi Sqrt(1 - x^2)), {x, 0, 1}] =20 0.63662 which is 2/Pi, as desired.=20 Unfortunately, I don't see any way to keep MeanDist all that general. = Mathematica has no way of guessing what the domain of your function is = in order to perform the integration in the correct range. =20 Cheers, Tomas Garza Cerrada de Cortes 31 Mexico 01040, D.F.

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**MORE: Statistics`ContinousDistributions` (integrating over a UniformDistribution)**

**Re: MORE: Statistics`ContinousDistributions` (integrating over a UniformDistribution)**