Is the following snippet of use?
In [1]: from sympy.stats import *
In [2]: n = 5
In [3]: mus = [Symbol('mu'+str(i), bounded=True) for i in range(n)]
In [4]: sigmas = [Symbol('sigma'+str(i), positive=True) for i in range(n)]
In [5]: Xs = [Normal('X'+str(i), mu, sigma) for i, (mu, sigma) in
enumerate(zip(mus, sigmas))]
In [6]: E(prod(Xs))
<< Big Scary Thing >>
In [7]: simplify(E(prod(Xs)))
Out[7]: μ₀⋅μ₁⋅μ₂⋅μ₃⋅μ₄
Please note that you shouldn't have to switch down to numerics for this
problem. There is a clean analytic form.
On Thu, Aug 29, 2013 at 2:01 PM, Janwillem van Dijk
<[email protected]>wrote:
> I have problems lambdifying a symbolic function that needs array arguments.
> The problem I want to solve is to find the statistics of the product of
> Normal distributions. Attached is an example that works and even yields the
> correct answers. However, I think that the part where the symbolic versions
> are lambdified into Python callable functions can't win a beauty contest.
> For example a part of the class :
>
> self.MeanNN = E(NN)
>
> self.meanNN = lambdify([mu, s], self.MeanNN)
>
> def meanOfNs(self, mu, s):
>
> mus = mu + s
>
> return self.meanNN(*mus)
>
> Than called as:
>
> Nn = ProductOfNormals(n)
>
> print('Mean: ', Nn.meanOfNs(mu, s))
>
> Where mu and s are lists of length n
>
>
> In case the arguments are spelled out it looks nicer but I want the
> flexibility of variable number of distributions in the product.
>
> e.g.:
>
> self.MeanNN = E(NN)
>
> self.meanNN = lambdify([mu1, s1, mu2, s2], self.MeanNN)
>
> Than:
>
> NN = ProductNormalNormal()
>
> print('Mean: ', NN.meanNN(mu0, s0, mu1, s1))
>
> Where all parms are scalar
>
>
> Any tips on how to get a more elegant solution are welcome!!
>
> The complete example is attached. I am using SymPy 0.7.3 on Python 2.7.4
> on Linux.
>
> Cheers, Janwillem
>
>
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