Re: Evaluating integrals
- To: mathgroup at smc.vnet.net
- Subject: [mg66224] Re: Evaluating integrals
- From: Paul Abbott <paul at physics.uwa.edu.au>
- Date: Thu, 4 May 2006 05:21:28 -0400 (EDT)
- Organization: The University of Western Australia
- References: <e39jvr$clu$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
In article <e39jvr$clu$1 at smc.vnet.net>, "masha" <mshunko at gmail.com>
wrote:
> I am new to Mathematica and am tryong to figure out how to work
> efficently with integrals and run into the following issue. Let's say I
> want to integrate a simple function like Exp[-r b c], where c is a
> normal random variable. Doing this by hand, I can simply complete the
> square and end up with a simple result: Exp[(r^2 b^2 sc^2 )/ 2].
Computing the ExpectedValue of a function with respect to a specified
distribution is available in the Statistics` packages.
After loading the package stubs,
<<Statistics`
you can compute the ExpectedValue as follows:
ExpectedValue[Function[c, Exp[-r b c]], NormalDistribution[mu, s]]
Exp[(b^2 r^2 s^2)/2 - (b mu r)]
> However, if I run the following code in Mathematica:
>
> g[c_] := PDF[NormalDistribution[µc, sc], c]
> pi = Exp[-r b c]
> FullSimplify[Integrate[pi g[c], c, Assumptions -> {c > 0, r > 0, b >
> 0}]]
Because you need to integrate over {c, -Infinity, Infinity}.
g[c_] = PDF[NormalDistribution[mu, s], c]
Assuming[s > 0, Integrate[Exp[-r b c] g[c], {c, -Infinity, Infinity}]]
You will get exactly the same answer as above.
Cheers,
Paul
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