Re: Re: Write an expression in a specific form
- To: mathgroup at smc.vnet.net
- Subject: [mg108062] Re: [mg108014] Re: [mg107952] Write an expression in a specific form
- From: DrMajorBob <btreat1 at austin.rr.com>
- Date: Sun, 7 Mar 2010 05:08:43 -0500 (EST)
- References: <4977982.1267699026953.JavaMail.root@n11>
- Reply-to: drmajorbob at yahoo.com
Here's the same thing a TAD simpler. (I just trusted Integrate to do its
job.)
g[m_, s_, x_] = PDF[NormalDistribution[m, s]][x]
E^(-((-m + x)^2/(2 s^2)))/(Sqrt[2 \[Pi]] s)
c = Integrate[g[m1, s1, x] g[m2, s2, x], {x, -Infinity, Infinity},
Assumptions -> {s1 > 0, s2 > 0}]
E^(-((m1 - m2)^2/(2 (s1^2 + s2^2))))/(Sqrt[2 \[Pi]] Sqrt[s1^2 + s2^2])
Checking that we have a PDF:
Integrate[1/c g[m1, s1, x] g[m2, s2, x], {x, -Infinity, Infinity},
Assumptions -> {s1 > 0, s2 > 0}]
1
If that's the PDF, here's the mean:
m = Integrate[x/c g[m1, s1, x] g[m2, s2, x], {x, -Infinity, Infinity},
Assumptions -> {s1 > 0, s2 > 0}]
(m2 s1^2 + m1 s2^2)/(s1^2 + s2^2)
And here's the standard deviation:
s = Sqrt@Integrate[(x - m)^2/c g[m1, s1, x] g[m2, s2, x], {x, -Infinity,
Infinity}, Assumptions -> {s1 > 0, s2 > 0}] // PowerExpand
(s1 s2)/Sqrt[s1^2 + s2^2]
Verifying equality:
g[m1, s1, x] g[m2, s2, x] - c g[m, s, x] // Simplify
0
Bobby
On Sat, 06 Mar 2010 16:34:03 -0600, DrMajorBob <btreat1 at austin.rr.com>
wrote:
> OK, here's a simple solution:
>
> g[m_, s_, x_] = PDF[NormalDistribution[m, s]][x]
>
> E^(-((-m + x)^2/(2 s^2)))/(Sqrt[2 \[Pi]] s)
>
> c = Integrate[g[m1, s1, x] g[m2, s2, x], {x, -Infinity, Infinity},
> Assumptions -> {s1 > 0, s2 > 0}]
>
> E^(-((m1 - m2)^2/(2 (s1^2 + s2^2))))/(Sqrt[2 \[Pi]] Sqrt[s1^2 + s2^2])
>
> Checking that we have a PDF:
>
> Integrate[1/c g[m1, s1, x] g[m2, s2, x], {x, -Infinity, Infinity},
> Assumptions -> {s1 > 0, s2 > 0}]
>
> 1
>
> If that's the PDF, here's the mean:
>
> m = Integrate[x/c g[m1, s1, x] g[m2, s2, x], {x, -Infinity, Infinity},
> Assumptions -> {s1 > 0, s2 > 0}]
>
> (m2 s1^2 + m1 s2^2)/(s1^2 + s2^2)
>
> And here's the standard deviation:
>
> s = Sqrt@Integrate[(x - m)^2/c g[m1, s1, x] g[m2, s2,
> x], {x, -Infinity, Infinity}, Assumptions -> {s1 > 0, s2 > 0}]
>
> Sqrt[(s1^2 s2^2)/(s1^2 + s2^2)]
>
> But can we verify it?
>
> 1/c g[m1, s1, x] g[m2, s2, x] - g[m, s, x] // Simplify
>
> (E^(-((m2 s1^2 + m1 s2^2 - (s1^2 + s2^2) x)^2/(
> 2 s1^2 s2^2 (s1^2 + s2^2)))) (-s1 s2 +
> Sqrt[(s1^2 s2^2)/(s1^2 + s2^2)] Sqrt[s1^2 + s2^2]))/(Sqrt[
> 2 \[Pi]] s1 s2 Sqrt[(s1^2 s2^2)/(s1^2 + s2^2)])
>
> That needs to be zero for all x, and that's true if and only if:
>
> -s1 s2 + Sqrt[(s1^2 s2^2)/(s1^2 + s2^2)] Sqrt[s1^2 + s2^2] == 0
>
> That's a condition on s1 and s2 alone... and it's an easy one:
>
> -s1 s2 + Sqrt[(s1^2 s2^2)/(s1^2 + s2^2)] Sqrt[
> s1^2 + s2^2] // PowerExpand
>
> 0
>
> Done.
>
> Bobby
>
> On Fri, 05 Mar 2010 03:33:11 -0600, David Park <djmpark at comcast.net>
> wrote:
>
>> I suppose that someone will get this with some simple command but, in
>> case
>> not, here is my somewhat laborious solution using Presentations. I won't
>> show all the output lines, but just the results. Most of the steps are
>> done
>> on the argument of Exp. In the final steps I substituted f for Exp to
>> keep
>> Mathematica from recombining the products, and then put the final
>> answer in
>> Subexpressions.
>>
>> Needs["Presentations`Master`"]
>>
>> G[m_, s_, x_] := Exp[-((x - m)^2/(2 s^2))]/(s Sqrt[2 \[Pi]])
>>
>> initial = G[m1, s1, x]*G[m2, s2, x]
>> Part[initial, 2, 2]
>> step1 = CompleteTheSquare[%, x]
>> step2 = step1 // MapLevelParts[Simplify, {{1, 2, 3}}]
>> rule1 = (expr1 = Part[step2, 2, 1]) -> (expr2 =
>> Factor[Together[expr1]])
>> rule2 = -2 expr2 -> 1/s^2
>> step3 = step2 /. rule1 /. rule2
>> step4 = MapAt[Simplify, step3, {{2, 3, 1, 1}}]
>> rule3 = -Part[step4, 2, 3, 1, 1] -> m
>> step5 = step4 /. rule3
>> msrules = {rulem = Reverse[rule3], rules2 = Reverse[1/# & /@ rule2],
>> rules = PowerExpand[Sqrt[#]] & /@ rules2}
>> (Times @@ f /@ step5)/(2 \[Pi] s1 s2)
>> MapAt[FactorOut[Sqrt[2 \[Pi]] s, CreateSubexpression], %, {6}];
>> MapAt[FactorOut[Sqrt[2 \[Pi]] s1 s2/s, CreateSubexpression], %, {5}];
>> final = % /. f -> Exp
>>
>> The final answer, with Subexpressions converted to HoldForm:
>>
>> HoldForm[E^(-((-m + x)^2/(2 s^2)))/(Sqrt[2 \[Pi]] s)] HoldForm[(
>> E^(-((m1 - m2)^2/(2 (s1^2 + s2^2)))) s)/(Sqrt[2 \[Pi]] s1 s2)]
>>
>> With the rules:
>>
>> {m -> (m2 s1^2 + m1 s2^2)/(s1^2 + s2^2),
>> s^2 -> (s1^2 s2^2)/(s1^2 + s2^2), s -> (s1 s2)/Sqrt[s1^2 + s2^2]}
>>
>> This tests that the results are the same
>>
>> (final // ReleaseSubexpressions[]) /. msrules // Simplify;
>> initial // ExpandAll // Simplify;
>> %% == %
>> True
>>
>>
>> David Park
>> djmpark at comcast.net
>> http://home.comcast.net/~djmpark/
>>
>>
>> From: Ares Lagae [mailto:ares.lagae at sophia.inria.fr]
>>
>> Hi all,
>>
>> I am a beginner in Mathematica, and I have the following "problem": How
>> can
>> I write an expression in a specific form?
>>
>> For example:
>>
>> - Define a Gaussian:
>>
>> G[m_, s_, x_] := (1/(s*Sqrt[2*Pi]))*Exp[-(x - m)^2/(2*s^2)];
>>
>> - Product of two Gaussians:
>>
>> G[m1, s1, x] * G[m2, s2, x]
>>
>> - How can I get Mathematica to write the result in terms of c * G[m_,
>> s_,
>> x_]? I.e., get the values for c, m and s.
>>
>> Thanks,
>>
>> Ares Lagae
>>
>>
>>
>>
>
>
--
DrMajorBob at yahoo.com