Re: Fourier Transform and convolution.
- To: mathgroup at smc.vnet.net
- Subject: [mg26601] Re: [mg26583] Fourier Transform and convolution.
- From: Tomas Garza <tgarza01 at prodigy.net.mx>
- Date: Thu, 11 Jan 2001 10:39:12 -0500 (EST)
- References: <200101090651.BAA00223@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
No problem. Just bear in mind that you want to use the "characteristic
function" and *not* the Fourier transform of the N(0, 1) independent random
variables in order to obtain the distribution of their sum. The Fourier
transform differs from the ch. f. by a constant 1/Sqrt[2 Pi]. Then proceed
as follows:
In[1]:=
<< Statistics`ContinuousDistributions`
In[2]:=
ndist = NormalDistribution[0, 1];
Now square of the characteristic function of the normal (0, 1) distribution
and obtain its InverseFourierTransform, adequately multiplied by the
constant referred to above:
In[3]:=
distSum = (1/Sqrt[2
Pi])*InverseFourierTransform[CharacteristicFunction[ndist, t]^2, t, x]
Out[3]=
(1/(2 Sqrt[Pi]))*Exp[(-x^2)/4]
This is what you are looking for. If you integrate this from -6 to 6 you get
approximately 1, as expected. You may look at the density functions:
In[4]:=
gr1 = Plot[PDF[ndist, x], {x, -6, 6}];
In[5]:=
gr2 = Plot[(1/Sqrt[2 Pi])*
InverseFourierTransform[CharacteristicFunction[ndist, t]^2, t,
x], {x, -6, 6}];
In[6]:=
Show[gr1, gr2];
What went wrong with what you did? You say "we perform
a Fourier Transform of the pdf n(0,1) and then square each point". I don't
know what you mean by this. What you have to do is obtain (not perform,
please) the Fourier Transform (multiplied by the adequate constant) and
square the transform, not each point. Then obtain the inverse. That's all.
Tomas Garza
Mexico City
----- Original Message -----
From: <Adil.Reghai at dresdnerkb.com>
To: mathgroup at smc.vnet.net
Subject: [mg26601] [mg26583] Fourier Transform and convolution.
> Hi,
>
> We wanted to perform the distribution of a sum of two independent
variables
> X and Y which are both N(0,1).
> The way We wanted to do it is to use the convolution property. In order to
> compute this convolution we perform
> a Fourier Transform of the pdf n(0,1) and then square each point. After
that
> we performed an inverse Fourier transform.
> Instead of obtaining a N(0,sqrt(2)) we obtain a distribution which is
> shifted in the form of a U.
>
> Could you please tell us what went wrong?
>
> Regards
>
> Adil,
>
>
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- References:
- Fourier Transform and convolution.
- From: Adil.Reghai@dresdnerkb.com
- Fourier Transform and convolution.