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From: Richard Hayden on 26 Apr 2010 09:35
Hi,

Suppose we have Z = X + Y as non-negative real random variables, where
X and Y are independent.

Assume that we know the marginal pdfs for Z and Y, f_z(t) and f_y(t)
respectively. I want to compute the marginal pdf for X, say f_x(t).
Now I know f_z(t) = (f_x * f_y)(t), where * is the convolution
operator, so I need to invert the convolution to get f_x(t). I can see
how this can be done by taking Laplace transforms and then performing
a Laplace inversion (numerically is fine), but am I missing something?
Is there an easier way? Laplace inversion seems like a sledgehammer
for what looks like a very simple problem?

Any guidance gratefully appreciated.

Thanks,

Richard.
From: danheyman on 26 Apr 2010 11:00
On Apr 26, 9:35 am, Richard Hayden <r.hay...(a)gmail.com> wrote:
> Hi,
>
> Suppose we have Z = X + Y as non-negative real random variables, where
> X and Y are independent.
>
> Assume that we know the marginal pdfs for Z and Y, f_z(t) and f_y(t)
> respectively. I want to compute the marginal pdf for X, say f_x(t).
> Now I know f_z(t) = (f_x * f_y)(t), where * is the convolution
> operator, so I need to invert the convolution to get f_x(t). I can see
> how this can be done by taking Laplace transforms and then performing
> a Laplace inversion (numerically is fine), but am I missing something?
> Is there an easier way? Laplace inversion seems like a sledgehammer
> for what looks like a very simple problem?
>
> Any guidance gratefully appreciated.
>
> Thanks,
>
> Richard.

For some special cases you can dispense with Laplace transforms [e.g.
if Z and Y are both normal or both Poisson], but in general transforms
are the way to go. The methods for solving integral equations seem to
need some structure to get closed-form solutions, but I haven't
investigated them in detail.
From: karl on 26 Apr 2010 14:05
Richard Hayden schrieb:
> Hi,
>
> Suppose we have Z = X + Y as non-negative real random variables, where
> X and Y are independent.
>
> Assume that we know the marginal pdfs for Z and Y, f_z(t) and f_y(t)
> respectively. I want to compute the marginal pdf for X, say f_x(t).
> Now I know f_z(t) = (f_x * f_y)(t), where * is the convolution
> operator, so I need to invert the convolution to get f_x(t). I can see
> how this can be done by taking Laplace transforms and then performing
> a Laplace inversion (numerically is fine), but am I missing something?
> Is there an easier way? Laplace inversion seems like a sledgehammer
> for what looks like a very simple problem?
>
> Any guidance gratefully appreciated.
>
> Thanks,
>
> Richard.
You calculate the convolution integral.

Ciao

Karl
From: Ray Vickson on 26 Apr 2010 17:24
On Apr 26, 11:05 am, karl <oud...(a)nononet.com> wrote:
> Richard Hayden schrieb:
>
> > Hi,
>
> > Suppose we have Z = X + Y as non-negative real random variables, where
> > X and Y are independent.
>
> > Assume that we know the marginal pdfs for Z and Y, f_z(t) and f_y(t)
> > respectively. I want to compute the marginal pdf for X, say f_x(t).
> > Now I know f_z(t) = (f_x * f_y)(t), where * is the convolution
> > operator, so I need to invert the convolution to get f_x(t). I can see
> > how this can be done by taking Laplace transforms and then performing
> > a Laplace inversion (numerically is fine), but am I missing something?
> > Is there an easier way? Laplace inversion seems like a sledgehammer
> > for what looks like a very simple problem?
>
> > Any guidance gratefully appreciated.
>
> > Thanks,
>
> > Richard.
>
>  You calculate the convolution integral.

No, No, No. He has to "invert" the convolution integral, which is
equivalent to solving an integral equation.

R.G. Vickson

>
> Ciao
>
> Karl

From: Stephen Montgomery-Smith on 26 Apr 2010 23:07
Richard Hayden wrote:
> Hi,
>
> Suppose we have Z = X + Y as non-negative real random variables, where
> X and Y are independent.
>
> Assume that we know the marginal pdfs for Z and Y, f_z(t) and f_y(t)
> respectively. I want to compute the marginal pdf for X, say f_x(t).
> Now I know f_z(t) = (f_x * f_y)(t), where * is the convolution
> operator, so I need to invert the convolution to get f_x(t). I can see
> how this can be done by taking Laplace transforms and then performing
> a Laplace inversion (numerically is fine), but am I missing something?
> Is there an easier way? Laplace inversion seems like a sledgehammer
> for what looks like a very simple problem?
>
> Any guidance gratefully appreciated.
>
> Thanks,
>
> Richard.

I would use the Fourier transform instead of the Laplace transform. But
otherwise, I think the sledgehammer is needed.


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