5
$\begingroup$

I have the following transform:

$$F(y) = \int_{0}^{\infty} y\exp{\left[-\frac{1}{2}(y^2 + x^2)\right]} I_0\left(xy\right)f(x)\;\mathrm{d}x$$

with the following conditions:

  • $f(x)$ and $F(y)$ must be real and positive (they are continuous probability distributions)
  • $x,y$ must be real and positive (they are magnitudes)

@Carlo suggested the following approach. Re-arrange such that we have: $$\frac{F(y)}{y}\exp{\left[\frac{1}{2}y^2\right]} = \int_{0}^{\infty} \left(\frac{f(x)}{x}\exp{\left[-\frac{1}{2} x^2\right]}\right) I_0\left(xy\right)x\;\mathrm{d}x$$

Now define:

$$ G(y) = \frac{F(y)}{y}\exp{\left[\frac{1}{2}y^2\right]} \quad\quad g(x) = \frac{f(x)}{x}\exp{\left[-\frac{1}{2} x^2\right]}$$

We can now write: $$G(y) = \int_{0}^{\infty} g(x) I_0\left(xy\right)x\;\mathrm{d}x$$

Make the substitution $y = iq$ along with the fact that $I_0(ixq) = J_0(-xq) = J_0(xq)$ and we have:

$$G(iq) = \int_{0}^{\infty} g(x) J_0\left(xq\right)x\;\mathrm{d}x$$

This is a Hankel transform, which has an inverse: $$g(x)=\int_{0}^{\infty}G(iq)J_{0}(xq)q\;\mathrm{d}q$$

Substituting in $q = \frac{1}{i}y$ and $\mathrm{d}q = \frac{1}{i}\mathrm{d}y$ we obtain:

$$g(x)= -\int_{0}^{i\infty}G(y)I_{0}(xy)y\;\mathrm{d}y$$

Substituting out $G(y), \;g(x)$:

$$\frac{f(x)}{x}\exp{\left[-\frac{1}{2} x^2\right]} = -\int_{0}^{i\infty} \frac{F(y)}{y}\exp{\left[\frac{1}{2}y^2\right]} I_{0}(xy)y\;\mathrm{d}y$$

and finally solve for $f(x)$:

$$f(x) = -\int_{0}^{i\infty} x\exp{\left[\frac{1}{2}(y^2 + x^2)\right]} I_0\left(xy\right)F(y)\;\mathrm{d}y$$

  • Is this inversion valid given the aforementioned conditions?

  • How can I perform this integral numerically?

I'm directly measuring $F(y)$ (by normalising the histogram of a large set of values) and so I need to calculate $f(x)$ by doing this integral numerically. However I don't understand how to do this when the data is all real but the integral is over the imaginary axis. Can anyone help?

Thank you!

$\endgroup$

1 Answer 1

6
$\begingroup$

define $G(y)=y^{-1}\exp(y^2/2)(Tf)(y)$ and $g(x)=x^{-1}\exp(-x^2/2)f(x)$, then you seek the solution to the integral equation

$$G(iq)=\int_{0}^{\infty}g(x)J_{0}(xq)xdx$$

This is a Hankel transform. The inverse is

$$g(x)=\int_{0}^{\infty}G(iq)J_{0}(xq)qdq$$

$\endgroup$
7
  • $\begingroup$ Thanks for your answer! although in the two integrals you have $f(x)$ when i think you mean $g(x)$? I've edited my question to include the derivation you suggest, along with the extra steps needed to get $f(x)$. Would you mind checking my logic is sound? Thanks again. $\endgroup$
    – CBowman
    Oct 11, 2013 at 1:45
  • $\begingroup$ thanks for spotting the typo, I corrected it; with respect to your update: when you change variables in the integral from $q$ to $y=iq$, the integral over $y$ would then run along the positive imaginary axis. $\endgroup$ Oct 11, 2013 at 7:27
  • $\begingroup$ Ah I see, I've never studied integration in the complex plane; what is the appropriate notation to denote that it is along the positive imaginairy axis? thanks $\endgroup$
    – CBowman
    Oct 11, 2013 at 9:43
  • $\begingroup$ the way to think about this, is that the Hankel transform with $J_0$ is like a Fourier transform, whereas your transform with $I_0$ is like a Laplace transform. The inversion formula for the Laplace transform also involves an integration along the imaginary axis $\int_{0}^{i\infty}dy$ $\endgroup$ Oct 11, 2013 at 9:46
  • $\begingroup$ That's a help picture to have in mind, thanks again! I'll change the upper limit to $i\infty$ where needed. $\endgroup$
    – CBowman
    Oct 11, 2013 at 9:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.