1
$\begingroup$

We know that the first order Marcum Q-function can be represented as $$Q_1(y, b)=\int_{b}^{\infty} x \exp{(-(x^2+y^2)/2)} I_0(y x) \, dx ,$$ where $I_0(\cdot)$ is the modified Bessel function of the first kind. In my case, $y$ and $b$ are positive.

I am trying to compute the following integral: $\int_0^{\infty} Q_1(y,b) \frac{y}{\sigma^2} \exp{(-y^2/(2\sigma^2))} \, dy$.
Note that $f(y) = \frac{y}{\sigma^2} \exp{(-y^2/(2\sigma^2))} $ is the PDF of the Rayleigh distribution.

I have no clue how to derive it. Any idea ?

$\endgroup$
  • $\begingroup$ the result seems to be $\exp(-b^2/(2+2\sigma^2))$. Change the order of integrations and do the $y$-integral first. $\endgroup$ – Pierpaolo Vivo Feb 2 '16 at 19:12
  • $\begingroup$ @PierpaoloVivo Could you please detail your comment in an answer so that I can accept it. $\endgroup$ – din Feb 2 '16 at 20:48
  • $\begingroup$ cannot right now, but will do $\endgroup$ – Pierpaolo Vivo Feb 2 '16 at 21:11
1
$\begingroup$

The integral can be written as $$ \frac{1}{\sigma^2}\int_b^\infty dx\ x\ e^{-x^2/2}\int_0^\infty dy\ y e^{-y^2 \left(\frac{1}{2}+\frac{1}{2\sigma^2}\right)}I_0(xy)\ . $$ Now we can use the following formula 6.633.4 of Gradshteyn-Ryzhik $$ \int_0^\infty dy\ y\ e^{-\alpha y^2}I_\nu (\beta y)J_\nu(\gamma y)=\frac{1}{2\alpha}\exp\left(\frac{\beta^2-\gamma^2}{4\alpha}\right)J_\nu\left(\frac{\beta\gamma}{2\alpha}\right)\qquad \mathrm{Re}\ \alpha>0,\quad\mathrm{Re}\ \nu\ , >-1$$ to perform the $y$-integral (noting that $J_0(0)=1$), resulting in $$ \frac{1}{\sigma^2}\frac{\sigma^2}{1+\sigma^2}\int_b^\infty\ dx\ x\ e^{-x^2/2}e^{\frac{\sigma^2 x^2}{2+2\sigma^2}}\ , $$ which is elementary, and gives eventually $$ \frac{1}{\sigma^2}\frac{\sigma^2}{1+\sigma^2}\left(\sigma ^2+1\right) e^{-\frac{b^2}{2 \sigma ^2+2}}=\exp(-b^2/(2\sigma^2+2))\ . $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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