1
$\begingroup$

I'm trying to calculate hit probabilities on a dart board if the dart thrower has some Gaussian angle distribution function with width $\Delta$ and some systematic angle offsets $\phi_0, \theta_0$. This is a problem related to my work.

The distribution function is

$p(\theta, \phi) = \frac{1}{2 \pi \Delta^2} e^{-\frac{(\phi-\phi_0)^2}{2\Delta^2}}e^{-\frac{(\theta-\theta_0)^2}{2\Delta^2}}$

To get the hit probability on the dart boad, this distribution function must be integrated over a circle centered around 0 with some radius $R$ at a far away distance $D$.

So far I turned in into an integral in circle coordinates and now I'm stuck with this bad boy:

$p_{tot} = e^{-\rho_0^2} \int_0^{\frac{R}{\sqrt{2} D \Delta}} d\rho \rho e^{-\rho^2} I_0(2\rho \rho_0)$

where $\rho = \sqrt{\phi^2 + \theta^2}$ and $\rho_0$ equivalently. $I_0$ is the 0th modified Bessel function.

Any ideas on how to properly evaluate this? I'd be happy about a reasonable approximation. Taylor expansions don't really work because the function in the integral has a peak that shifts to the right with increasing $\rho_0$. The area under the peak will grow like $e^{\rho_0^2}$ which can lead to some numerical issues but I guess I can approximate it with a delta distribution for really large $\rho_0$.

$\endgroup$
10
  • $\begingroup$ (i) What is actually your integral in terms of $\phi$ and $\theta$? (ii) Why does $D$ not enter your expression for $p_{tot}$? (iii) What do you mean by "far away distance $D$? That $D\to\infty$ (iv) Can you state your question formally, without darts, " far away distance", ...? $\endgroup$ Commented Jun 23, 2023 at 11:44
  • $\begingroup$ (v) How can angles, with values (say) in $[0,2\pi)$, be normally distributed? (vi) Is $\Delta$ small? $\endgroup$ Commented Jun 23, 2023 at 12:12
  • $\begingroup$ (i) The integral is over $\int \int p(\theta,\phi) d\theta d\phi$ and the limits are a circle with radius $R/D$ around (0,0). (ii) D is in the upper limit of the integral. (iii) R/D <<1 (iv) sorry, physicist (v) and (vi) $\Delta << 1$ $\endgroup$
    – mzw
    Commented Jun 23, 2023 at 18:26
  • $\begingroup$ (I) Can it be assumed that $R/D$ is much greater than $\Delta^2$ (and $R/D$ is much less than $1$)? (II) Can $\rho_0>0$ be assumed fixed? (III) Your expression for $p_{tot}$ is incorrect. (IV) Does your symbol $<<$ mean "much less"? $\endgroup$ Commented Jun 23, 2023 at 19:56
  • $\begingroup$ (i) No. (ii) No, I need a range of values for $\rho_0$ (iii) There may be a factor missing that's not really relevant for the Bessel integral. (iv) yes. $\endgroup$
    – mzw
    Commented Jun 23, 2023 at 20:24

1 Answer 1

1
$\begingroup$

Q: The OP seeks a "reasonable approximation" for large or small $\rho_0$ of the function $$P(x)= \int_0^{x} e^{-\rho^2-\rho_0^2} \rho I_0(2\rho \rho_0)\,d\rho,\;\;x\geq 0.$$


Consider the kernel $$p_{\rm exact}(\rho)=e^{-\rho^2-\rho_0^2} \rho I_0(2\rho \rho_0).$$ For small $\rho_0$ a Taylor series in powers of $\rho_0$ is accurate, $$p_{\rm small}(\rho)=e^{-\rho^2} \rho (1+\rho^2 \rho_0^2-\rho_0^2).$$ For large $\rho_0$ an asymptotic expansion of the Bessel function gives $$p_{\rm large}(\rho)=\frac{1}{2\sqrt{\pi}}e^{-\rho^2-\rho_0^2}e^{2\rho\rho_0}.$$

In the plots below I compare $\int_0^x p(\rho)\,d\rho$ for the exact expression (blue) and the approximation (orange). It turns out the approximations $p_{\rm small}$ and $p_{\rm large}$ are already quite accurate for $\rho_0\lesssim 0.5$ and $\rho_0\gtrsim 3$, respectively.

Left plot: comparison of $\int_0^x p_{\rm exact}(\rho)\,d\rho$ and $\int_0^x p_{\rm small}(\rho)\,d\rho$ for $\rho_0=0.5$. Right plot: comparison of $\int_0^x p_{\rm exact}(\rho)\,d\rho$ and $\int_0^x p_{\rm large}(\rho)\,d\rho$ for $\rho_0=3$. The approximations (orange) are almost indistinguishable from the exact answer (blue).

$\endgroup$
9
  • $\begingroup$ Do you have a bound on the accuracy of your approximation of the integral? $\endgroup$ Commented Jun 23, 2023 at 11:43
  • $\begingroup$ I feel like there is a mistake somewhere here. I0 scales like exp(x)/sqrt(2pi x) which still leaves me with a non-analytic integral. $\endgroup$
    – mzw
    Commented Jun 23, 2023 at 18:20
  • $\begingroup$ the asympotic expansion of $\rho I_0(2\rho\rho_0)$ gives the exponential of $ 2\rho\rho_0 +(1/2)\log\rho$, and for large $\rho_0$ the log can be neglected. $\endgroup$ Commented Jun 23, 2023 at 21:03
  • $\begingroup$ Also, as noted in my other comment, the expression for $p_{tot}$ in the OP is incorrect. $\endgroup$ Commented Jun 23, 2023 at 21:04
  • $\begingroup$ @CarloBeenakker : According to this comment by the OP, we cannot assume that $\rho_0$ is large. $\endgroup$ Commented Jun 23, 2023 at 21:06

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .