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I'm trying to plot a graph for the following expectation

$$\mathbb{E}\left[ a \mathcal{Q} \left( \sqrt{b } \gamma \right) \right]=a 2^{-\frac{\kappa }{2}-1} b^{-\frac{\kappa }{2}} \theta ^{-\kappa } \left(\frac{\, _2F_2\left(\frac{\kappa }{2}+\frac{1}{2},\frac{\kappa }{2};\frac{1}{2},\frac{\kappa }{2}+1;\frac{1}{2 b \theta ^2}\right)}{\Gamma \left(\frac{\kappa }{2}+1\right)}-\frac{\kappa \, _2F_2\left(\frac{\kappa }{2}+\frac{1}{2},\frac{\kappa }{2}+1;\frac{3}{2},\frac{\kappa }{2}+\frac{3}{2};\frac{1}{2 b \theta ^2}\right)}{\sqrt{2} \sqrt{b} \theta \Gamma \left(\frac{\kappa +3}{2}\right)}\right)$$ where $a$ and $b$ are constant values, $\mathcal{Q}$ is the Gaussian Q-function, which is defined as $\mathcal{Q}(x) = \frac{1}{\sqrt{2 \pi}}\int_{x}^{\infty} e^{-u^2/2}du$ and $\gamma$ is a random variable with Gamma distribition, i.e., $f_{\gamma}(y) \sim \frac{1}{\Gamma(\kappa)\theta^{\kappa}} y^{\kappa-1} e^{-y/\theta} $ with $\kappa > 0$ and $\theta > 0$.

This equation was also found with Mathematica, so it seems to be correct.

Follows some examples, where I have checked the analytical results against the simulated ones.

When $\kappa = 12.85$, $\theta = 0.533397$, $a=3$ and $b = 1/5$ it returns the correct value $0.0218116$.

When $\kappa = 12.85$, $\theta = 0.475391$, $a=3$ and $b = 1/5$ it returns the correct value $0.0408816$.

When $\kappa = 12.85$, $\theta = 0.423692$, $a=3$ and $b = 1/5$ it returns the value $-1.49831$, which is negative. However, the correct result should be a value around $0.0585$.

When $\kappa = 12.85$, $\theta = 0.336551$, $a=3$ and $b = 1/5$ it returns the value $630902$. However, the correct result should be a value around $0.1277$.

Therefore, the issue happens as $\theta$ decreases. For values of $\theta > 0.423692$ the analytical matches the simulated results. The issue only happens when $\theta <= 0.423692$.

I'd like to know if that is an accuracy issue or if I'm missing something here and if there is a way to correctly plot a graph that matches the simulation.

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    $\begingroup$ This might be a bug in Mathematica kernel (I have already found one such issue with the $_2F_1$ function — Wolfram corrected it within a few months after I contacted them), but I suspect simple rounding error here: you seem to be subtracting huge values here, the hypergeometric function is of the order $10^11$ to $10^16$ for the given problematic parameters. $\endgroup$ – Mateusz Kwaśnicki Jul 28 at 7:33
  • $\begingroup$ Maybe using a higher PrecisionGoal value helps. But this might also increase the calculation time considerably. $\endgroup$ – Johannes Trost Jul 28 at 8:50
  • $\begingroup$ @MateuszKwaśnicki, I have also seen the same problem plotting with matlab... perhaps there is a way to simplify or derive the above expression another way with other functions.... $\endgroup$ – Felipe Augusto de Figueiredo Jul 28 at 12:20
  • $\begingroup$ @FelipeAugustodeFigueiredo: This is perhaps better suited to mathematica.se then. Anyway, have you tried working with higher precision? This can be done by writing SetPrecision[12.85, 50] etc. for all numeric constants. $\endgroup$ – Mateusz Kwaśnicki Jul 28 at 12:40
  • $\begingroup$ @MateuszKwaśnicki, thanks for your suggestion, with that I was able to correctly get the expected results. $\endgroup$ – Felipe Augusto de Figueiredo Jul 28 at 14:59
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You just need to use a higher precision, using exact numbers when possible, as shown in the Mathematica work below (with $k:=\kappa$ and $t:=\theta$). (However, this question is indeed better suited for Mathematica SE.)

enter image description here

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The accuracy issue with the evaluation of the hypergeometric function can be avoided for integer $\kappa$, since then the full expression reduces to an error function (see my answer to your previous question).

I tried this out for $\kappa=5$. Then $$\mathbb{E}\left[ a \mathcal{Q} \left( \sqrt{b } \gamma \right) \right]=\frac{a}{48 b^4 {\theta}^8} \left[\sqrt{\frac{2b}{\pi }} {\theta} \left(b {\theta}^2 \left(4 b {\theta}^2 \left(3 b {\theta}^2 \left(\sqrt{2 \pi b} {\theta}-2\right)+1\right)+1\right)+1\right)-e^{\frac{1}{2 b {\theta}^2}} \left(b {\theta}^2 \left(3 b {\theta}^2 \left(8 b^2 {\theta}^4-4 b {\theta}^2+1\right)+2\right)+1\right) \text{erfc}\left(\frac{1}{\sqrt{2b} {\theta}}\right)\right].$$

In the plot, for $a=3,b=1/5$ as a function of $\theta$, I compare the above expression with the error function (gold) with a numerical evaluation of the original expression with the hypergeometric function (blue). You see that the two evaluations agree for sufficiently large $\theta$, but then upon reducing $\theta$ the latter becomes numerically unstable while the former does not.

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