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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. I've got the same plotting issue with Matlab.

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. Perhaps there is another way to derive the above equation with other functions or there might be a way to simplify it and get more accurate results.

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    $\begingroup$ In the last case, simply changing 12.85 to 11.85 causes a massive change in the expectation and produces a large negative value. I suspect a numerical issue. Using NExpectation gives a better result: With[{κ = 12.85, θ = 0.336551, a = 3, b = 1/5}, G = GammaDistribution[κ, θ]; a * NExpectation[q[Sqrt[b] y], y \[Distributed] G] ] (* 0.127737 *) $\endgroup$
    – flinty
    Jul 28, 2020 at 13:26
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    $\begingroup$ It's impolite of you to expect an answer without providing the code to reproduce. $\endgroup$ Jul 28, 2020 at 14:11

1 Answer 1

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It can be solved by increasing the precision with SetPrecision. See the excerpt below. This question has also been solved here: https://mathoverflow.net/questions/366754/inaccurate-results-for-the-analytical-expression-of-mathbbe-left-a-mathcal/366791#366798

prec = 100;
m = 16;
a = 4 (1 - (1/Sqrt[m]));
b = 3/(m - 1);

pdb = -12;
p = 10^(pdb/10);

betag = (1.5)^2;

betah = (2.3)^2;

bits = 8;

Q = 2^bits;

n = 8;

ef = SetPrecision[
   n p (betag  betah) (1 + ((n - 1)/16) (Q^2) (Sin[Pi/Q]^2)), prec];

ef2 = SetPrecision[
   n (p^2) (1/
      256) (((betag^2) (betah^2))) (512 (n + 
         1) + (32 (n - 1) (Q^2))/(Pi^2) + ((n - 
           1) (Q^2) (-32 Cos[(4 Pi)/Q] + 
           Pi (Sin[
               Pi/Q]^2) (16 (Pi + 4  n  Pi) + (n - 
                 2) Q ((n - 3) Pi Q (Sin[Pi/Q]^2) + 
                 16 Sin[(2 Pi)/Q]))))/(Pi^2)), prec];

a1 = SetPrecision[(ef2 - (ef^2)), prec];
a2 = SetPrecision[(ef2 - 5 (ef^2)), prec];
a3 = SetPrecision[-6 (ef^2), prec];

κ = SetPrecision[(-a2 + Sqrt[(a2^2) - 4 a1 a3])/(2 a1), prec];

θ = SetPrecision[Sqrt[ef/(κ (κ + 1))], prec];

f1 = SetPrecision[
   2^(-1 - κ/2) a b^(-κ/2) θ^-κ, prec];
f2 = SetPrecision[(κ HypergeometricPFQ[{1/2 + κ/2, 
      1 + κ/2}, {3/2, 3/2 + κ/2}, 1/(2 b θ^2)])/(
   Sqrt[2] Sqrt[b] θ Gamma[(3 + κ)/2]), prec];
f3 = SetPrecision[
   HypergeometricPFQ[{1/2 + κ/2, κ/2}, {1/2, 
     1 + κ/2}, 1/(2 b θ^2)]/Gamma[1 + κ/2], prec];

SetPrecision[f1 (f3 - f2), prec]
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