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I'm trying to analytically find the following expectation

$$\mathbb{E}\left[ a \mathcal{Q} \left( \sqrt{b } \gamma \right) \right],$$ where $a$ and $b$ are constant values, $\mathcal{Q}$ is the Gaussian Q-function, which is defined as $\mathcal{Q}(x) = \frac{1}{2 \pi}\int_{x}^{\infty} e^{-u^2/2}du$ 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} $.

By using Mathematica, I've found the following solution:

$$\mathbb{E}\left[ a \mathcal{Q} \left( \sqrt{b } \gamma \right) \right] = a 2^{-\frac{\kappa }{2}-3} b^{-\frac{\kappa }{2}-\frac{1}{2}} \theta ^{-\kappa -1} \left(2 \sqrt{2} \sqrt{b} \theta \, _2\tilde{F}_2\left(\frac{\kappa +1}{2},\frac{\kappa }{2};\frac{1}{2},\frac{\kappa +2}{2};\frac{1}{2 b \theta ^2}\right)-\kappa \, _2\tilde{F}_2\left(\frac{\kappa +1}{2},\frac{\kappa +2}{2};\frac{3}{2},\frac{\kappa +3}{2};\frac{1}{2 b \theta ^2}\right)\right),$$

however, I'd like to know the steps to find this solution or to find another one.

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$\newcommand\Ga\Gamma$ Without loss of generality $a=1$. Let then $Q:=\mathcal Q$, $k:=\kappa>0$, and $t:=\theta\sqrt b>0$, so that $\sqrt b\,\gamma$ has the gamma distribution with parameters $k,t$. Let also $c:=\Ga(k)t^k$. Then, letting $f$ denote the standard normal pdf, we have
$$c\,EaQ(\sqrt b\,\gamma)=\int_0^\infty dy\,y^{k-1} e^{-y/t}Q(y) =\sum_{j=0}^\infty\frac{(-1/t)^j}{j!}I_j,\tag{1}$$ where $$I_j:=\int_0^\infty dy\,y^{k+j-1}Q(y)=\frac1{k+j}\,\int_0^\infty dy\,y^{k+j}f(y) \\ =\frac{2^{(j+k)/2-1} \Ga((j+k+1)/2)}{(k+j)\sqrt{\pi }};\tag{2}$$ the penultimate equality here is obtained by integration by parts and the last equality is obtained by the substitution $w=y^2/2$. Using now the identity $\Ga(x+1)=x\Ga(x)$ about $j/2$ times to reduce $\Ga((j+k+1)/2)$ to $\Ga((k+1)/2)$ for even $j$ and to $\Ga(k/2)$ for odd $j$, from (1) and (2) we get $$c_1\,EQ(\sqrt b\,\gamma) \\ = \sqrt{2} (k+1) t \Gamma \left(\frac{k+1}{2}\right) \, _2F_2\left(\frac{k}{2}+\frac{1}{2},\frac{k}{2};\frac{1}{2},\frac{k}{2}+1;\frac{1}{2 t^2}\right) \\ -k^2 \Gamma \left(\frac{k}{2}\right) \, _2F_2\left(\frac{k}{2}+\frac{1}{2},\frac{k}{2}+1;\frac{3}{2},\frac{k}{2}+\frac{3}{2};\frac{1}{2 t^2}\right),$$ where $c_1:=c\,\sqrt{\pi } k (k+1) t\,2^{(3-k)/2}$.

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  • $\begingroup$ Dear, I'm trying to find the equation below starting from your findings, but the terms multiplying the hypergeometric functions do not match the ones below. I'm not sure if I'm doing something wrong or if there is something missing in your answer. $\endgroup$ – Felipe Augusto de Figueiredo Jul 20 at 17:57
  • $\begingroup$ @FelipeAugustodeFigueiredo : what do you mean by "the equation below"? $\endgroup$ – Iosif Pinelis Jul 20 at 18:47
  • $\begingroup$ Sorry, I meant the equation found by Carlo,i.e., the other answer to the question. $\endgroup$ – Felipe Augusto de Figueiredo Jul 20 at 18:49
  • $\begingroup$ @FelipeAugustodeFigueiredo : I have just verified my result numerically for several instances of $(k,t)$. This, together with the above proof, makes be believe that my answer is correct. $\endgroup$ – Iosif Pinelis Jul 20 at 19:11
  • $\begingroup$ Dear, perhaps you could help me with a question related to your answer. Please, check the link mathoverflow.net/questions/366754/… $\endgroup$ – Felipe Augusto de Figueiredo Jul 28 at 11:37
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Mathematica gives an answer in terms of a hypergeometric function:

$$\frac{a 2^{-\frac{\kappa}{2}-\frac{3}{2}} b^{-\frac{\kappa}{2}} {\theta}^{-k} \, _2F_2\left(\frac{\kappa}{2}+\frac{1}{2},\frac{\kappa}{2};\frac{1}{2},\frac{\kappa}{2}+1;\frac{1}{2 b {\theta}^2}\right)}{\sqrt{\pi } \Gamma \left(\frac{\kappa}{2}+1\right)}-\frac{a 2^{-\frac{\kappa}{2}-2} \kappa b^{-\frac{\kappa}{2}-\frac{1}{2}} {\theta}^{-k-1} \, _2F_2\left(\frac{\kappa}{2}+\frac{1}{2},\frac{\kappa}{2}+1;\frac{3}{2},\frac{\kappa}{2}+\frac{3}{2};\frac{1}{2 b {\theta}^2}\right)}{\sqrt{\pi } \Gamma \left(\frac{\kappa}{2}+\frac{3}{2}\right)}$$

For integer $\kappa$ this reduces to an error function, for example, for $\kappa=1$ one has $a \left(1-e^{\frac{1}{2 b \theta^2}} \text{erfc}\left(\frac{1}{ \sqrt{2b} \theta}\right)\right)(2 \sqrt{2 \pi })^{-1}$. For larger integer values of $\kappa=1,2,3,4,5,6$, taking $a=\sqrt{2\pi}$, $b=1/2$, $\theta=1$ for ease of notation, one has $$\frac{1}{2} (1-e \,\text{erfc}(1)),\frac{1}{2} \left(e\, \text{erfc}(1)-\frac{2}{\sqrt{\pi }}+1\right),\frac{1}{2}-e\, \text{erfc}(1),\frac{1}{6} \left(4 e \,\text{erfc}(1)-\frac{8}{\sqrt{\pi }}+3\right),\frac{1}{12} \left(-11 e \,\text{erfc}(1)-\frac{2}{\sqrt{\pi }}+6\right),\frac{1}{60} \left(23 e \,\text{erfc}(1)-\frac{70}{\sqrt{\pi }}+30\right)$$

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  • $\begingroup$ Dear, could you, please, post the Mathematica code? I've found a similar answer however, mine is in terms of HypergeometricPFQRegularized functions. $\endgroup$ – Felipe Augusto de Figueiredo Jul 20 at 13:21
  • $\begingroup$ just apply FunctionExpand to your output $\endgroup$ – Carlo Beenakker Jul 20 at 13:36
  • $\begingroup$ Thanks Carlo. Onother question, how did you figured out that it can be reduced to an error function for integer values of $\kappa$? Is it possible to have a expression involving the error function for any integer value of $\kappa$? $\endgroup$ – Felipe Augusto de Figueiredo Jul 20 at 14:14
  • $\begingroup$ sorry, no general expression for arbitrary integer $\kappa$. $\endgroup$ – Carlo Beenakker Jul 20 at 14:44
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    $\begingroup$ done: mathoverflow.net/a/366791/11260 $\endgroup$ – Carlo Beenakker Jul 28 at 13:42

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