# How to find the expectation $\mathbb{E}\left[ a \mathcal{Q} \left( \sqrt{b } \gamma \right) \right]$?

I'm trying to 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., $$f_{\gamma}(y) \sim \frac{1}{\Gamma(\kappa)\theta^{\kappa}} y^{\kappa-1} e^{-y/\theta}$$.

I have tried to solve that with Mathematica, however, it says it does not converge.

• Define: q[x_] := 1/(2 π) Integrate[Exp[-u^2/2], {u, x, ∞}] G = GammaDistribution[α, β, γ, μ]; gpdf = PDF[G, g];. Then Expectation[a q[Sqrt[b] g], g \[Distributed] G] , but this will not work, neither will Integrate[a q[Sqrt[b] g]*gpdf, {g, 0,∞}]. It's too complicated for Mathematica - however if you have values for the gamma distribution parameters, then this is quite easy even with unknown constants a,b. – flinty Jul 20 '20 at 1:52

## 1 Answer

It's not too hard to set this up. I had no luck with the more complex four parameter GammaDistribution but fortunately you only mentioned the two parameter version:

q[x_] = 1/(2 π) Integrate[Exp[-u^2/2], {u, x, ∞}]
G = GammaDistribution[κ, θ];
gpdf = PDF[G, y]
result = a*Expectation[q[Sqrt[b] y], y \[Distributed] G]


Result:

2^(-3 - κ/2) a b^(-(1/2) - κ/
2) θ^(-1 - κ) (2 Sqrt Sqrt[
b] θ HypergeometricPFQRegularized[{(1 + κ)/
2, κ/2}, {1/2, (2 + κ)/2}, 1/(
2 b θ^2)] - κ HypergeometricPFQRegularized[{(
1 + κ)/2, (2 + κ)/2}, {3/2, (3 + κ)/2}, 1/(
2 b θ^2)])


Let's do a quick check to make sure it returned something reasonable. We'll generate some gamma distributed random numbers, apply the function and get the mean. Then also use these constants with our result obtained earlier to confirm it's a close match to this numerical experiment:

qn[x_?NumericQ] := 1/(2 π) NIntegrate[Exp[-u^2/2], {u, x, ∞}]
With[{κ = 1, θ = 2, a = 4, b = 3},
rands = RandomVariate[GammaDistribution[κ, θ], 5000];
Mean[a*qn[Sqrt[b]*#] & /@ rands]
]
(* result: 0.155478 *)

N[result /. {κ -> 1, θ -> 2, a -> 4, b -> 3}]
(* result: 0.15502 *)


Looks about right!

Your version of the Q-function has 1/(2 π). I think this should be 1/Sqrt[2 π] instead. In which case the result changes:

q[x_] = 1/Sqrt[2 π] Integrate[Exp[-u^2/2], {u, x, ∞}]
G = GammaDistribution[κ, θ];
gpdf = PDF[G, y]
result = a*Expectation[q[Sqrt[b] y], y \[Distributed] G]


Result:

2^(-(5/2) - κ/2) a b^(-(1/2) - κ/
2) Sqrt[π] θ^(-1 - κ) (2 Sqrt Sqrt[
b] θ HypergeometricPFQRegularized[{(1 + κ)/
2, κ/2}, {1/2, (2 + κ)/2}, 1/(
2 b θ^2)] - κ HypergeometricPFQRegularized[{(
1 + κ)/2, (2 + κ)/2}, {3/2, (3 + κ)/2}, 1/(
2 b θ^2)])

• +1. Also, dist = TransformedDistribution[a q[Sqrt[b] y], y \[Distributed] GammaDistribution[\[Kappa], \[Theta]]]; Mean[dist]] works, too. – JimB Jul 20 '20 at 3:27
• +1 Recommend that the definition of q use Set rather than SetDelayed so that the integral is only done once. Also, in the check with random variates you can use q rather than qn. – Bob Hanlon Jul 20 '20 at 3:50
• For whatever it's worth: $q(x)=\frac{\text{erfc}\left(\frac{x}{\sqrt{2}}\right)}{2 \sqrt{2 \pi }}$. – JimB Jul 20 '20 at 4:11