1
$\begingroup$

I have a random variable $X \sim \operatorname{Gamma}(\alpha, \beta)$.

How can I compute or approximate $\mathbb{E} \sin(X)$ very quickly? Iterative quadrature would be too slow, I need some closed form expression.

One idea I considered was to use the Gamma moments and Tyler approximation, but it would take too many terms, since $X$ has large standard deviation (in the order of 10–50) and so the mass is not tightly concentrated.

Improve this question
$\endgroup$
4
  • 2
    $\begingroup$ MathOverflow is for mathematicians to ask each other questions about their research. See Math.StackExchange to ask general questions in mathematics. $\endgroup$
    – M. Dus
    Commented Oct 17, 2018 at 16:27
  • $\begingroup$ It seems that sometimes $X\sim\operatorname{Gamma}(\alpha,\beta)$ means $$ \Pr(X\in S) = \frac 1 {\Gamma(\alpha)} \int_S \left( \frac x \beta \right) e^{-x/\beta} \, \left( \frac{dx} \beta\right) \quad \text{for } S\subseteq [0,+\infty) $$ and sometimes it means $$ \Pr(X\in S) = \frac 1 {\Gamma(\alpha)} \int_S (\beta x) e^{-\beta x} (\beta\,dx) \quad \text{for } S\subseteq[0,+\infty). $$ To say which you have in mind might be convenient. $\qquad$ $\endgroup$
    – Michael Hardy
    Commented Oct 18, 2018 at 2:05
  • 2
    $\begingroup$ I would first try writing $(\sin x) e^{-x/\beta}$ as $\operatorname{Im} \left( e^{-x/\beta + ix} \right). \qquad$ $\endgroup$
    – Michael Hardy
    Commented Oct 18, 2018 at 2:11
  • $\begingroup$ This seems a perfectly legitimate question for this site, unless one believes approximation theory and error analysis are somehow not "mathematical research" $\endgroup$
    – Yemon Choi
    Commented Nov 17, 2018 at 20:12

1 Answer 1

Reset to default
6
$\begingroup$

You can use the characteristic function. From Wikipedia, we have $$E[e^{itX}] = \left(1 - \frac{it}{\beta} \right)^{-\alpha}.$$

You have to be a bit careful when $\alpha$ isn't an integer, as you have a branch cut.

This means that $E[\sin(X)] = \operatorname{Im}\left[ \left(1 - \beta^{-1} i \right)^{-\alpha} \right]$. Writing $1 - \beta^{-1}i = r e^{i \theta}$ with $r = (1 + \beta^{-2})^{1/2}$ and $\theta = -\arctan(\beta^{-1})$ gives \begin{align} \operatorname E[ \sin(X)] & = r^{-\alpha} \operatorname{Im}(e^{-i\theta \alpha}) \\[8pt] & = (1 + \beta^{-2})^{-\alpha/2} \sin\left(\alpha \arctan(\beta^{-1}) \right). \end{align}

Improve this answer
$\endgroup$

You must log in to answer this question.

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