Value of $\mathbb{E}(\frac{1}{aX+b})$? There are rules e.g. $\mathbb{E}(aX+bY)=a\mathbb{E}(X)+b\mathbb{E}(Y)$, where $X$, $Y$ are random variables and $a$ and $b$ are constants.
But what about $\mathbb{E}(\frac{1}{aX+b})$?
 A: If $\left|\frac{a}{b}X\right|<1$ (almost surely) and $X$ has all its moments such that the following converges, then:
$$E[\frac{1}{aX+b}]=\frac{1}{b}(1-E[aX/b]+E[(aX/b)^2]-E[(aX/b)^3]+\cdots),$$
which can be calculated if you know the moments of $X$. On the other hand, occasionally it's a reasonable approximation to cutoff the series, say:
$$E[\frac{1}{aX+b}]\approx\frac{1}{b}(1-E[aX/b]),$$
which holds when $X$ is small. 
Such techniques are actually quite useful in other areas, especially when you know one random variable is dominated by another in size. See here.
A: OP has declined to revise the question to reflect the elaborations and questions s/he has posed in the comments. I think it's entirely sufficient to indicate that the answer to this question is simply to apply the Law of the Unconscious Statistician. The Law states that in the discrete case,
$$
E(g(X))=\sum_A g(x)p_X(x)
$$
where $g$ is some function and $p_X(x)$ is the pmf of $x$ and $A=\{x:p_X(x)>0\}.$ Note that you must also check that the sum (or integral) is absolutely convergent, cf Riemann's rearrangement theorem.
Downvoters are invited to describe why this answer is incorrect.
A: You cannot move the expectation in the denominator if that is what you are asking. Due to Jensen's inequality you would end up either understimating or overestimating the expectation, depending on the value of $\alpha$. 
What I would do is try to first find the distribution of the linear transformation then of its reciprocal and finally compute the expectation of the final distribution. Equivalently, you can just use the Law of the Unconsious Statistican and avoid all this hustle.
