Suppose I am interested in calculating $\mathbb{E}(X|A)$, where $A$ is a certain fixed event and $X\geq{0}$. I am motivated by an application in which $0<P(A)=1-\varepsilon$, where $\varepsilon$ is small (i.e. conditioning on a highly probable event).
Intuitively, it seems to me that if $P(A)$ gets arbitrarily close to 1, the effect of conditioning should vanish and $\mathbb{E}(X|A)$ should get close to $\mathbb{E}X$. I can formalise this by giving an upper bound: $$\mathbb{E}(X|A)\leq\frac{\mathbb{E}X}{P(A)}.$$
However, can a lower bound (or, alternatively, some approximation) be given?
1) On the one hand, the best lower-bound of the type you are looking for is 0, since for any value $E[X]>0$ and any $\epsilon>0$ you can define $$X =\left\{ \begin{array}{ll} E[X]/\epsilon &\mbox{ with prob $\epsilon$} \\ 0 & \mbox{ with prob $1-\epsilon$} \end{array} \right.$$
2) On the positive side, you can say: $$E[X|A] \geq \inf_{B: P[B]\geq P[A]}E[X|B] \geq E[X|X\leq \theta]$$ where $\theta$ is any value for which $P[X\leq \theta] \leq P[A]$. I actually used this idea in paper once. The intuition is that, assuming there is a $\theta$ for which $P[X\leq \theta] = P[A]$, the "infimizing event" is indeed the event $\{X \leq \theta\}$.
Here is a formal proof of the above inequality. Let $X$ be a random variable, let $A$ be an event such that $P[A]>0$, and let $\theta$ be a real number such that $0<P[X\leq \theta] \leq P[A]$.
Claim: $E[X|A] \geq E[X|X\leq \theta]$.
Proof: Define indicator functions: \begin{align} 1_A &= \left\{ \begin{array}{ll} 1 &\mbox{ if $A$ occurs} \\ 0 & \mbox{ otherwise} \end{array} \right.\\ 1_{X\leq\theta} &= \left\{ \begin{array}{ll} 1 &\mbox{ if $X\leq \theta$} \\ 0 & \mbox{ otherwise} \end{array} \right. \end{align} We have: $$ X1_A = X1_{X\leq \theta} + X(1_A-1_{X\leq \theta}) \quad (1)$$ Now note that: $$ X(1_A-1_{X\leq \theta}) \geq \theta (1_A-1_{X\leq \theta}) \quad (2) $$ This can be seen by considering the cases $X>\theta$ and $X\leq \theta$. Now substitute (2) into (1):
$$X1_A \geq X1_{X\leq \theta} + \theta (1_A-1_{X\leq\theta}) $$ Taking expectations of the above gives: \begin{align} E[X|A]P[A] &\geq E[X|X\leq \theta]P[X\leq\theta] + \theta(P[A]-P[X\leq\theta])\\ &=E[X|X\leq \theta]P[A] + \underbrace{(\theta- E[X|X\leq\theta])}_{\geq 0}\underbrace{(P[A]-P[X\leq\theta])}_{\geq 0}\\ &\geq E[X|X\leq \theta]P[A] \end{align} Dividing by $P[A]$ gives the result. $\Box$
Years ago I used a similar result to prove Theorem 8 in a "Capacity and Delay Tradeoffs" paper: http://www-bcf.usc.edu/~mjneely/pdf_papers/neely_capacity_delay_tradeoff_it.pdf