Bounds on conditional expectation in terms of unconditional? 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? 
 A: 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
