Let X be a positive r.v, and let A an event with positive measure. For the law of total expectation it is possible to express the expectation as:
\begin{equation}
\mathbb{E}[X]=\mathbb{E}[X|A]\mathbb{P}(A)+\mathbb{E}[X|A^c]\mathbb{P}(A^c)
\end{equation}
Assume now B to be another event such that $A \subset B $. Conditioning $X$ on $B$, I should obtain:
\begin{split}
\mathbb{E}[X | B]=\mathbb{E}[X|A,B]\mathbb{P}(A|B)+\mathbb{E}[X|A^c,B]\mathbb{P}(A^c|B)= \\
\mathbb{E}[X | B]=\mathbb{E}[X|A]\mathbb{P}(A)/\mathbb{P}(B)+\mathbb{E}[X|A^c,B]\mathbb{P}(A^c|B) \\
\end{split}
Assume now,
- $\mathbb{E}[X|A]>\mathbb{E}[X]$
- $\mathbb{E}[X|A^c,B]<\mathbb{E}[X|A^c]<\mathbb{E}[X]$
I want to prove whether $\mathbb{E}[X|B]>\mathbb{E}[X]$. This seems logical to me, since $\mathbb{E}[X|A]$ is bigger than $\mathbb{E}[X]$ and we increase the probability of this event. At the same time, however, $\mathbb{E}[X|A^c,B]$ decreases. In addition, if this is not the case, is there a lower bound
for $\mathbb{E}[X|A^c,B]$ such that $\mathbb{E}[X|B]>\mathbb{E}[X]$?
Best Answer
As you noted:
$$\begin{align}\mathbf{E}[X\mid B]&=\mathbf{E}[X\mid B\cap A]\mathbf{P}[A\mid B]+\mathbf{E}[X\mid B\cap A^c]\mathbf{P}[A^c\mid B]\\ &=\mathbf{E}[X\mid A]\mathbf{P}[A]+\mathbf{E}[X\mid B\cap A^c]\mathbf{P}[A^c\mid B]\end{align}$$
Now as you may notice, the condition $\mathbf{E}[X\mid A]>\mathbf{E}[X]$ is not sufficient to prove that $\mathbf{E}[X\mid B]>\mathbf{E}[X]$ (just pick some event $A$ ridiculously small). Also, the additional assumption that $\mathbf{E}[X\mid B\cap A^c]$ is small won't help in order to prove that $\mathbf{E}[X\mid B]$ is large!
If you want to have $\mathbf{E}[X\mid B]>\mathbf{E}[X]$, you need to have:
$$\mathbf{E}[X\mid B\cap A^c]>\frac{\mathbf{E}[X]-\mathbf{E}[X\mid A]\mathbf{P}[A]}{\mathbf{P}[A^c\mid B]}$$
And if you already know that $\mathbf{E}[X\mid A]>\mathbf{E}[X]$, a sufficient condition for $\mathbf{E}[X\mid B]>\mathbf{E}[X]$ to hold is:
$$\mathbf{E}[X\mid B\cap A^c]>\mathbf{E}[X]\frac{1-\mathbf{P}[A]}{\mathbf{P}[A^c\mid B]}=\frac{\mathbf{E}[X]\mathbf{P}[A^c]\mathbf{P}[B]}{\mathbf{P}[A^c\cap B]}$$
You can't do much better than that in general.