# Converting a Conditional Expectation into an Unconditional one

Consider two random variables $$V_1$$ and $$V_2$$, where $$V_1$$ is continuous and $$V_2$$ is binary.

In a paper, I found that the following equality holds: \begin{align*} \mathbb{E}\left[V_1|V_2=1\right]=\mathbb{E}\left[\frac{V_2}{\mathbb{E}[V_2]}V_1\right] \end{align*} I remember that I have proved that this equality holds without any assumption about two years ago.

But, I cannot come up with the proof again.

So, does that equality holds always? or other assumptions are needed?

• You should include reference or link to that paper! Commented Jul 10, 2023 at 12:26

In general, for any random variable $$X$$ and an event $$A$$ with $$P(A) > 0$$, $$E[X|A]$$ is defined as (see this question, this question and this question for related discussions): \begin{align} E[X|A] = \frac{E[XI_A]}{P(A)}. \tag{1}\label{1} \end{align}
Now taking $$X = V_1$$ and $$A = [V_2 = 1]$$ in $$\eqref{1}$$, and using the fact that $$V_2$$ is binary (so that $$I_{[V_2 = 1]} = V_2$$ and $$P(V_2 = 1) = E[V_2]$$) completes the proof.
• @Zhanxiong Dear Zhanxiong, I have an additional question. As you explained $E\left[V_1|V_2=1\right]=E\left[\frac{V_2}{E[V_2]}V_1\right]$ hold. Then, is there any way to convert the conditional mean to an unconditional one? For example, if we multiply $E[V_2]$ into $V_1$, then $E\left[V_1|V_2=1\right]=E\left[V_2V_1\right]$, which is still not $E[V_1]$. Commented Jul 11, 2023 at 12:03
• @MinChulPark Isn't $E[V_2V_1]$ already an unconditional expectation? Of course you cannot expect $E[V_1|V_2 = 1] = E[V_1]$ in general (which holds when $V_1$ and $V_2$ are independent). Commented Jul 11, 2023 at 13:37