# Conditional expectation conditioned on an Indicator variable

Suppose I have a random variable $$u$$ that is standard uniformly distributed.

And I have an indicator variable $$S_{i}=1\left(V_{i}>0.5\right)$$.

Now I am interested in the following conditional expectation. $$\mathbb{E}\left[u\mid S_{i}=1\right]$$.

Now $$u$$ is continuous whereas $$S_i$$ is discrete.

I do know that when X and Y are both continoious $$E[X \mid Y=y]=\int_{-\infty}^{\infty} x f_{X \mid Y}(x \mid y) d x$$. And if X and Y are both discrete $$E[X \mid Y=y]=\sum_{x} x f_{X \mid Y}(x \mid y)$$

However, I am not sure how to write $$\mathbb{E}\left[u\mid S_{i}=1\right]$$ as either a summation or integral.

I am trying to find

$$\mathrm{ATET}=-\delta_0\cdot \mathbb{E}\left[u^2\mid S_{i}=1\right]+\delta_1\cdot \mathbb{E}\left[u\mid S_{i}=1\right]+\delta_2$$

Where I know the joint distribution $$$$F_{U, V}(u, v)=u v+\theta u(1-u) v(1-v)$$$$ And u and v are uniformly distributed on [0,1]

When I follow your proposition I obtain $$\frac{1}{2}\int_0^1u*1du=\frac{1}{4}$$

However this does not seem to be a right solution to me

• What is the joint distribution of $(U,V)$? Mar 31 at 19:28
• The joint distribution is $F_{U, V}(u, v)=u v+\theta u(1-u) v(1-v)$
– WHN
Apr 1 at 11:45
• Is this a joint CDF? What is $\theta$? Apr 1 at 16:07
• This is indeed the joint CDF, $\theta\in R$ is just a parameter
– WHN
Apr 1 at 22:17
• $E[U\mid V>0.5]=\frac{E[UI(V>0.5)]}{P(V>0.5)}$. The numerator equals $\iint uI(v>0.5)\,f_{U,V}(u,v)\,du\,dv$ where $f_{U,V}$ is the joint pdf of $(U,V)$. Apr 2 at 10:14

"$$S_i=1$$" is exactly the event $$E = \{V_i>0.5\}$$ so the conditional expectation is given by $$\mathbb{E}\left[u\mid S_{i}=1\right] = \frac{1}{\mathbb P(E)}\int_E xf(x)dx$$ Where $$f$$ is the pdf of $$u$$.