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 \begin{equation} F_{U, V}(u, v)=u v+\theta u(1-u) v(1-v) \end{equation} 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

  • $\begingroup$ What is the joint distribution of $(U,V)$? $\endgroup$ Commented Mar 31, 2022 at 19:28
  • $\begingroup$ The joint distribution is $F_{U, V}(u, v)=u v+\theta u(1-u) v(1-v)$ $\endgroup$
    – WHN
    Commented Apr 1, 2022 at 11:45
  • $\begingroup$ Is this a joint CDF? What is $\theta$? $\endgroup$ Commented Apr 1, 2022 at 16:07
  • $\begingroup$ This is indeed the joint CDF, $\theta\in R$ is just a parameter $\endgroup$
    – WHN
    Commented Apr 1, 2022 at 22:17
  • $\begingroup$ $E[U\mid V>0.5]=\frac{E[UI(V>0.5)]}{P(V>0.5)}$. The numerator equals $\iint uI(v>0.5)\,f_{U,V}(u,v)\,du\,dv$ where $f_{U,V}$ is the joint pdf of $(U,V)$. $\endgroup$ Commented Apr 2, 2022 at 10:14

1 Answer 1


"$S_i=1$" 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 $u$.

  • $\begingroup$ Thank you for your response. However, using this I do not get the desired expression. I will extend my question to show what I am looking for precisely. $\endgroup$
    – WHN
    Commented Mar 30, 2022 at 23:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.