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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

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  • $\begingroup$ What is the joint distribution of $(U,V)$? $\endgroup$ Mar 31 at 19:28
  • $\begingroup$ The joint distribution is $F_{U, V}(u, v)=u v+\theta u(1-u) v(1-v)$ $\endgroup$
    – WHN
    Apr 1 at 11:45
  • $\begingroup$ Is this a joint CDF? What is $\theta$? $\endgroup$ Apr 1 at 16:07
  • $\begingroup$ This is indeed the joint CDF, $\theta\in R$ is just a parameter $\endgroup$
    – WHN
    Apr 1 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 $(U,V)$. $\endgroup$ Apr 2 at 10:14

1 Answer 1

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"$S_i=1$" 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 $u$.

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  • $\begingroup$ 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
    Mar 30 at 23:55

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