Finding the expectation of the given expression

Question

Suppose we have a random variable $\theta= U[0,1]$ and we have

X= 2 if $\theta\leq 0.25$ and

X=1 if $ 0.25\leq\theta \leq 1$

How would we compute the Expectation of X in this case?

Answer 1

It's a simple problem and it's hard to see the usage of what you wrote in the comment :

$\begin{array}{ll}E[X] & =E[X|\theta\leq0.25].P(0\leq\theta\leq0.25)+E[X|\theta\leq0.25].P(0.25\leq\theta\leq1)\\ & = E[2].\frac{0.25}{1} + E[1].\frac{0.75}{1}= 2(0.25)+1(0.75) \end{array}$

Now, consider this problem :

$X(\theta)=\left\{ \begin{array}{ll} 2 & ;\theta\in(0,0.25]\\ \theta^2 & ;\theta\in(0.25,1) \end{array}\right.$

Now again by conditioning we have :

$$E[X]=E[X(\theta)|\theta\leq0.25].P(0\leq\theta\leq0.25)+E[X(\theta)|\theta\leq0.25].P(0.25\leq\theta\leq1)$$ Therefore, to calculate $E[X(\theta)|\theta\leq0.25]$, you should use truncated distribution of $\theta$ with the condition $\theta\leq0.25$. Here is the where the statement $\frac{1}{0.25}$ emerges !

If you have questions, comment it below.