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I have a random variable $X$ and a constant $c\geq 0$. I define the r.v. $Y = \min(X, c)$ and I want to calculate $E[Y]$.

I have seen different posts on similar topics, so I am trying to pull all together from the point of view of someone that has only basic math and probability notions.

I know that

$ Y= \min(X, c)= \begin{cases} X & X \leq c \\ c & X > c \end{cases} $

By the Law of Total Expectation I can write

\begin{equation} \begin{split} E[Y] & = & E[Y|X\leq c]\cdot P(X\leq c)+ E[Y|X > c]\cdot P(X > c) \\ & = & E[X|X\leq c]\cdot P(X\leq c)+ E[c|X > c]\cdot P(X > c) \\ & = & \int_{-\infty}^{c}\frac{x\cdot f(x) dx}{P(X\leq c)}\cdot P(X\leq c)+c\cdot[{1-F(c)}] \\ & = & E_{-\infty}^{c}[X]+c\cdot[{1-F(c)}] \end{split} \end{equation}

where $E_{-\infty}^{c}[X]$ is the patial moment of $X$.

I am not sure about the correctness of the formula above, so my questions are:

  1. Is my reasoning correct?
  2. If correct, how do I calculate $E_{-\infty}^{c}[X]$?
  3. If $X$ is truncated on the left (that is, $X\geq d$ and $d < c$) should I change the limit of integration from $-\infty$ to $d$ or it is enough to know that $f(X) = 0$ for $X<d$?

Thanks

Libra
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    Correctly, your random variable is not truncated at $c$, but censored at $c$. Censoring a continuous random variable above at $c$ gives rise to $P(Y=c)=P(X>c)$ i.e. discrete mass at $c$; truncation does not. – wolfies Jul 20 '15 at 17:29
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    Notationally it's easier just to write $\text{E}(Y) = \text{E}(Y ; X \leq c) + \text{E}(Y ; X > c) = \text{E}(X ; X \leq c) + c P(X > c)$. We don't know if $X$ is continuous or discrete so the first term is either an integral or a sum. – dsaxton Jul 20 '15 at 17:30
  • @dsaxton so if $X$ is continuous we have $E(X;X\leq c) = \int_{-\infty}^{c}x\cdot f(x) dx$ whereas if $X$ is defined over the interval $[d, c]$ we have $E(X;X\leq c) = \int_{d}^{c}x\cdot f(x) dx$, right? – Libra Jul 20 '15 at 22:18
  • @wolfies I was not aware of the difference between truncated and censored, thanks for the clarification. – Libra Jul 20 '15 at 22:20