Indicator variables in Expectation

Question

I've come across this result a lot but never had the concept explained in a little depth.

Basically its that

$E[X\times1_{X>c}]=\int_{c}^\infty x\ f_X(x)dx$

where $1_{X>c}$ is the indicator variable taking 1 with probability $P(X>c)$ , 0 otherwise.

I think my confusion stems from trying to show how the left hand side differs from $E[X\ |X>c \ ]$ particularly because I looked at this:

https://www.quora.com/What-is-the-expected-value-of-Y-given-that-Y-is-greater-than-C

but got confused a bit in Curt Clemens's 3 steps proof.

Answer 1

The conditional PDF $f_{X|X>c}(x)$ is the scaled unconditional PDF over smaller domain $$f_{X|X>c}(x)=\frac{f_{X}(x)}{P(X>c)}$$ because the conditional PDF should integrate to $1$ over $[c..\infty]$, and unconditional PDF should integrate to $1$ over $[-\infty..\infty]$.

The use of indicator variable is just a quick notation for the above.

Answer 2

First a quibble that may just have been sloppiness, not misunderstanding on your part: $1_{X>c}$ is an indicator variable that takes the value $1$ whenever $X>c$ and zero otherwise. When you say it takes the value 1 with probability $P(X>c)$ that is true but is not a full definition since it could for instance take the value $1$ with this probability but be independent from $X.$

$X1_{X>c}$ is a random variable that is a function of the random variable $X.$ It takes the value $X$ whenever it happens that $X>c$ and takes the value zero when it happens that $X\le c.$

You want to take the expectation value of this function of a random variable $X$. In order to do that you integrate the function against the density function of $X$. The function is piecewise $x$ for $x>c$ and $0$ for $x<c.$ So the expectation is $$ E(X1_{X>c}) = \int_{-\infty}^\infty x1_{x>c} f_X(x)dx = \int_c^\infty xf_X(x)dx$$ where the 2nd equality just uses the fact that $1_{x>c}$ is zero when $x\le c$ and $1$ otherwise.

This is different from $E(X|X>c)$ but is related by $$ E(X|X>c) = \frac{E(X1_{X>c})}{P(X>c)}.$$

Think of it this way: You get the conditional expectation by integrating $x$ against the conditional probability given $X>c.$ $E(X1_{X>c})$ integrates over the right region but against the wrong distribution. To get the conditional distribution you would need to also normalize by $P(X>c)$ since the pie (i.e. the possible range for $X$) is smaller now.