Expected value of truncated distribution

Question

This definition is taken from the Wikipedia page for truncated distribution:

Let $X$ be a random variable with a continuous distribution, $f(x)$ be its probability density function and $F(x)$ be its cumulative distribution function.

Now the truncated distribution where the top of the distribution has been removed is as follows: $$f(x|X\le y)=\frac{g(x)}{F(y)}$$ where $g(x)=f(x)$ for all $x\le y$ and $g(x)=0$ everywhere else.

I am trying to derive the expected value of the $f(x|X\le y)$ and I am thinking about using the conditional expectation the following way for the case when $x\le y$. $$E(X|X\le y)=\int^y_{-\infty} x_{X|X\le Y}f(x|x\le y)dx=\int^y_{-\infty}x\frac{f(x)}{F(y)}dx$$ but I don't know how to continue from here since I am not sure what happens to $F(y)$.

Answer 1

Notice that $y$ is a constant, hence $F(y)$ is a constant.

That is

$$E(X|X \le y) = \frac{\int_{-\infty}^y xf(x) \, dx}{F(y)}$$

This is consistent with the formula given on the wikipedia page

$$\frac{\int_a^b xf(x) \, dx}{F(b)-F(a)}$$ since we have

$$\lim_{t\to -\infty}F(t)=0$$