I am trying to understand conceptually what does the following give me or tell me:
$$\int f(x) \cdot g(x) \, dx$$
where $f(x)$ is any continuous function of $x$ and $g(x)$ is the probability density function for a random variable, for example a normal distribution's PDF is:
$$ g(x) = \frac1 {2 \pi \sigma^2} \exp\left(\frac{- (x – \mu)^2 }{ 2 \sigma ^2}\right) $$
I understand the integral of a PDF gives me the CDF. So:
$$\int_{-\infty}^0 g(x) \, dx$$
Gives me the probability of $x$ being less than $0$. However, what happens when you multiply $g(x)$ by another function $f(x)$ and take the integral? I heard it gives you the expected payoff assuming $f(x)$ is a function of payoffs and you take an integral from -infinity to +infinity. This, if true, I conceptually understand. sum of payoffs times the probabilities is the expected value of whatever game you are playing.
I start getting confused when the boundaries of the integral are not $\pm \ infty$. I'm not sure in that case what integral conceptually means. For example:
$$\int_{-\infty}^0 f(x) g(x) \, dx$$
What does that tell me?
Best Answer
Suppose $g$ is the pdf of random variable $X$, then
$$E[f(X)|X \in A]= \frac{\int_A f(x)g(x) \, dx}{\int_A g(t) \, dt}$$
Hence $$\int_A f(x) g(x) \, dt = Pr(X \in A) E[f(X)|X \in A],$$
it gives you the product of the conditional expectation of $f(X)$ given that $X \in A$ and the probability that $X$ is in $A$.
I think $E[f(X)|X \in A]$ is a more interesting quantity, and I would divide your integral with $Pr(X \in A)$ to recover it.