# Solved – What does the integral of a function times a function of a random variable represent, conceptually

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:

$$g(x) = frac1 {2 pi sigma^2} expleft(frac{- (x – mu)^2 }{ 2 sigma ^2}right)$$

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?

Contents

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.

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