I am having trouble getting the conditional PDF from Bayes' Rule for the following problem:
Fred wants to sell his car, after moving back to Blissville (where he is happy with the bus system). He decides to sell it to the first person to offer at least \$15,000 for it. Assume that the offers are independent Exponential random variables with mean \$10,000. Find the expected amount of money that Fred gets for the car.
So the way I solved this was using the memoryless property, which is simply,
$$E(X \mid X > 15000)=15000+EX=\$25000$$
Now, I noticed that the solution provides a different way to do it that I am not so familiar with but would like to understand it better. Specifically, it uses Bayes' Rule to get the conditional PDF and then integrate $x$ multiplied by the PDF to get the expected value. Concretely, we see:
$$f(x \mid X>a)=\frac{P(X>a \mid X=x)f(x)}{P(X>a)}=\frac{\lambda e^{-\lambda x}}{e^{-\lambda a}}=\lambda e^{-\lambda (x-a)}$$
In this solution, I don't understand how we get from
$$\frac{P(X>a \mid X=x)f(x)}{P(X>a)}=\frac{\lambda e^{-\lambda x}}{e^{-\lambda a}}$$
What happens to $P(X>a \mid X=x)$? I feel like in the solution this term equals $1$. But doesn't this term only equal $1$ when $x>a$? If $x<a$, then wouldn't the term be $0$? How do we know what $x$ is?
You are correct, in the solution we require $x>a$. It is legitimate to replace $P(X>a\mid X=x)$ with $1$ only if $x>a$; otherwise $P(X>a\mid X=x)$ equals zero, and the resulting conditional density is zero as well. As a check, the conditional density integrates to 1 provided it is zero when $x\le a$.