Expected value, CDF

Question

Suppose we have a positive, continuous probability variable X, the following expression for the variable does not depend on s:

E (X − s | X ≥ s), where E( ) Is the expected value.

What is the cumulative distribution function for this problem?

Answer 1

Ooh, this is good. It's got probability and differential equations!

So writing out the expectation:

[∫(x-s)f(x)dx] / ∫f(x)dx = c (>= 0)

Here the bounds of both integrals are [s, ∞].

Expanding and using the definition of the CDF (F(x)):

∫xf(x)dx = (s + c) * (1 - F(s))

Taking the derivative of both sides with respect to s yields;

-sf(s) = 1 - F(s) -sf(s) - cf(s)

Because s can be anything x can be, we can substitute x for s:

F(x) + cf(x) - 1 = 0

This is a differential equation with solution:

F(x) = Aexp(-x/c) + 1

To determine A, we use the initial condition F(0) = 0 which says A = -1. Thus:

F(x) = 1 - exp(-x/c)

where c is any positive constant. In fact, this is the CDF of the exponential distribution:

f(x) = 1/c * epx(-x/c) for x >= 0