Calculating the expectation of a random variable using its cumulative distribution function

Question

I am working through the following problem from a past exam:

Let $(\Omega, \mathcal{A}, P)$ be a probability space, and let $X$ be a random variable such that $P(X > t) = t^{−3}$ for $t > 1$ and $P(X > t) = 1$ for $t \leq 1$. Compute the expectation $\mathbb{E}(X)$.

My plan is to calculate the Radon-Nikodym derivative of $\rho_X$ wrt. Lebesgue measure, and thus to calculate $\mathbb{E}(X) = \int X d \rho_X$ from a Lebesgue integral. I can find the Radon-Nikodym derivative $f$ by writing the cumulative distribution function $F_X$ in the following form:

$F_X(t) = \int_{-\infty}^t f (s) ds$.

So far, I have found that $F_X(t) = (1-t^{−3})\chi_{(1,\infty)}(t)$. But I am not sure how to find an integral which evaluates to this.

Also, I have looked at the given solutions, which say that

$F_X(t) = 1 − t^{−7}$

which can then be written nicely as

$F_X(t) = \int_{-\infty}^t 7s^{−8}χ_{[1,∞)}(s)ds$.

But I am almost certain that this must be incorrect...

Can anyone tell me the correct expression for $F_X$ and how to find an integral (of the correct form) that evaluates to it?

Answer 1

Alternative way can be as follows:
As given the random variable is non-negative so $$E[X]=\int_0^\infty P(X>=t)dt=1+\int_1^\infty P(X>=t)dt=1+\int_1^\infty t^{-3}dt=3/2$$

Answer 2

So it should be pointed out that this calculation is really in the realm of "elementary probability". You can think of this as Radon-Nikodym derivatives if you want, but generally "Radon-Nikodym derivative with respect to Lebesgue measure" is just abbreviated to "density", and it works the same in elementary probability as it does in measure-theoretic probability. Getting too bogged down in measure-theoretic details can sometimes obscure what you are doing.

With that in mind $P(X>t)=t^{-3}$, so the density is the negative of the derivative of that (the same as the derivative of the CDF), so $3t^{-4}$ for $t>1$ and zero elsewhere. So the expectation is $\int_1^\infty t \cdot 3t^{-4} dt$.