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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?

IIM
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2 Answers2

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

user1131274
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  • How did you deduce that $E[X]=\int_0^\intfy P(X \geq t)dt$? – IIM Dec 20 '16 at 16:26
  • Also, I would still be interested to find a solution using the original method, because I would like to learn how to use Radon-Nikodym derivatives for this type of problem. – IIM Dec 20 '16 at 16:27
  • @IIM For nonnegative random variables, this is actually a rather routine exercise. If you don't understand it, it may help to start by gaining intuition through looking at the discrete case before moving on to the continuous case. The idea in the discrete case is simple enough: when the r.v. takes on the value $x$ with probability $p$, $P(X \geq t)$ contains that $p$ for $t$ between $0$ and $x$, which overall contributes $px$ to the expectation as you would expect. – Ian Dec 20 '16 at 16:28
  • Its standard result for non-negative random variable. For proof check this out https://en.wikipedia.org/wiki/Expected_value#Continuous_distribution_taking_non-negative_values or http://math.stackexchange.com/questions/172841/integral-of-cdf-equals-expected-value – user1131274 Dec 20 '16 at 16:31
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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 $\int_1^\infty t \cdot 3t^{-4} dt$.

Ian
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