Find expected value using CDF

Question

I'm trying to solve the following problem:

Let $X$ have the CDF $F(x) = 1 - x^{-\alpha}, x\ge1$.
Find $E(X)$ for those values of $\alpha$ for which $E(X)$ exists.

How can I determine which values of $\alpha$ exist? What do I do with the CDF?

There are formulas for finding the expected value when you have a frequency function or density function. Wikipedia says the CDF of $X$ can be defined in terms of the probability density function $f$ as follows:

$$F(x) = \int_{-\infty}^x f(t)\,dt$$

This is as far as I got. Where do I go from here?

Answer 1

Note that $F(1)=0$ in this case so the distribution has probability $0$ of being less than $1$, so $x \ge 1$, and you will also need $\alpha > 0$ for an increasing cdf.

If you have the cdf then you want the anti-integral or derivative which with a continuous distribution is like this:

$$f(x) = \frac{dF(x)}{dx}$$

and in reverse $F(x) = \int_{1}^x f(t)\,dt$ for $x \ge 1$.

Then to find the expectation you need to find

$$E[X] = \int_{1}^{\infty} x f(x)\,dx$$

providing that this exists. I will leave the calculus to you.

Answer 2

Usage of the density function is not necessary

Integrate 1 minus the CDF

When you have a random variable $X$ that has a support that is non-negative (that is, the variable has nonzero density/probability for only positive values), you can use the following property:

$$ E(X) = \int_0^\infty \left( 1 - F_X(x) \right) \,\mathrm{d}x $$

A similar property applies in the case of a discrete random variable.

Proof

Since $1 - F_X(x) = P(X\geq x) = \int_x^\infty f_X(t) \,\mathrm{d}t$,

$$ \int_0^\infty \left( 1 - F_X(x) \right) \,\mathrm{d}x = \int_0^\infty P(X\geq x) \,\mathrm{d}x = \int_0^\infty \int_x^\infty f_X(t) \,\mathrm{d}t \mathrm{d}x $$

Then change the order of integration:

$$ = \int_0^\infty \int_0^t f_X(t) \,\mathrm{d}x \mathrm{d}t = \int_0^\infty \left[xf_X(t)\right]_0^t \,\mathrm{d}t = \int_0^\infty t f_X(t) \,\mathrm{d}t $$

Recognizing that $t$ is a dummy variable, or taking the simple substitution $t=x$ and $\mathrm{d}t = \mathrm{d}x$,

$$ = \int_0^\infty x f_X(x) \,\mathrm{d}x = \mathrm{E}(X) $$

Attribution

I used the Formulas for special cases section of the Expected value article on Wikipedia to refresh my memory on the proof. That section also contains proofs for the discrete random variable case and also for the case that no density function exists.

Answer 3

The result extends to the $k$th moment of $X$ as well. Here is a graphical representation:

Answer 4

I think you actually mean $x\geq 1$, otherwise the CDF is vacuous, as $F(1)=1-1^{-\alpha}=1-1=0$.

What you "know" about CDFs is that they eventually approach zero as the argument $x$ decreases without bound and eventually approach one as $x \to \infty$. They are also non-decreasing, so this means $0\leq F(y)\leq F(x)\leq 1$ for all $y\leq x$.

So if we plug in the CDF we get:

$$0\leq 1-x^{-\alpha}\leq 1\implies 1\geq \frac{1}{x^{\alpha}}\geq 0\implies x^{\alpha}\geq 1 > 0\implies x\geq 1 \>.$$

From this we conclude that the support for $x$ is $x\geq 1$. Now we also require $\lim_{x\to\infty} F(x)=1$ which implies that $\alpha>0$

To work out what values the expectation exists, we require:

$$\newcommand{\rd}{\mathrm{d}}E(X)=\int_{1}^{\infty}x\frac{\rd F(x)}{\rd x}\rd x=\alpha\int_{1}^{\infty}x^{-\alpha} \rd x$$

And this last expression shows that for $E(X)$ to exist, we must have $-\alpha<-1$, which in turn implies $\alpha>1$. This can easily be extended to determine the values of $\alpha$ for which the $r$'th raw moment $E(X^{r})$ exists.

Answer 5

The Answer requiring change of order is unnecessarily ugly. Here's a more elegant 2 line proof.

$$\int udv = uv - \int vdu$$

Now take $du = dx$ and $v = 1- F(x)$

\begin{align} \int_{0}^{\infty} [ 1- F(x)] dx &= [x(1-F(x))]_{0}^{\infty} + \int_{0}^{\infty} x f(x)dx \\ &= 0 + \int_{0}^{\infty} x f(x)dx \\ &= \mathbb{E}[X] \qquad \blacksquare \end{align}

Answer 6

In case when a conditional expectation using only CDF is needed, we can formulate two cases,

$\mathbb{E}\left(x|x\geq y\right)=y+\frac{\int_{y}^{\infty}\left(1-F(x)\right)dx}{\left(1-F(y)\right)}$

$\mathbb{E}\left(x|x\leq y\right)=y-\frac{\int_{-\infty}^{y}F(x)dx}{F(y)}$

The derivation leverages on previous post such that we first define following integral, $\int_{y}^{\infty} [ 1- F(x)] dx = [x(1-F(x)) ]_{y}^{\infty} + \int_{y}^{\infty} x f(x)dx$

$\int_{y}^{\infty} x f(x)dx=\mathbb{E}\left(x|x\geq y\right)(1-F(y))$

Then using this definition and some algebra we arrive at first result. The second result can be obtained in the same way.

Answer 7

Alternative derivation of $EX = \int_0^\infty \left(1-F_X(x)\right)\mathrm d x$ (for a positive r.v. $X$):

For any $x \ge 0$ one has that $x=\int_0^x 1 \mathrm dt = \int_0^\infty \mathbf I_{t\le x}\mathrm dt$, where $\mathbf I$ is the indicator function. Let $x=X$, take expectations on both sides, and use Fubini's theorem to exchange the integrals to obtain $EX = \int_0^\infty P(X\ge t)\mathrm dt$.