We have to prove the following equation:
$$E(X)=\int_0^{\infty}[1-F_X(x)]dx $$ for positive random variable $X$.
I found this equation proved in every book using the concept of double integrals. However, I am not that good at multivariable calculus so I tried to prove the equation by the following method. I proceeded as follows:
$E[X]= \int_{0}^{\infty} x f_X(x)dx$
By using the definition $f(x) = {F^{'}_X}(x)$, we get
$ \int_{0}^{\infty} x F^{'}_X(x)dx$
Now, By using Integration by parts, we get
$\Bigg[ x (1-F_X(x)) \Bigg]_{x=0}^{x \rightarrow \infty} + \int_0^{\infty}[1-F_X(x)]dx $
But now when I am trying to put the limit in the equation, I am getting confused in evaluating the following limit.
$\lim\limits_{h\rightarrow \infty} h (1-F_X(h)) \ + \int_0^{\infty}[1-F_X(x)]dx $
Why the first expression in the above equation actually vanishes?
We know the property of the CDF that $$\lim_{x\to\infty} F_X(x) =1$$ , Do we here mean that the value of limit is precisely 1? (Keeping in mind, the increasing nature of $F_X(x)$ )
I think that if this is true, then $(1-F_X(h))$ will be precisely equal to zero and hence, $\lim\limits_{h\rightarrow \infty} h (1-F_X(h)) $ should be precisely equal to zero.
Is this step correct to prove the value of the above limit equal to zero?
Also, is there any reason why this method (if it is valid) is not mentioned in books as it looks relatively simple to me?
Best Answer
Suppose the limit $\lim_{x\to+\infty}x(1-F(x))$ is not zero. Then there exists a constant $C>0$ such that $1-F(x)\geq C/x$ for $x$ sufficiently large. Then the integral $\int_0^{+\infty} (1-F(x))\mathrm dx$ would not converge...