Verify Alternative Formula for Expected Value I am studying for the first actuarial exam (Exam P) and came across a formula in my ACTEX prep manual that I had never seen before:
$$E[X] = a + \int_{a}^{b}{[1-F(x)]dx}$$
And the text said this was true as long as $x$ was continuously defined on the interval, and as long as $b\lt \infty$. True for continuous and discrete! I tried it out with a few different, very straightforward functions and could not get it to equal an expected value answer I found in the typical manner. Am I missing something in the application? Has anyone seen this before?
Any help is appreciated!
 A: It is true, and it is well known.
$$\begin{align} & \text{Continuous R.V.}&&\text{Discrete Integer-Valued R.V.}
\\[2ex]\hline&a+\int_a^b (1-F_X(x))\operatorname d x 
&& a+\sum_{k=a}^{b} (1-F_X(k))
\\[1ex]=~& a+\int_0^{b-a} (1-F_X(u+a))\operatorname d u 
&~=~& a+ \sum_{j=0}^{b-a} (1-F_X(j+a))
\\[1ex]=~& a+\int_0^{b-a} \int_{u+a}^b f_X(v)\operatorname d v\operatorname d u
&~=~& a+\sum_{j=0}^{b-a}\;\sum_{i=1+j+a}^{b} p_X(i)
\\[1ex]=~& a+\iint_{\{0\leq u\leq v-a\leq b-a\}}  f_X(v)\operatorname d (u,v)
&~=~& a+ \underset{0\leq j \color{navy}{~<~} i-a\leq b-a}{\sum\sum}p_X(i)
\\[1ex]=~& a+\int_{a}^{b}  f_X(v)\int_{0}^{v-a}\operatorname d u\operatorname d v
&~=~& a + \sum_{i=a}^{b} p_X(i)\;\sum_{j=0}^{i-a-1} 1
\\[1ex]=~& a\int_{a}^{b}f_X(v)\operatorname d v+\int_{a}^{b} f_X(v)\,(v-a) \operatorname d v
&~=~& a+ \sum_{i=a}^b p_X(i)(i-a)
\\[1ex]=~& \int_{a}^{b} v f_X(v)\operatorname d v
&~=~& \sum_{i=a}^b i~p_X(i)
\\[2ex]=~& \mathsf E(X)
&~=~& \mathsf E(X)
\end{align}$$
$\blacksquare$
A: Let $F(x)=\int_a^x f(x) dx$ where $f$ is  probability density function of $X$. Then by integrating by parts
$$E[x]=\int_{a}^{b} xf(x) dx = \int_a^b xd(F(x)) = [xF(x)]_a^b -\int_a^bF(x)dx 
$$
$$= bF(b)-aF(a)- \int_a^b F(x)dx = b- \int_a^b F(x)dx$$
which is equal to the RHS of your formula because
$$a + \int_{a}^{b}(1-F(x))dx=a + \int_{a}^{b}dx-\int_{a}^{b}F(x)dx
=b-\int_a^bF(x)dx.$$
A: For a continuous random variable
$$
E[X]=\int_a^b dx\ x\ f_X(x)\ ,
$$
where $f_X(x)$ is the probability density function. But
$$
f_X(x)=F'(x)\ ,
$$
where $F$ is the cumulative distribution function. 
Substituting in the integral above, and integrating by parts we have
$$
E[X]=\int_a^b dx\ x\ F'(x)=x F(x)\Big|_a^b -\underbrace{\int_a^b F(x)dx}_{\star}\ .
$$
Now, note that
$$
\int_a^b [1-F(x)]dx=(b-a)-\underbrace{\int_a^b F(x)dx}_{\star}\ .
$$
Hence
$$
E[X]=x F(x)\Big|_a^b -\int_a^b dx\ F(x)=b F(b)-a F(a)+\int_a^b [1-F(x)]dx-(b-a)
$$
$$
=a+\int_a^b [1-F(x)]dx\ ,
$$
using the fact that $F(a)=0$ and $F(b)=1$ (by definition of cumulative distribution function for a density supported on [a,b]).
