Show that E(X) is equal to this formula, on the interval [a, b]:

$$E(X)=a+\int_a^b[1-F(x)]dx$$
E(X) is the expectation of X, F(x) is the cdf of X

Question

Show that E(X) is equal to this formula, on the interval [a, b]:

$$E(X)=a+\int_a^b[1-F(x)]dx$$
E(X) is the expectation of X, F(x) is the cdf of X

kirbykirby · Answer

X is just some random variable. It can be continuous, discrete or mixed

hartnn · Answer

CDF = F(x)=P(X $\le $x)
so, 1-F(x) =P(X $\ge $x) =$\int \limits_x^{\infty }f(x)dx$ 
you can use this.

anonymous · Answer

i forget but i think there is some gimmick here where you take 
$$\int_0^{\infty} xf(x)dx$$and rewrite it as 
$$-\int_0^{\infty}x(-f(x))dx$$ then use parts but i could be wrong

hartnn · Answer

now, change the order of double integration....

kirbykirby · Answer

Oh i think I might have figured it out after all ._. the integration by parts triggered smtg

hartnn · Answer

sorry, that should be $\int \limits_x^{\infty }f(t)dt$
oh...okk.

kirbykirby · Answer

hm i think double int would work too . thanks !

zarkon · Answer

double integral is the easy way

hartnn · Answer

if you are comfortable with changing the order of integration, it'll give u d answer in few steps...