Is the following succession correct? Would it not make more sense if the last inequality was reversed?
$|E[X I_{X > t}]| = E[X] - E[XI_{X < t}] \geq E[X] - t$?
Here $I_{X>t}$ is the indicator function ($1$ if $X>t$ else $0$) and $E[X] > 0$ is given.
thank you in advance
I think the answer is no.
let $P(X=1)=P(X=2)=P(X=3)=\frac{1}{3}$
$$E(X1_{X>2})=3*P(X=3)=1$$ $$E(X1_{X<2})=1*P(X=1)=\frac{1}{3}$$ $$E(X)=2$$
Maybe following inequalities help you: (I think they are hold) $$E(X) \leq E(X 1_{\{ X> t \}} ) + t P(X\leq t) \hspace{.5cm} (1)$$
$$E(X) \geq t P(X\geq t)+ E(X 1_{\{ X< t \}} ) \hspace{.5cm} (2) $$ It is depend on $t>0$ or not you can use them.
Proof (1):let $A=\{ X>t\}$ It is obvious (see proof (4))
$$E(X)=E(X 1_{\{ X> t \}} ) +E(X 1_{\{ X \leq t \}} ) $$
and
$$E(X 1_{\{ X \leq t \}} )=\int_{-\infty}^{t} x f_X(x) dx $$ $$\leq \int_{-\infty}^{t} t f_X(x) dx=t P(X\leq t)$$
so
$$E(X)=E(X 1_{\{ X> t \}} ) +E(X 1_{\{ X \leq t \}} ) $$ $$\leq E(X 1_{\{ X> t \}} ) + t P(X\leq t) \hspace{.5cm} (3)$$
In (3) depend on $t\geq 0$ or $t<0$ you can use the fact $P(X\leq t)\in [0,1]$
for example if $t>0$
$$ E(X 1_{\{ X> t \}} ) + t P(X\leq t) \leq E(X 1_{\{ X> t \}} ) + t $$
Proof (2)
$$E(X)=E(X 1_{\{ X\geq t \}} ) +E(X 1_{\{ X < t \}} ) $$
$$=\int_t^{-\infty} xf_X(x) dx \, +E(X 1_{\{ X < t \}} ) $$ $$\geq tP(X\geq t) +E(X 1_{\{ X < t \}} ) $$
Proof (4) simply for continues variables $$E(X)=\int_{-\infty}^{t} xf_X(x) dx +\int_{t}^{+\infty} xf_X(x) dx$$ $$=\int_{-\infty}^{+\infty} x 1_{x\leq t}f_X(x) dx +\int_{-\infty}^{+\infty} 1_{x>t} xf_X(x) dx$$ $$=E(X 1_{\{ X \leq t \}} ) +E(X 1_{\{ X> t \}} ) $$
for all type of random variables
$$E(X)=E(X|A) P(A)+E(X|A^{c})P(A^{c})$$
$$=E(X|\{ X>t \} ) P(\{ X>t \} )+E(X|\{ X \leq t\})P(\{ X\leq t\})$$
$$=\frac{E(X 1_{\{ X>t \}} )}{P(\{ X>t \} )} P(\{ X>t \} ) + \frac{E(X 1_{\{ X\leq t \}} )}{P(\{ X\leq t \} )} P(\{ X\leq t \} ) $$
$$=E(X 1_{\{ X> t \}} ) +E(X 1_{\{ X \leq t \}} )$$ so