the expected value is a non-negative random variable is always greater than the expected value of its subset? Let's say $X$ is a non-negative random variable. And say $c$ is a positive constant. Why is $$E(X)\geq E\left(X_{\chi (X \geq c) }\right)$$
I don't know why the expected value is a non-negative random variable is always greater than the expected value of its subset?
The second inequality that confused me is
$$E\left(X_{\chi (X \geq c) }\right) \geq c P(X \geq c)$$
Cany anyone help me explain these 2 inequalities?
 A: I think you mean to write $X\chi_{(X\ge c)}$, i.e. the product of the random variable $X$ and the random variable $\chi_{(X\ge c)}$. In that case $\chi_A$ represents the same thing as the entity $I_A$ in @Bey's answer: it's the indicator of event $A$.
To prove your first result it's enough to show that the random variable $X$ is at least as big as the random variable $X\chi_{(X\ge c)}$. This is true, by considering cases:

*

*If $X\ge c$, then $\chi_{(X\ge c)}=1$, so $X\chi_{(X\ge c)}=X$.

*If $X< c$, then $\chi_{(X\ge c)}=0$, so $X\chi_{(X\ge c)}=0$.

In both cases we have $$X\ge X\chi_{(X\ge c)}.$$ (Remember that $X$ is non-negative.) Hence the expectation of the LHS exceeds the expectation of the RHS, which proves the first result.
As for your second result, first establish the following inequality between random variables, again by arguing cases:
$$X\chi_{(X\ge c)}\ge c\chi_{(X\ge c)}.$$ Hence
$$E(X\chi_{(X\ge c)})\ge E(c\chi_{(X\ge c)}).$$ To finish, use linearity: $E(cY)=cE(Y)$ for any $c$ and $Y$; and the fact that $E(\chi_A)=P(A)$ for any event $A$.
A: As @lulu said, you should explain all notation because you are using it in a very non-standard way.
My answer attempts to show the machinery of expectation (hence the integrals). A more elegant, integral-free way to show this is given by @grand_chat's answer.

This can be made clearer using indicator notation:
$$1_{S}(x) = 1 \iff x\in S,\;\;0\text{ otherwise}$$
$$E[X] = \int_{\mathbb{R^+}} xdP = \int_{[0,r)}xdP + \int_{[r,\infty)}xdP$$
$$=\int_{\mathbb{R^+}}x1_{[0,r)}(x)dP + \int_{\mathbb{R^+}}x1_{[r,\infty)}(x)dP $$
Now, if $X\geq 0$ and $c>0$ then:
$$E[X1_{[c,\infty)}]=\int_{\mathbb{R^+}}x1_{[c,\infty)}(x)dP = E[X] - \int_{\mathbb{R^+}}x1_{[0,c)}(x)dP$$
Since $X\geq 0$,
$$\int_{\mathbb{R^+}}x1_{[0,c)}(x)dP \geq 0 \implies E[X1_{[c,\infty)}] \leq E[X]$$
For $E[X1_{>c}]>cP(X>c)$ you can see that:
$$\int_{\mathbb{R^+}} x1_{>c} dP = \int_c^{\infty} x dP > \int_c^{\infty} c dP$$
