# let $X$ be a random variable , $A\subset B$ is that $\mathbb{E} [X|A] \leq \mathbb{E}[X|B]$ true in general?

1.let $X$ be a random variable , $A\subset B$ is that $\mathbb{E} [X|A] \leq \mathbb{E}[X|B]$ true in general ?

2 what about If $X$ is Gaussian random variable , $A\subset B$ is that $\mathbb{E} [X|X \leq 0] \leq \mathbb{E}[X| X\leq c]$ where $c>0$?

The first point is not true.

Consider $X$ being a uniform rv on $\{0,1\}$, then $$1=E(X|\{X \in \{1\}\})>E(X|\{X \in \{0,1\}\})=\frac 1 2.$$

The second one is true for $c \ge 0$ since $$\mathbb{E} [X|X \leq 0] = \int_{-\infty}^0 xf(x)dx \leq \int_{-\infty}^cxf(x)dx=\mathbb{E}[X| X\leq c]$$

This follows from the fact that : $xf(x)>0$ on $(0,c)$.

Just add as a slight generalization of the problem:

Assume $\mathbb{P}\{0 < X \leq c\} > 0$. Consider

\begin{align} \mathbb{E}[X \mid X \leq c] &= \mathbb{E}[X \mid X \leq 0, X\leq c]\mathbb{P}\{X \leq 0 \mid X \leq c\} \\ &~~~~+ \mathbb{E}[X \mid 0 < X \leq c, X\leq c]\mathbb{P}\{0 < X \leq c \mid X \leq c\} \\ & = \mathbb{E}[X \mid X \leq 0] \frac {\mathbb{P}\{X \leq 0\}} {\mathbb{P}\{X \leq c\}} + \mathbb{E}[X \mid 0 < X \leq c] \frac {\mathbb{P}\{0 < X \leq c\}} {\mathbb{P}\{X \leq c\}} \\ & > \mathbb{E}[X \mid X \leq 0] \frac {\mathbb{P}\{X \leq 0\}} {\mathbb{P}\{X \leq c\}} + \mathbb{E}[X \mid X \leq 0] \frac {\mathbb{P}\{0 < X \leq c\}} {\mathbb{P}\{X \leq c\}} \\ & = \mathbb{E}[X \mid X \leq 0] \frac {\mathbb{P}\{X \leq 0\} + \mathbb{P}\{0 < X \leq c\}} {\mathbb{P}\{X \leq c\}} \\ & = \mathbb{E}[X \mid X \leq 0] \end{align}

The first line is due to the law of probability.

Since $\mathbb{P}\{X \leq 0 \mid X \leq 0\} = 1$ and $\mathbb{P}\{X > 0 \mid 0 < X \leq c \} = 1$, we have $$\mathbb{E}[X \mid X \leq 0] \leq 0 < \mathbb{E}[X \mid 0 < X \leq c]$$ and this results in the inequality in the 3rd line.

The constant $0$ here is relatively arbitrary, and in general we have $$\mathbb{E}[X \mid X \leq b] \leq \mathbb{E}[X \mid X \leq c]$$ whenever $b < c$, and $\mathbb{P}\{b < X \leq c\} > 0$.